INTRODUCTION.

1. Algebra is a general method of reasoning concerning the relations which magnitudes of every kind bear to each other in respect of quantity. It is sometimes called universal arithmetic; its first principles and operations being similar to those of common arithmetic. The symbols which it employs to denote magnitudes are, however, more general and more extensive in their application than those employed in that science; hence, and from the great facility with which the various relations of magnitudes to one another may be expressed, by means of a few signs or characters, the application of algebra to the resolution of problems is much more extensive than that of common arithmetic.

2. There are various opinions as to the etymology of the name algebra. It is pretty certain, however, that the word is Arabic, and that from the Arabians the name, as well as the art itself, is derived. Lucas de Burgo, the first European author whose treatise on algebra was printed, calls it by the Arabic name Algebr a e Almucabala, which is explained to denote the art of refutation and comparison, or opposition and comparison, or refutation and equation, all which agree well enough with the nature of this art. Besides this etymology of the name algebra, several others have been imagined; that, however, which we have just now given seems to be the most probable of any hitherto assigned.

3. The origin of algebra, as well as that of most other branches of mathematical sciences, is involved in obscurity; there are indeed traces of it to be found in the works of some of the earliest philosophers and mathematicians, the subject of whose writings must necessarily have led them to the discovery, and, in some measure, to the application of this science.

4. The oldest treatise of algebra, which has come down to the present times, was written by Diophantus of Alexandria, who flourished about the year 350 after Christ, and who wrote 13 books on algebra or arithmetic in the Greek language: though only five of these have hitherto been printed, and one book, which is imperfect, on multangular numbers. It was not, however, from this author, but from the Moors or Arabians, that this, as well as most other sciences was received in Europe; and some writers are of opinion, that they again received it from the Greeks, while others suppose that they had it from the Persians; and that these last derived algebra, as well as the arithmetical method of computing by ten characters or digits, from the Indians.

5. The Arabians themselves say, that it was invented by Mahomet ben Mula or Ion of Moles, who it seems flourished about the 8th or 9th century. It seems more probable that Mahomet was not the inventor, but only a person well skilled in the art; and that the Arabians received their knowledge of it from Diophantus, or other Greek writers, as they did that of geometry and some other sciences, which they improved and translated into their own language.

6. However this may be, it seems to be pretty certain, that the science was first brought to Europe about the beginning of the 15th century, by Leonardus Pisanus, who travelled into Arabia and other eastern countries for the purpose of acquiring mathematical knowledge; and, in a short time, it began to be cultivated in Italy, where it was called l'Arte Maggiore, "the greater art," to distinguish it from common arithmetic, which was called l'Arte Minore, "the lesser art." It was also known in that country by the name Regola de la Cofa, or "rule of the thing," where by Cofa, or the thing, was meant the first or simple power of the unknown quantity.

7. Between the years 1470 and 1487, Lucas Pacioli, or Lucas de Burgo, a Cordelier, or Minorite friar, published several treatises on arithmetic, algebra, and geometry; and, in 1494, his principal work, entitled Summa de Arithmetica Proporioni et Proportionalita was printed. The part of this work which relates to algebra, and which he calls l'Arte Maggiore; ditta dal vulgo la Regola de la Cofa over Algebra e Almucabala, may be considered as exhibiting a pretty accurate state of the science, as it was then known in Europe; and probably it was much the same in Africa and Asia, from whence the Europeans derived the knowledge of it. It appears from this work, that their knowledge extended no farther than quadratic equations, of which they used only the positive roots; that they used only one unknown quantity; that they used no marks nor signs for either quantities or operations, excepting a few abbreviations of the words or names themselves; and that the art was only employed in the resolution of certain numerical problems. So that either the Africans had not carried algebra beyond quadratic equations; or else (what indeed is not improbable) the Europeans had not learned the whole of the art, as it was then known to the former.

8. After the publication of the books of Lucas de Burgo, algebra became more generally known and improved, especially in Italy; for about the year 1505, Scipio Ferreus, who was then professor of mathematics at Bononia, found out a rule for resolving one case of a compound cubic equation; but, as it appears to have been the custom of the times with respect to such matters, he kept the rule a profound secret from his contemporaries. The same thing was afterwards discovered in 1535 by Nicolas Tartalea, who then resided in Venice, and who had five years before found the solution of two other cases of cubic equations.

9. The next work upon algebra which was printed after the books of Lucas de Burgo, was written by Hieronymus Cardan, of Bononia, a very learned man, who published in 1539 his arithmetical writings, in nine books, at Milan, where he practised physic, and read public lectures on mathematics. The same author in 1545 published a tenth book, containing the whole doctrine of cubic equations, which had been in part communicated to him under an oath of secrecy. by Tartalea, but which, notwithstanding this circumstance, Cardan thought proper to publish, alleging (not altogether without reason) that he had made so many additions to Tartalea's discovery as to render it in a manner his own. Accordingly we find, that even to the present times, the common rule for resolving cubic equations is generally known by the name of Cardan's rule, although it would certainly be more just to attribute it to its first inventor, Tartalea.

10. Equations of the fourth order appear to have been first resolved by Lewis Ferrari, a disciple of Cardan's; and different methods of resolution were afterwards given by Descartes and others. This indeed is the greatest length that mathematicians have been able to carry the resolution of equations; for, with respect to those of the fifth, and all higher degrees, all attempts to resolve them, except in particular cases, have hitherto been found impracticable.

11. After this period, writers on algebra became more numerous; and many improvements were gradually made, both in the notation and in the theory of the science. Among other writers who cultivated it with success may be reckoned Bombelli, another Italian mathematician; Stifelius and Scheubelius, both of Germany; Robert Recorde, an English mathematician; and many others.

12. Among the mathematicians to whom algebra is particularly indebted, it is proper to mention Francis Vieta, a native of France, who wrote about the year 1600. Among various improvements in all parts of the science, he first introduced the general use of the letters of the alphabet, to denote indefinite given quantities, which, before his time, had only been done in some particular cases. The English mathematician, Harriot, deserves also to be particularly mentioned. His algebra, which was published after his death, in 1631, shows that he cultivated that science with great success. For, besides improving the notation, so as to render it nearly the same as it is at present, he first explained clearly a most important proposition in the theory of equations, namely, that an equation of any degree may be considered as produced by the continual multiplication of as many simple equations as there are units in the exponent of the highest power of the unknown quantity in that equation: Hence he showed the relation which subsists between the coefficients of the terms of an equation and its roots.

13. Without mentioning all the writers on algebra who flourished about this time, and who feverishly contributed more or less to its improvement, we proceed to observe, that nothing has contributed more to the advancement of every branch of mathematical knowledge than the happy application which the celebrated philosopher Descartes made of algebra to the science of geometry; for his geometry, first published in 1637, may be considered rather as the application of algebra to geometry than as either algebra or geometry taken by itself as a science. Besides this happy union effected between the two sciences, Descartes contributed much to the improvement of both; and indeed he may be considered as having paved the way for all the discoveries which have since been made in mathematics.

14. After the publication of Descartes's Geometry, the science of algebra may be considered as having attained some degree of perfection. It has, however, received many improvements from later writers, who, pursuing the paths struck out by Harriot and Descartes, having produced many new and beautiful theories, both in algebra and geometry. The writers upon algebra from this time become too numerous, and the respective improvements made by each too minute, to be particularly noticed in this introduction. It is, however, necessary to mention another mathematician, to whom algebra lies under considerable obligations, namely, M. Fermat, who may be considered as the rival of Descartes; for it appears that he was in possession of the method of applying algebra to the improvement of geometry before the publication of the celebrated work of the latter philosopher. Besides, Fermat appears to have been deeply versed in the theory of indeterminate problems; and he republished the oldest and most esteemed treatise upon that subject which is known, namely, Diophantus's Arithmetic, to which he added many valuable notes of his own.

15. Having now given a brief account of the origin of algebra, and of the writers who contributed the most to bring it to the state of perfection it had attained about the middle of the 16th century, which indeed was considerable, we shall conclude this introduction, by observing, that although its progress has since been very gradual, it has been upon the whole considerably improved; particularly by the labours of the foreign mathematicians, Schooten, Hudde, Van-Heurcat, De Witte, Slusius, Huygens, &c. As to the algebraical writers of our own country, those whose labours have been most conspicuous were Wallis, and more especially Sir Isaac Newton, to whom, among other things, we owe the invention of the binomial theorem: also Pell, Barrow, Kersey, Halley, Raphson, and many others. We now proceed to explain the science itself.

Notation and Explanation of the Signs.

16. In arithmetic there are ten characters, which being variously combined, according to certain rules, serve to denote all magnitudes whatever. But this method of expressing quantities, although of the greatest utility in every branch of the mathematics (for we must always have recourse to it in the different applications of that science to practical purposes), is yet found to be inadequate, taken by itself, to the more difficult cases of mathematical investigation; and it is therefore necessary, in many inquiries concerning the relations of magnitude, to have recourse to that more general mode of notation, and more extensive system of operations, which constitute the science of algebra.

17. In algebra quantities of every kind may be denoted by any characters whatever, but those commonly used are the letters of the alphabet: And as in every mathematical problem, there are certain magnitudes given, in order to determine other magnitudes, which are unknown, the first letters of the alphabet a, b, c, &c., are used to denote known quantities, while those to be found are represented by x, y, z, &c., the last letters of the alphabet.

18. The sign + (plus) denotes that the quantity before which it is placed is to be added to some other quantity. Thus \(a + b\) denotes the sum of \(a\) and \(b\); \(3 + 5\) denotes the sum of 3 and 5, or 8.

19. The sign − (minus) signifies that the quantity before Notation. before which it is placed is to be subtracted. Thus \(a - b\) denotes the excess of \(a\) above \(b\); \(6 - 2\) is the excess of 6 above 2, or 4.

20. Quantities which have the sign \(+\) prefixed to them are called positive or affirmative; and such as have the sign \(-\) are called negative.

When quantities are considered abstractedly, the terms positive and negative can only mean that such quantities are to be added or subtracted; for as it is impossible to conceive a number less than 0, it follows, that a negative quantity by itself is unintelligible. But, in considering the affections of magnitude, it appears, that in many cases, a certain opposition may exist in the nature of quantities. Thus, a person's property may be considered as a positive quantity, and his debts as a negative quantity. Again, any portion of a line drawn to the right hand may be considered as positive, while a portion of the same line, continued in the opposite direction, may be taken as negative.

When no sign is prefixed to a quantity, \(+\) is always understood, or the quantity is to be considered as positive.

21. Quantities which have the same sign, either \(+\) or \(-\), are said to have like signs. Thus, \(+a + b\) have like signs, but \(+a - c\) and \(-a + c\) have unlike signs.

22. A quantity which consists of one term, is said to be simple; but if it consist of several terms, connected by the signs \(+\) or \(-\), it is then said to be compound. Thus \(+a + c\) are simple quantities; and \(b + c\), also \(+a + b - d\), are compound quantities.

23. To denote the product arising from the multiplication of quantities; if they be simple, they are either joined together, as if intended to form a word, or else the quantities are connected together, with the sign \(\times\) interposed between every two of them. Thus \(ab\), or \(a \times b\), denotes the product of \(a\) and \(b\); also \(abc\), or \(a \times b \times c\), denotes the product of \(a\), \(b\), and \(c\); the latter method is used when the quantities to be multiplied are numbers. If some of the quantities to be multiplied be compound, each of them has a line drawn over it called a vinculum, and the sign \(\times\) is interposed between as before. Thus \(a \times (c + d) \times e - f\) denotes that \(a\) is to be considered as one quantity, the sum of \(c\) and \(d\) as a second, and the difference between \(e\) and \(f\) as a third; and that these three quantities are to be multiplied into one another. Instead of placing a line over such compound quantities as enter a product, it is now common among mathematical writers to enclose each of them between two parentheses, so that the last product may be otherwise expressed thus, \(a(c + d)(e - f)\), or thus, \(a \times (c + d) \times (e - f)\).

24. A number prefixed to a letter is called a numerical coefficient, and denotes how often that quantity is to be taken. Thus, \(3a\) signifies that \(a\) is to be taken three times. When no number is prefixed, the coefficient is understood to be unity.

25. The quotient arising from the division of one quantity by another is expressed by placing the dividend above a line, and the divisor below it. Thus \(\frac{12}{3}\) denotes the quotient arising from the division of 12 by 3, or 4; \(\frac{b}{a}\) denotes the quotient arising from the division of \(b\) by \(a\). This expression of a quotient is also called a fraction.

26. The equality of two quantities is expressed by putting the sign \(=\) between them. Thus \(a + b = c - d\) denotes that the sum of \(a\) and \(b\) is equal to the excess of \(c\) above \(d\).

27. Simple quantities, or the terms of compound quantities, are said to be like, which consist of the same letter or letters. Thus \(+ab\) and \(-5ab\) are like quantities; but \(+ab\) and \(+abb\) are unlike.

There are some other characters which will be explained when we have occasion to use them; and in what follows we shall suppose that the operations of common arithmetic are sufficiently understood; for algebra, being an extension of that science, ought not to be embarrassed by the demonstration of its elementary rules.

Sect. I. Fundamental Operations.

28. The primary operations in algebra are the same as in common arithmetic, namely, addition, subtraction, multiplication, and division; and from the various combinations of these four, all the others are derived.

Problem I. To Add Quantities.

29. In addition there may be three cases: the quantities to be added may be like, and have like signs; or, they may be like, and have unlike signs; or, lastly, they may be unlike.

Case 1. To add quantities which are like, and have like signs.

Rule. Add together the coefficients of the quantities, prefix the common sign to the sum, and annex the letter, or letters, common to each term.

Examples.

| Add together | Add together | |--------------|--------------| | \(+7a\) | \(-2ax\) | | \(+3a\) | \(-ax\) | | \(+a\) | \(-5ax\) | | \(+2a\) | \(-12ax\) |

Sum, \(+13a\) Sum, \(-20ax\)

Case 2. To add quantities which are like, but have unlike signs.

Rule. Add the positive coefficients into one sum, and the negative ones into another; then subtract the least of these sums from the greatest, prefix the sign of the greatest to the remainder, and annex the common letter, or letters, as before.

Examples.

| Add together | Add together | |--------------|--------------| | \(+2ax\) | \(+6ab + 7\) | | \(+ax\) | \(+4ab + 9\) | | \(+3ax\) | \(+ab - 5\) | | \(+9ax\) | \(+7ab - 13\)|

Sum of the pos. \(+11ax\) Sum of the neg. \(-4ab - 18\)

Sum required, \(+7ax\) Sum required, \(+10ab - 2\) Cafe II. To multiply compound quantities.

Rule. Multiply every term of the multiplicand by all the terms of the multiplier, one after another, by the preceding rule, and collect their products into one sum, which will be the product required.

Examples.

Multiply \(4a - 2b + c\) By \(3a\) Product \(12aa - 6ab + 3ac\)

\(2x + y\) \(x - 2y\)

\(2xx + xy\) \(-4xy - 2yy\)

\(2xx - 3xy - 2yy\)

\(aa - ab + bb\) \(a + b\)

\(a - b + c\) \(a + b - c\)

\(aaa - aab + abb\) \(+aab - abb + bbb\)

\(aa - ab + ac\) \(+ab - bb + bc\)

\(-ac + bc - cc\)

\(aaa * * bbb\)

\(aa * * -bb + 2bc - cc\).

34. The reason of the rules for the multiplication of quantities may be explained in the following manner: Let it be required to multiply \(a - b\) by \(c - d\); because multiplication is a repeated addition of the multiplicand as often as the multiplier contains unity, therefore, \(a - b\) is to be taken as often as there are units in \(c - d\), and the sum will be the product required. Now if \(a - b\) be taken as often as there are units in \(c\), the result will evidently exceed the product required, and that by a quantity equal to \(a - b\), taken as often as there are units in \(d\). But, from the nature of addition \(a - b\) taken as often as there are units in \(c\), is \(ca - cb\), and for the same reason, \(a - b\) taken as often as there are units in \(d\) is \(da - db\); therefore, to obtain the product required, we must subtract \(da - db\) from \(ca - cb\): but from what has been shewn in subtraction, the remainder will be \(ca - cb - da + db\); therefore the product arising from the multiplication of \(a - b\) by \(c - d\) is \(ca - cb - da + db\); hence the reason of the general rule for the signs, as well as the other rules, is manifest.

35. When several quantities are multiplied together so as to constitute a product, each of them is called a factor of that product; thus \(a, b,\) and \(c\) are factors of the product \(abc\); also \(a + x,\) and \(b - x,\) are factors of the product \((a + x)(b - x)\).

36. The products arising from the continual multiplication of the same quantity are called powers of that quantity, which is called the root. Thus \(aa, ana,\) \(aaaa,\) &c. are powers of the root \(a.\) These powers are commonly expressed, by placing above the root, towards the right hand, a figure, denoting how often the root is repeated. This figure serves to denote the power, and is called its index or exponent. Thus, the quantity \(a\) being considered as the root, or as the first power of \(a,\) we have \(aa\) or \(a^2\) for its second power. Division. power, aaa or $a^3$ for its third power, aaaa or $a^4$ for its fourth power, and so on.

37. The second and third powers of a quantity are generally called its square and cube; and the fourth, fifth, and sixth powers are sometimes respectively called its biquadrate, surfoldid, and cubocube.

38. By considering the notation of powers, and the rules for multiplication, it appears that powers of the same root are multiplied by adding their exponents. Thus $a \times a^2 = a^3$, also $x^3 \times x^4 = x^7$; and in general $a^m \times a^n = a^{m+n}$.

**Prob. IV. To Divide Quantities.**

39. General Rule for the Signs.—If the signs of the divisor and dividend be like, the sign of the quotient is +; but if they be unlike, the sign of the quotient is −.

This rule is easily derived from the general rule for the signs in multiplication, by considering that the quotient must be such a quantity as when multiplied by the divisor shall produce the dividend, with its proper sign.

40. The quotient arising from the division of one quantity by another may be expressed by placing the dividend above a line and the divisor below it, (§ 25); but it may also be often expressed in a more simple manner by the following rules:

**Case 1.** When the divisor is simple, and a factor of every term of the dividend.

**Rule.** Divide the coefficient of each term of the dividend by the coefficient of the divisor, and exchange out of each term the letter or letters in the divisor: the result is the quotient.

**Ex. 1.** Divide $12abc$ by $3ac$.

From the method of notation, the quotient may be expressed thus, $\frac{12abc}{3ac}$; but the same quotient, by the rule just given, is more simply expressed thus, $4b$.

**Ex. 2.** Divide $16a^2xy - 28a^2wx^2 + 4a^2w^2$ by $4a^2x$.

The quotient is $4ay - 7wx^2 + x^2$.

If the divisor and dividend be powers of the same quantity, the division will evidently be performed by subtracting the exponent of the divisor from that of the dividend. Thus $a^5$, divided by $a^3$, has for a quotient $a^5 - a^3 = a^2$.

**Case 2.** When the divisor is simple, but not a factor of the dividend.

**Rule.** The quotient is expressed by a fraction, of which the numerator is the dividend, and the denominator the divisor.

Thus the quotient of $3ab^2$, divided by $2mbc$, is the fraction $\frac{3ab^2}{2mbc}$.

It will sometimes happen, that the quotient found thus may be reduced to a more simple form, as shall be explained when we come to treat of fractions.

**Case 3.** When the divisor is compound.

**Rule 1.** The terms of this dividend are to be arranged according to the powers of some one of its letters, and those of the divisor according to the powers of the same letter.

2. The first term of the dividend is to be divided by the first term of the divisor, observing the general rule for the signs; and this quotient being set down for a part of the quotient wanted, is to be multiplied by the whole divisor, and the product subtracted from the dividend. If nothing remain, the division is finished; but if there be a remainder, it is to be taken for a new dividend.

3. The first term of the new dividend is next to be divided by the first term of the dividend, as before, and the quotient joined to the part already found, with its proper sign. The whole divisor is also to be multiplied by this part of the quotient, and the product subtracted from the new dividend; and thus the operation is to be carried on till there be no remainder, or till it appear that there will always be a remainder.

To illustrate this rule, let it be required to divide $8a^2 + 2ab - 15b^2$ by $2a + 3b$, the operation will stand thus:

$$\begin{align*} &2a + 3b)8a^2 + 2ab - 15b^2(4a - 5b \\ &8a^2 + 12ab \\ &- 10ab - 15b^2 \\ &- 10ab - 15b^2 \end{align*}$$

Here the terms of the divisor and dividend are arranged according to the powers of the quantity $a$. We now divide $8a^2$, the first term of the dividend, by $2a$ the first term of the divisor; and thus get $4a$ for the first term of the quotient. We next multiply the divisor by $4a$, and subtract the product $8a^2 + 12ab$ from the dividend; we thus get $-10ab - 15b^2$ for a new dividend.

By proceeding in all respects as before, we find $-5b$ for the second term of the quotient, and no remainder; the operation is therefore finished, and the whole quotient is $4a - 5b$.

The following examples will also serve to illustrate the manner of applying the rule.

**Ex. 1.**

$$\begin{align*} 3a - b)3a^3 - 12a^2 + ab - 2b^2(a^2 - 4a + 2b \\ 3a^2 - a^2b \\ - 12a^2 + 10ab \\ - 12a^2 + 4ab \\ + 6ab - 2b^2 \\ + 6ab - 2b^2 \end{align*}$$

**Ex. 2.**

$$\begin{align*} a + b)a^3 + b^3(a^3 - ab + b^3 \\ a^3 + a^2b \\ - ab + b^3 \\ - ab^2 - ab^2 \\ + ab^2 + b^3 \\ + ab^2 + b^3 \end{align*}$$

**Ex. 3.** 46. The reciprocal of a fraction is another fraction, having its numerator and denominator respectively equal to the denominator and numerator of the former.

Thus \( \frac{b}{a} \) is the reciprocal of the fraction \( \frac{a}{b} \).

47. The following proposition is of great importance in the operations relating to fractions.

If the numerator and denominator of a fraction be either both multiplied, or both divided, by the same quantity, the value of that fraction is the same as before.

For let any fraction \( \frac{b}{a} = c \); then because \( c \) is the quotient arising from the division of \( b \) by \( a \), it follows that \( b = ac \); and multiplying both by any quantity \( n \), we have \( nb = nc \): let these equals be both divided by the same quantity \( na \), and the quotients will be equal, that is \( \frac{nb}{na} = \frac{c}{a} \); hence the truth of the proposition is manifest.

48. From this proposition, it is obvious that a fraction may be very differently expressed, without changing its value, and that any integer may be reduced to the form of a fraction, by placing the product arising from its multiplication by any assumed quantity as the numerator, and the assumed quantity as the denominator of the fraction. It also appears that a fraction very complex in its form may often be reduced to another of the same value, but more simple, by finding a quantity which will divide both the numerator and denominator, without leaving a remainder. Such a common measure, or common divisor, may be either simple or compound; if it be simple, it is readily found by inspection, but if it be compound, it may be found as in the following problem.

49. Prob. I. To find the greatest common Measure of two Quantities.

Rule 1. Range the quantities according to the power of some one of the letters, as taught in division, leaving out the simple divisors of each quantity.

2. Divide that quantity which is of most dimensions by the other one, and if there be a remainder, divide it by its greatest simple divisor; and then divide the last compound divisor by the resulting quantity, and if any thing yet remain, divide it also by its greatest simple divisor, and the last compound divisor by the resulting quantity; proceed in this way till nothing remain, and the last divisor shall be the common measure required.

Note. It will sometimes be necessary to multiply the dividends by simple quantities in order to make the divisions succeed.

Ex. 1. Required the greatest common measure of the quantities \( ax^3 - x^3 \) and \( a^3 - 2ax + ax^2 \). The simple divisor \( x \) being taken out of the former of these quantities, and \( a \) out of the latter, they are reduced to \( a^2 - x^2 \), and \( a^2 - 2ax + x^2 \), and as the quantity \( a \) rises to the same dimensions in both, we may take either of them as the first divisor; let us take that which consists of fewest terms, and the operation will stand thus: \[ \frac{a^2 - x^2}{a^2 - x^2} = \frac{a^2 - ax + x^2}{a^2 - x^2} \]

which divided by \(2x\) is \(a-x)(a+x)\)

\[ \frac{a^2 - ax}{a^2 - x^2} + ax - x^2 \]

Hence it appears that \(a-x\) is the greatest common measure required.

Ex. 2. Required the greatest common measure of \(8a^2b^2 - 15ab^3 + 2b^4\) and \(9a^4b - 9a^3b^2 + 3a^2b^3 - 3ab^4\).

It is evident, from inspection, that \(b\) is a simple divisor of both quantities; it will therefore be a factor of the common measure required. Let the simple divisors be now left out of each quantity, and they are reduced to \(4a^2 - 5ab + b^2\) and \(3a^3 - 3a^2b + ab^2 - b^3\); but as the second of these is to be divided by the first, it must be multiplied by 4 to make the division succeed, and the operation will stand thus:

\[ \begin{align*} 4a^2 - 5ab + b^2 & \quad 12a^3 - 12a^2b + 4ab^2 - 4b^3 \\ & \quad (3a^2 + ab - 4b^2) \\ & \quad 12a^3 - 15ab + 3b^2 \end{align*} \]

This remainder is to be divided by \(b\), and the new dividend multiplied by 3, to make the division again succeed, and the work will stand thus:

\[ \begin{align*} 3a^2 + ab - 4b^2 & \quad 12a^3 - 15ab + 3b^2 \\ & \quad (4a^2 + 4ab - 16b^2) \\ & \quad 19ab + 19b^2 \end{align*} \]

This remainder is to be divided by \(-19b\), which being done, and the last divisor taken as a dividend as before, the rest of the operation will be as follows:

\[ \begin{align*} a-b & \quad 3a^2 + ab - 4b^2(3a + 4b) \\ & \quad 3a^2 - 3ab \\ & \quad + 4ab - 4b^2 \\ & \quad + 4ab - 4b^2 \end{align*} \]

from which it appears that the compound divisor sought is \(a-b\), and remarking that the quantities proposed have also a simple divisor \(b\), the greatest common measure which is required will be \(b(a-b)\).

50. The reason of the rule given in this problem may be deduced from the following considerations:

1. If two quantities have a compound divisor common to both, and they be either multiplied or divided by any simple quantities, the results will each have the same compound divisor. Thus the quantities \(p(a-x)\) and \(q(a-x)\) have the common divisor \(a-x\), and the quantities \(n(p(a-x), r(q(a-x))\) have each the very same divisor.

2. In the operation of division, whatever quantity measures both the divisor and dividend, the same will also measure the remainder. For let \(x\) be such a quantity, then the divisor and dividend may be represented by \(ax\) and \(bx\); let \(q\) be the quotient, and the remainder will evidently be \(bx-qux\), which is evidently divisible by \(x\).

3. Whatever quantity measures both the divisor and remainder, the same will also measure the dividend. For let the divisor be \(ax\), and the remainder \(rx\), then, \(q\) denoting the quotient, the dividend will be \(aqx+rx\), which, as well as the divisor and dividend, is divisible by \(x\).

51. Let us apply these observations to the last example. From the first observation, the reason for leaving out the simple quantities in the course of the operation, as well as for multiplying by certain other quantities, to make the divisions succeed, is obvious; and from the second observation it appears, that whatever quantity measures \(a^2-5ab+b^2\), and \(12a^3-12a^2b+4ab^2-4b^3\), the same must measure \(3a^2+ab-4b^2\), the first remainder, as also \(19ab+19b^2\), the second remainder; but the only compound divisor which this last quantity can have is \(a-b\), which is also found to be a divisor of \(3a^2+ab-4b^2\), or of \(3a^2+ab-4b^2\), the first remainder, therefore, by the third observation, \(a-b\) must also be a divisor of \(12a^3-15ab+3b^2\), or of \(4a^2-5ab+b^2\), the first divisor, and therefore also it must be a divisor of \(12a^3-12a^2b+4ab^2-4b^3\), the first dividend, so that \(a-b\) is the greatest common measure, as was required.

52. Prob. II. To Reduce a Fraction to its lowest Terms.

Rule. Divide both numerator and denominator by their greatest common measure, which may be found by prob. I.

Ex. 1. Reduce \(\frac{56a^2bc}{240abc}\) to its lowest terms.

It appears from inspection, that the greatest common measure is \(8ac\), and dividing both numerator and denominator by this quantity, we have \(\frac{56a^2bc}{240abc} = \frac{7ab}{3dc}\).

Ex. 2. Reduce \(\frac{a^2x-x^3}{a^2-2a^2x+ax^2}\) to its lowest terms.

We have already found in the first example of prob. I, that the greatest common measure of the numerator and denominator is \(a-x\); and dividing both by this quantity we have

\[ \frac{a^2x-x^3}{a^2-2a^2x+ax^2} = \frac{ax+x^2}{a^2-ax} \]

In like manner we find \(\frac{2a^2b-9a^3b^2+3a^2b^3-3ab^4}{8a^2b^2-10ab^3+2b^4} = \frac{9a^2+3ab^2}{8ab-2b^2}\); the common measure being \(b(a-b)\) as was shown in example 2, problem I.

53. Prob. III. To Reduce a mixed Quantity to an improper Fraction.

Rule. Multiply the integer by the denominator of the fraction, and to the product add the numerator, and the denominator being placed under this sum will give the improper fraction required. Ex. 1. Let \( x + \frac{x^2}{a} \) and \( x - \frac{a^2 - x^2}{x} \) be reduced to improper fractions.

First \( x + \frac{x^2}{a} = \frac{ax + x^2}{a} \), the answer.

And \( x - \frac{a^2 - x^2}{x} = \frac{x^2 - a^2 + x^2}{x} = \frac{2x^2 - a^2}{x} \), Ans.

Ex. 2. Reduce \( a - x + \frac{x^2}{a + x} \) to an improper fraction.

\( a - x + \frac{x^2}{a + x} = \frac{(a + x)(a - x) + x^2}{a + x} = \frac{a^2}{a + x} \), Ans.

54. Prob. IV. To Reduce an Improper Fraction to a whole or mixed number.

Rule. Divide the numerator by the denominator for the integral part, and place the remainder, if any, over the denominator, and it will be the mixed quantity required.

Ex. 1. Reduce \( \frac{ax + a^2}{x} \) to a whole or mixed quantity.

\( ax + a^2 = a + \frac{a^2}{x} \) the answer required.

Ex. 2. Reduce \( \frac{ax + 2x^2}{a + x} \), also \( \frac{x^2 - y^2}{x - y} \), to whole or mixed quantities.

First \( \frac{ax + 2x^2}{a + x} = x + \frac{x^2}{a + x} \) the answer.

And \( \frac{x^2 - y^2}{x - y} = x + y \) a whole quantity, which is the answer.

55. Prob. V. To Reduce Fractions of different Denominators to others of the same value which shall have a common Denominator.

Rule. Multiply each numerator separately into all the denominators except its own for the new numerators, and all the denominators together for the common denominator.

Ex. 1. Reduce \( \frac{a}{b}, \frac{c}{d}, \frac{e}{f} \) to fractions of equal value which have a common denominator.

\[ \begin{align*} a \times d \times f &= adf \\ c \times b \times f &= cbf \\ e \times b \times d &= ebd \\ b \times d \times f &= bdf \end{align*} \]

New numerators.

Hence we find \( \frac{a}{b} = \frac{adf}{bdf}, \frac{c}{d} = \frac{cbf}{bdf}, \frac{e}{f} = \frac{ebd}{bdf} \) where the new fractions have a common denominator, as was required.

Ex. 2. Reduce \( \frac{ax}{a - x} \) and \( \frac{a^2 - x^2}{a + x} \) to fractions of equal value and having a common denominator.

\[ \begin{align*} ax(a + x) &= a^3 + ax^2 \\ (a^2 - x^2)(a + x) &= a^3 - ax^2 - ax^2 + x^3 \\ a - x)(a + x) &= a^2 - x^2 \end{align*} \]

the common denominator.

Hence \( \frac{ax}{a - x} = \frac{a^2 + ax^2}{a^2 - x^2} \) and \( \frac{a^2 - x^2}{a + x} = \frac{a^2 - ax^2 - ax^2 + x^3}{a^2 - x^2} \).

56. Prob. VI. To Add or Subtract Fractions.

Rule. Reduce the fractions to a common denominator, and add or subtract their numerators, and the sum or difference placed over the common denominator, is the sum or remainder required.

Ex. 1. Add together \( \frac{a}{b}, \frac{c}{d}, \frac{e}{f} \).

\[ \begin{align*} \frac{a}{b} &= \frac{adf}{bdf} \\ \frac{c}{d} &= \frac{bcf}{bdf} \\ \frac{e}{f} &= \frac{bde}{bdf} \end{align*} \]

Hence \( \frac{a}{b} + \frac{c}{d} + \frac{e}{f} = \frac{adf + bcf + bde}{bdf} \) the sum required.

Ex. 2. From \( \frac{a + x}{a} \) subtract \( \frac{a + x}{a + x} \).

\[ \begin{align*} \frac{a + x}{a} &= \frac{a^2 + 2ax + x^2}{a^2 + ax} \\ \frac{a}{a + x} &= \frac{a^2 + ax}{a^2 + ax} \end{align*} \]

Hence \( \frac{a + x}{a} - \frac{a}{a + x} = \frac{2ax + x^2}{a^2 + ax} \).

Ex. 3. Add together \( \frac{x + 2}{3}, \frac{x}{4}, \frac{x - 5}{2} \).

\[ \frac{x + 2}{3} + \frac{x}{4} + \frac{x - 5}{2} = \frac{8x + 16 + 6x + 12x - 60}{24} = \frac{13x - 22}{12} \]

If it be required to add or subtract mixed quantities, they may either be reduced to the form of fractions by prob. 3, and then added or subtracted, or else these operations may be performed first on the integer quantities, and afterwards on the fractions.

57. Prob. VII. To Multiply Fractions.

Rule. Multiply the numerators of the fractions for the numerator of the product, and the denominators for the denominator of the product.

Ex. 1. Multiply \( \frac{b}{a} \) by \( \frac{d}{c} \).

\[ \frac{b}{a} \times \frac{d}{c} = \frac{bd}{ac} \text{ the product required.} \]

Ex. 2. Multiply \( \frac{a + b}{c} \) by \( \frac{a - b}{d} \).

\[ \frac{a + b}{c} \times \frac{a - b}{d} = \frac{a^2 - b^2}{cd}, \text{ the product.} \]

If it be required to multiply an integer by a fraction, the integer may be considered as having unity for a denominator. Thus \( (a + x) + \frac{3d}{c} = \frac{a + x}{1} \times \frac{3d}{c} = \frac{3ad + 3dx}{c} \). Fractions. Mixed quantities may be multiplied after being reduced to the form of fractions by prob. 3. Thus

\[ (b + \frac{bx}{a}) \times \frac{a}{x} = \frac{ab + bx}{a} \times \frac{a}{x} = \frac{a^2b + abx}{ax} = \frac{ab + bx}{x}. \]

58. The reason of the rule for multiplication may be explained thus. If \( \frac{a}{b} \) is to be multiplied by \( c \), the product will evidently be \( \frac{ac}{b} \); but if it is only to be multiplied by \( \frac{c}{d} \) the former product must be divided by \( d \), and it becomes \( \frac{ac}{bd} \) which is the product required.

Or let \( \frac{a}{b} = m \), and \( \frac{c}{d} = n \), then \( a = bm \) and \( c = dn \) and \( ac = bdmn \); hence \( mn \) or \( \frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd} \).

59. Prob. VIII. To Divide Fractions.

Rule. Multiply the denominator of the divisor by the numerator of the dividend for the numerator of the quotient. Then multiply the numerator of the divisor by the denominator of the dividend for the denominator of the quotient.

Or, multiply the dividend by the reciprocal of the divisor, the product will be the quotient required.

Ex. 1. Divide \( \frac{a}{b} \) by \( \frac{c}{d} \).

\[ \frac{c}{d} \left( \frac{ad}{bc} \right) \text{ the quotient required, or } \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc} \text{ as before.} \]

Ex. 2. Divide \( \frac{a^2 + ab}{2x} \) by \( \frac{3a^2}{a - b} \).

\[ \frac{3a^2}{a - b} \left( \frac{a^2 + ab}{2x} \right) = \frac{a^2 - b}{6ax} \text{ the quotient.} \]

If either the divisor or dividend be an integer quantity, it may be represented as a fraction, by placing unity for a denominator; or if it be a mixed quantity, it may be reduced to a fraction by prob. 3. and the operation of division performed agreeably to the rule.

60. The reason of the rule for division may be explained thus, let it be required to divide \( \frac{c}{d} \) by \( \frac{a}{b} \). If \( \frac{c}{d} \) is to be divided by \( a \), the quotient is \( \frac{c}{ad} \); but if it is to be divided by \( \frac{a}{b} \), then the last quotient must be multiplied by \( b \); thus we have \( \frac{cb}{ad} \) for the quotient required. Or let \( \frac{a}{b} = m \), and \( \frac{c}{d} = n \), then \( a = bm \) and \( c = dn \); also \( ad = bdm \) and \( bc = bdn \); therefore \( \frac{bdn}{bdm} = \frac{n}{m} = \frac{bc}{ad} \).

Sect. III. Of Involution and Evolution.

61. In treating of multiplication, we have observed, that when a quantity is multiplied by itself any number of times, the product is called a power of that quantity, while the quantity itself, from which the powers are formed, is called the root (\$ 36.) Thus \( a \), \( a^2 \), and \( a^3 \) are the first, second, and third powers of the root \( a \); and in like manner \( \frac{1}{a} \), \( \frac{1}{a^2} \), and \( \frac{1}{a^3} \), denote the same powers of the root \( \frac{1}{a} \).

62. But before considering more particularly what relates to powers and roots, it will be proper to observe, that the quantities \( \frac{1}{a} \), \( \frac{1}{a^2} \), \( \frac{1}{a^3} \), &c. admit of being expressed under a different form; for, like as the quantities \( a \), \( a^2 \), \( a^3 \), &c. are expressed as positive powers of the root \( a \), so the quantities \( \frac{1}{a} \), \( \frac{1}{a^2} \), \( \frac{1}{a^3} \), &c. may be respectively expressed thus, \( a^{-1} \), \( a^{-2} \), \( a^{-3} \), &c. and considered as negative powers of the root \( a \).

63. This method of expressing the fractions \( \frac{1}{a} \), \( \frac{1}{a^2} \), \( \frac{1}{a^3} \), as powers of the root \( a \), but with negative indices, is a consequence of the rule which has been given for the division of powers; for we may consider \( \frac{1}{a} \) as the quotient arising from the division of any power of \( a \) by the next higher power, for example from the division of the \( 2a \) by the \( 3a \), and so we have \( \frac{1}{a} = \frac{a^2}{a^3} \); but since powers of the same quantity are divided by subtracting the exponent of the divisor from that of the dividend (\$ 40.), it follows, that \( \frac{a^2}{a^3} = a^{2-3} = a^{-1} \); therefore the fraction \( \frac{1}{a} \) may also be expressed thus, \( a^{-1} \). By considering \( \frac{1}{a} \) as equal to \( \frac{a^2}{a^4} \), it will appear in the same manner that \( \frac{1}{a^2} = \frac{a^3}{a^4} = a^{-2} \); and proceeding in this way, we get \( \frac{1}{a^3} = \frac{a^4}{a^5} = a^{-3} \), \( \frac{1}{a^4} = \frac{a^5}{a^6} = a^{-4} \), &c. and so on, as far as we please. It also appears, that unity or 1 may be represented by \( a^0 \), where the exponent is a cypher, for \( 1 = \frac{a^0}{a^0} = a^{0-0} = a^0 \).

64. The rules which have been given for the multiplication and division of powers with positive exponents will apply in every case, whether the exponents be positive or negative; and this must evidently take place, for the mode of notation, by which we represent fractional quantities as the powers of integers, but with negative exponents, has been derived from those rules. Thus \( \frac{1}{a^2} \times a^3 \) or \( a^{-2} \times a^3 = a^{-2+3} = a^1 = \frac{1}{a^2} \times \frac{1}{a^3} \). 65. From this method of notation it appears, that any quantity may be taken from the denominator of a fraction, and placed in the numerator, by changing the sign of its exponent; and hence it follows, that every fraction may also be represented as an integer quantity. Thus \(\frac{a^2}{bc^3}\) denotes the same thing as \(\frac{a^2b^{-1}}{c^3}\) or as \(a^2b^{-1}c^{-1}\), also \(\frac{a^2}{(x-1)^3}\) may be otherwise expressed thus, \(a^2(x-1)^{-3}\).

Of Involution.

66. Involution is the method of finding any power of any assigned quantity, whether it be simple or compound; hence its rules are easily derived from the operation of multiplication.

Case 1. When the quantity is simple.

Rule. Multiply the exponents of the letters by the index of the power required, and raise the coefficient to the same power.

Note. If the sign of the quantity be +, all its powers will be positive: but if it be —, then all its powers, whose exponents are even numbers, are positive, and all its powers whose exponents are odd numbers are negative.

Ex. 1. Required the cube, or third power of \(2a^2x\). \((2a^2x)^3 = 2 \times 2 \times 2a^2x^3 = 8a^2x^3\), the answer.

Ex. 2. Required the fifth power of \(-3a^2x^3\). \((-3a^2x^3)^5 = -243a^{10}x^{15}\), the answer.

Ex. 3. Required the fourth power of \(-\frac{2ax^2}{3b^3y}\). \(\left(-\frac{2ax^2}{3b^3y}\right)^4 = \frac{16a^4x^8}{81b^6y^4}\), the answer.

Case 2. When the quantity is compound.

Rule. The powers must be found by a continual multiplication of the quantity by itself.

Ex. Required the first four powers of the binomial quantity \(a+x\).

\[ \begin{align*} a+x & \text{ the root, or first power} \\ a^2+2ax+x^2 & \text{ the square, or second power} \\ a^3+3a^2x+3ax^2+x^3 & \text{ the cube, or third power} \\ a^4+4a^3x+6a^2x^2+4ax^3+x^4 & \text{ the fourth power.} \end{align*} \]

If it be required to find the same powers of \(a-x\), it will be found, that

\[ \begin{align*} a-x & \text{ is the root or first power;} \\ a^2-2ax+x^2 & \text{ the square, or 2d power;} \\ a^3-3a^2x+3ax^2-x^3 & \text{ the cube, or 3d power;} \\ a^4-4a^3x+6a^2x^2-4ax^3+x^4 & \text{ the 4th power.} \end{align*} \]

Hence it appears, that the powers of \(a+x\) differ from the powers of \(a-x\), only in this respect, that in the former the signs of the terms are all positive, but in the latter, they are positive and negative alternately.

67. Besides the method of finding the powers of a compound quantity by multiplication, which we have just now explained, there is another, more general, as well as more expeditious, by which a quantity may be raised to any power whatever without the trouble of finding any of the inferior powers, namely, by means of what is commonly called the binomial theorem. This theorem may be expressed as follows. Let \(a+x\) be a binomial quantity, which is to be raised to any power denoted by the number \(n\), then \((a+x)^n = a^n + \frac{n(n-1)}{1 \cdot 2} a^{n-2}x^2 + \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} a^{n-3}x^3 + \cdots + \frac{n(n-1)(n-2)\cdots(n-4)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} a^{n-5}x^5 + \cdots + x^n\). This series will always terminate when \(n\) is any whole positive number, by reason of some one of the factors \(n-1, n-2, \ldots\), becoming zero; but if \(n\) be either a negative or fractional number, the series will consist of an infinite number of terms; as, however, we mean to treat in this section only of the powers of quantities when their exponents are whole positive numbers, we shall make no farther remarks upon any other; we shall afterwards give a demonstration of the theorem, and shew its application to fractional and negative powers in treating of infinite series. The \(n\)th power of \(a+x\) will not differ from the same power of \(a-x\), but in the signs of the terms which compose it, for it will stand thus: \((a-x)^n = a^n - \frac{n(n-1)}{1 \cdot 2} a^{n-2}x^2 + \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} a^{n-3}x^3 - \cdots - \frac{n(n-1)(n-2)\cdots(n-4)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} a^{n-5}x^5 + \cdots + x^n\), where the signs are + and — alternately.

Ex. 1. Let it be required to raise \(a+x\) to the fifth power.

Here \(n\) the exponent of the power being 5, the first term \(a^n\) of the general theorem will be equal to \(a^5\); the second \(na^{n-1}x = 5a^4x\), the third \(\frac{n(n-1)}{1 \cdot 2} a^{n-2}x^2 = \frac{5 \times 4}{1 \cdot 2} a^3x^2 = 10a^3x^2\), the fourth \(\frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} a^{n-3}x^3 = \frac{5 \times 4 \times 3}{1 \cdot 2 \cdot 3} a^2x^3 = 10a^2x^3\), the fifth \(\frac{n(n-1)(n-2)(n-3)}{1 \cdot 2 \cdot 3 \cdot 4} a^{n-4}x^4 = \frac{5 \times 4 \times 3 \times 2}{1 \cdot 2 \cdot 3 \cdot 4} ax^4 = 5ax^4\), and the sixth and last \(\frac{n(n-1)(n-2)(n-3)(n-4)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} a^{n-5}x^5 = \frac{5 \times 4 \times 3 \times 2 \times 1}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} a^0x^5 = x^5\); the remaining terms Evolution of the general theorem all vanish, by reason of the factor \( n - s = 0 \) by which each of them is multiplied, so that we get \((a + x)^5 = a^5 + 5a^4x + 10a^3x^2 + 10a^2x^3 + 5ax^4 + x^5\).

Ex. 2. It is required to raise \( 2d - \frac{x}{5} \) to the third power.

In this case \( n = 3 \), so that if we put \( a = 2d \) and \( x = \frac{x}{3} \) we have the first term of the general theorem, or \( a^n = 8d^3 \), the second \( \frac{n}{1} a^{n-1}x = 3 \times 4d^2 \times \frac{x}{2} = 4d^2x \), the third \( \frac{n(n-1)}{1 \cdot 2} a^{n-2}x^2 = 3 \times 2d \times \frac{x^2}{9} = \frac{2d^2x^2}{3} \), and the fourth and last term \( \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} a^{n-3}x^3 = \frac{2d^3x^3}{27} \), and since the signs of the terms of any power of \( a - x \) are \( + \) and \( - \) alternately, we have \((2d - \frac{x}{3})^3 = 8d^3 - 4d^2x + \frac{2d^2x^2}{3} - \frac{x^3}{27}\).

68. If the quantity to be involved consists of more than two terms, as if \( p + q - r \) were to be raised to the 2d power, put \( p = a \) and \( q - r = b \) then \((p + q - r)^2 = (a + b)^2 = a^2 + 2ab + b^2 = p^2 + 2p(q - r) + (q - r)^2 \), but \( 2p(q - r) = 2pq - 2pr \), and by the general theorem \((q - r)^2 = q^2 - 2qr + r^2 \), therefore, we get \((p + q - r)^2 = p^2 + 2pq - 2pr + q^2 - 2qr + r^2 \); and by a similar method of procedure a quantity consisting of four or more terms may be raised to any power.

Of Evolution.

69. Evolution is the reverse of involution, or it is the method of finding the root of any quantity, whether simple or compound, which is considered as a power of that root; hence it follows that its operations, generally speaking, must be the reverse of those of involution.

70. To denote that the root of any quantity is to be taken, the sign \( \sqrt{} \) (called the radical sign) is placed before it, and a small number placed over the sign to express the denomination of the root. Thus \( \sqrt{a} \) denotes the square root of \( a \), \( \sqrt[3]{a} \) its cube root, \( \sqrt[4]{a} \) its fourth root, and in general, \( \sqrt[n]{a} \) its \( n \)th root. The number placed over the radical sign is called the index or exponent of the root, and is usually omitted in expressing the square root, thus either \( \sqrt{a} \) or \( \sqrt{a} \) denotes the square root of \( a \).

71. Case 1. When roots of simple quantities are to be found.

Rule. Divide the exponents of the letters by the index of the root required, and prefix the root of the numeral coefficient; the result will be the root required.

Note 1. The root of any positive quantity may be either positive or negative, if the index of the root be an even number; but if it be an odd number, the root can be positive only.

2. The root of a negative quantity is also negative when the index of the root is an odd number.

3. But if the quantity be negative, and the index of the root even, then no root can be assigned.

Ex. 1. Required the square root of \( 36a^2x^4 \).

Here the index of the root is 2, and the root of the coefficient 6, therefore \( \sqrt{36a^2x^4} = \pm 6ax^2 \) or \( \sqrt{36a^2x^2} = \pm 6ax^2 \), for neither of these quantities, when multiplied by itself, produces \( 36a^2x^4 \); so that the root required is \( \pm 6ax^2 \), where the sign \( \pm \) denotes that the quantity to which it is prefixed may be considered either as positive or negative.

Ex. 2. Required the cube root of \( 125a^6x^9 \).

Here the index of the root is 3, and the root of the coefficient 5, therefore \( \sqrt[3]{125a^6x^9} = 5a^2x^3 \) the root required; and in like manner the cube root of \( -125a^6x^9 \) is found to be \( -5a^2x^3 \).

72. If it be required to extract the square of \( -a^2 \), it will immediately appear that no root can be assigned; for it can neither be \( +a \), nor \( -a \), seeing that each of these quantities, when squared, produces \( +a^2 \); the root required is therefore said to be impossible, and may be expressed thus: \( \sqrt{-a^2} \).

The root of a fraction is found by extracting that root out of both numerator and denominator. Thus the square root of \( \frac{4a^2x^4}{9b^2y^6} \) is \( \frac{2ax^2}{3by^3} \).

Case 2. When the quantity of which the root is to be extracted is compound.

73. 1. To extract the square root.

Range the terms of the quantity according to the powers of the letters, as in division.

Find the square root of the first term for the first part of the root sought, subtract its square from the given quantity, and divide the remainder by double the part already found, and the quotient is the second term of the root.

Add the second part to double the first, and multiply their sum by the second part, subtract the product from the remainder, and if nothing remain, the square root is obtained. But if there is a remainder, it must be divided by the double of the parts already found, and the quotient will give the third term of the root, and so on.

Ex. 1. Required the square root of \( a^2 + 2ax + x^2 \).

\[ \begin{align*} \frac{a^2 + 2ax + x^2}{a^2} & = a \\ \frac{2ax + x^2}{a} & = 2ax + x^2 \end{align*} \]

Ex. 2. Ex. 2. Required the square root of \( x^4 - 2x^3 + \frac{3}{2}x^2 - \frac{x}{2} + \frac{1}{16} \).

\[ \begin{align*} & x^4 - 2x^3 + \frac{3}{2}x^2 - \frac{x}{2} + \frac{1}{16} \\ \times & -x - 2x^3 + x^2 \\ = & 2x^4 - 2x^3 + \frac{1}{4}x^2 - \frac{x}{2} + \frac{1}{16} \\ \times & \frac{1}{4}x^3 - \frac{x}{2} + \frac{1}{16} \\ = & * * * \end{align*} \]

74. To understand the reason of the rule for finding the square root of a compound quantity, it is only necessary to involve any quantity, as \( a + b + c \) to the second power, and observe the composition of its square; for we have \((a + b + c)^2 = a^2 + 2ab + b^2 + 2ac + 2bc + c^2\); but \(2ab + b^2 = (2a + b)b\) and \(2ac + 2bc + c^2 = (2a + 2b + c)c\), therefore,

\[(a + b + c)^2 = a^2 + (2a + b)b + (2a + 2b + c)c;\]

and from this expression the manner of deriving the rule is obvious.

As an illustration of the common rule for extracting the square root of any proposed number, we shall suppose that the root of 59049 is required.

Accordingly we have \((a + b + c)^2 = 59049\), and from hence we are to find the values of \(a\), \(b\), and \(c\).

\[ \begin{align*} a^2 &= 200 \times 200 = 40000 \\ b^2 &= 400 \\ c^2 &= 19049 \\ 2a + b &= 440 \\ 2a + 2b &= 880 \\ 2a + 2b + c &= 483 \end{align*} \]

The same example when wrought by the common rule (see Arithmetic) will stand thus:

\[ \begin{align*} a^3 &= 13312053 \\ 3a^2 &= 1200000 \\ 3ab &= 18000 \\ b^2 &= 900 \\ 3a^2 + 3ab + b^2 &= 1389000 \\ 3(a + b)^2 &= 158700 \\ 3(a + b)c &= 4830 \\ c^2 &= 49 \\ 3(a + b)^2 + 3(a + b)c + c^2 &= 163579 \end{align*} \]

and by a comparison of the two operations, the reason of the common rule is obvious.

75. II. To extract the cube root.

Range the terms of the quantity according to the powers of some one of the letters.

Find the root of the first term, for the first part of the root sought; subtract its cube from the whole quantity, and divide the remainder by 3 times the square of the part already found, and the quotient is the second part of the root.

Add together, 3 times the square of the part of the root already found, 3 times the product of that part and the second part of the root, and the square of the second part; multiply the sum by the second part, and subtract the product from the first remainder, and if nothing remain, the root is obtained; but if there is a remainder, it must be divided by 3 times the square of the sum of the parts already found, and the quotient is a third term of the root, and so on, till the whole root is obtained.

Ex. Required the cube root of \( a^3 + 3a^2x + 3ax^2 + x^3 \).

\[ \begin{align*} a^3 &= 59049 \\ 3a^2 &= 19049 \\ 3ax &= 483 \\ x^3 &= 49 \end{align*} \]

Hence 243 is the root required.

76. The reason of the preceding rule is evident from the composition of a cube, for if any quantity as \( a + b + c \) be raised to the third power, we have \((a + b + c)^3 = a^3 + (3a^2 + 3ab + b^2)b + (3(a + b)^2 + 3(a + b)c + c^2)c\), and by considering in what manner the terms \(a\), \(b\), and \(c\) are developed from this expression for the cube of their sum, we also see the reason for the common rule for extracting the cube root in numbers. Let it be required to find the cube root of 13312053, where the root will evidently consist of three figures; let us suppose it to be represented by \(a + b + c\), and the operation for finding the numerical values of these quantities may stand as follows.

\[ \begin{align*} a^3 &= 13312053 \\ 3a^2 &= 1200000 \\ 3ab &= 18000 \\ b^2 &= 900 \\ 3a^2 + 3ab + b^2 &= 1389000 \\ 3(a + b)^2 &= 158700 \\ 3(a + b)c &= 4830 \\ c^2 &= 49 \\ 3(a + b)^2 + 3(a + b)c + c^2 &= 163579 \end{align*} \] The operation as performed by the common rule (see Arithmetic) will stand thus:

\[ \begin{array}{c} 13312053 \\ 8 \\ \hline 12 \cdot 5312 \\ 18 \cdot 9 \\ \hline 1389 \cdot 4167 \\ 1587 \cdot 1145053 \\ 383 \cdot 49 \\ \hline 163579 \cdot 1145053 \end{array} \]

77. III. To extract any other root.

Rule. Range the quantity of which the root is to be found, according to the powers of its letters, and extract the root of the first term, and that shall be the first member of the root required.

Involve the first member of the root to a power less by unity than the number that denominates the root required, and multiply the power that arises by the number itself; divide the second term of the given quantity by the product, and the quotient shall give the second member of the root required.

Find the remaining members of the root in the same manner by considering those already found as making one term.

Ex. Required the cube root of \(x^6 + 6x^5 - 40x^3 + 96x - 64\).

\[ (x^2)^3 = x^6 \\ 3x^4 \cdot 6x^5 \\ (x^2 + 2x)^3 = x^6 + 6x^5 + 12x^4 + 8x^3 \\ 3x^4 + \text{&c.} = 12x^4 \\ (x^2 + 2x - 4)^3 = x^6 + 6x^5 - 40x^3 + 96x - 64. \]

In this example, the cube root of \(x^6\), or \(x^2\), is the first member of the root, and to find a second member the first is raised to the power next lower, or to the second power, and also multiplied by 3, the index of the root required; thus we get \(3x^4\) for a divisor, by which the second term \(6x^5\) being divided, we find \(2x\) for the second member of the root. We must now consider \(x^2 + 2x\) as forming one term; accordingly having subtracted its cube from the quantity, of which the root is sought, we have \(-12x^4\), &c., for a new dividend; and having also raised \(x^2 + 2x\) to the second power, and multiplied the result by 3, we find \(3x^4\), &c., for a divisor. As it is only the terms which contain the highest powers of the dividend and divisor that we have occasion for, the remaining terms are expressed by &c. Having divided \(-12x^4\) by \(3x^4\), we find \(-4\).

for the third term of the root; and because it appears that \(x^2 + 2x - 4\), when raised to the third power, gives a result very same with the proposed power, we conclude \(x^2 + 2x - 4\) to be the root sought.

78. In the preceding examples, the quantities whose roots were to be found have been all such as could have their roots expressed by a finite number of terms; but it will frequently happen that the root cannot be otherwise affixed than by a series consisting of an infinite number of terms: the preceding rules, however, will serve to determine any number of terms of the series. Thus the square root of \(a^2 + x^2\) will be found to be \(a + \frac{x^2}{2a} + \frac{x^4}{8a^3} + \frac{x^6}{16a^5} + \frac{x^8}{128a^7} + \text{&c.}\) and the cube root of \(a^3 + x^3\) will stand thus, \(a + \frac{x^3}{3a^2} + \frac{x^6}{9a^5} + \frac{x^9}{24a^8} + \text{&c.}\); but as the extraction of roots in the form of series can be more easily performed by other methods, we shall refer the reader to sect. 19, which treats of series, where this subject is again resumed.

Sect. IV. Of Surds.

79. It has been already observed (71.), that the root of any proposed quantity is found by dividing the exponent of the quantity by the index of the root; and the rule has been illustrated by suitable examples, in all which, however, the quotient expressing the exponent of the result is a whole number; but there may be cases in which the quotient is a fraction. Thus if the cube root of \(a^2\) were required, it might be expressed, agreeably to the method of notation already explained, either thus \(\sqrt[3]{a^2}\), or thus \(a^{2/3}\).

80. Quantities which have fractional exponents are called surds, or imperfect powers, and are said to be irrational, in opposition to others with integral exponents, which are called rational.

81. Surds may be denoted by means of the radical sign, but it will be often more convenient to use the notation of fractional exponents; the following examples will show how they may be expressed either way.

\[ \sqrt[3]{a} = a^{1/3}, \quad \sqrt[4]{ab} = b^{1/4}, \quad \sqrt[4]{a^3b^2} = a^{3/4}b^{1/2}, \quad \sqrt{a^2 + b^2} = (a^2 + b^2)^{1/2}, \quad \sqrt{(a-b)^3} = (a-b)^{3/2}, \quad \sqrt{a+b} = (a+b)^{1/2}, \quad \sqrt{ab} = (ab)^{1/2}. \]

\(a^{-1/2}b^{-3/2}\).

82. The operations concerning surds depend on the following principle. If the numerator and denominator of a fractional exponent be either both multiplied, or both divided by the same quantity, the value of the power is the same. Thus \(a^{m/n} = a^{cm/cn}\). For let \(a^{m/n} = b\), then raising both to the power \(n\), \(a^m = b^n\), and farther raising both to the power \(c\) we get \(a^{cm} = b^{cn}\); let the root \(cn\) be now taken, and we find \(a^{cm} = b^{cn}\).

83. Prob. I. To Reduce a Rational Quantity to the Form of a Surd of any given denomination.

Rule. Reduce the exponent of the quantity to the form of a fraction of the same denomination as the given surd. Ex. 1. Reduce \(a^3\) to the form of the cube root.

Here the exponent 2 must be reduced to the form of a fraction having 3 for a denominator, which will be the fraction \(\frac{6}{3}\); therefore \(a^2 = a^{\frac{6}{3}} = \sqrt[3]{a^3}\).

Ex. 2. Reduce \(5\) to the form of the cube root, and \(3ab^4\) to the form of the square root.

First \(5 = 5^{\frac{1}{3}} = \sqrt[3]{5 \times 5 \times 5} = \sqrt[3]{125}\).

And \(3ab^4 = 3^{\frac{3}{2}}a^{\frac{1}{2}}b^{\frac{4}{2}} = (3^{\frac{3}{2}}a^{\frac{1}{2}}b^{\frac{4}{2}})^{\frac{1}{2}} = \sqrt{9a^3b^4}\).

84. Prob. II. To Reduce Surds of different denominations to others of the same value, and of the same denominations.

Rule. Reduce the fractional exponents to others of the same value, and having the same common denominator.

Ex. 1. Reduce \(\sqrt{a}\) and \(\sqrt{b^2}\), or \(a^{\frac{1}{2}}\) and \(b^{\frac{2}{3}}\) to other equivalent surds of the same denomination.

The exponents \(\frac{1}{2}\) and \(\frac{2}{3}\), when reduced to a common denominator, are \(\frac{3}{6}\) and \(\frac{4}{6}\); therefore, the surds required are \(a^{\frac{3}{6}}\) and \(b^{\frac{4}{6}}\), or \(\sqrt{a^3}\) and \(\sqrt{b^4}\).

Ex. 2. Reduce \(3^{\frac{3}{2}}\) and \(2^{\frac{2}{3}}\) to surds of the same denomination.

The new exponents are \(\frac{3}{2}\) and \(\frac{2}{3}\), therefore we have \(3^{\frac{3}{2}} = 3^{\frac{3}{2}} = \sqrt{3^3} = \sqrt{27}\), and \(2^{\frac{2}{3}} = 2^{\frac{2}{3}} = \sqrt{2^2} = \sqrt{4}\).

And in the same way the surds \(A^{\frac{3}{2}}\), \(B^{\frac{2}{3}}\) are reduced to these two \(\sqrt{A^3}\) and \(\sqrt{B^4}\).

85. Prob. III. To Reduce Surds to their most simple terms.

Rule. Reduce the surd into two factors, so that one of them may be a complete power, having its exponent divisible by the index of the surd. Extract the root of that power, and place it before the remaining quantities with the proper radical sign between them.

Ex. 1. Reduce \(\sqrt{48}\) to its most simple terms.

The number 48 may be resolved into the two factors 16 and 3, of which the first is a complete square; therefore \(\sqrt{48} = (4^2 + 3)^{\frac{1}{2}} = 4 \times 3^{\frac{1}{2}} = 4 \sqrt{3}\).

Ex. 2. Reduce \(\sqrt{98a^3x}\) and \(\sqrt{24a^3x + 40a^3x^2}\), each to its most simple terms.

First \(\sqrt{98a^3x} = (7a^4 \times 2x)^{\frac{1}{2}} = 7a^2 \times (2x)^{\frac{1}{2}} = 7a^2 \sqrt{2x}\).

Also \(\sqrt{24a^3x + 40a^3x^2} = (2^3 a^3 (3x + 5x^2))^{\frac{1}{2}} = 2a \sqrt{3x + 5x^2}\).

86. Prob. IV. To Add and Subtract Surds.

Rule. If the surds are of different denominations, reduce them to others of the same denomination, by prob. 2.; and then reduce them to their simplest terms by last problem. Then, if the surd part be the same in them all, annex it to the sum, or difference of the rational parts, with the sign of multiplication, and it will give the sum, or difference required. But if the surd part be not the same in all the quantities, they can only be added or subtracted by placing the signs + or — between them.

Ex. 1. Required the sum of \(\sqrt{27}\) and \(\sqrt{48}\).

By prob. 3., we find \(\sqrt{27} = 3 \sqrt{3}\) and \(\sqrt{48} = 4 \sqrt{3}\), therefore \(\sqrt{27} + \sqrt{48} = 3 \sqrt{3} + 4 \sqrt{3} = 7 \sqrt{3}\).

Ex. 2. Required the sum of \(3^{\frac{3}{2}}\) and \(5^{\frac{3}{2}}\).

\(3^{\frac{3}{2}} = 3^{\frac{3}{2}} = \frac{3}{2} \sqrt{2}\) and \(5^{\frac{3}{2}} = 5^{\frac{3}{2}} = \frac{5}{2} \sqrt{2}\), therefore \(3^{\frac{3}{2}} + 5^{\frac{3}{2}} = \frac{3}{2} \sqrt{2} + \frac{5}{2} \sqrt{2} = \frac{8}{2} \sqrt{2} = 4 \sqrt{2}\).

Ex. 2. Required the difference between \(\sqrt{80a^3x}\) and \(\sqrt{20a^3x}\).

\(\sqrt{80a^3x} = (4^2a^4 \times 5x)^{\frac{1}{2}} = 4a^2 \sqrt{5x}\), and \(\sqrt{20a^3x} = (2^2a^2 \times 5x)^{\frac{1}{2}} = 2ax \sqrt{5x}\); therefore \(\sqrt{80a^3x} - \sqrt{20a^3x} = (4a^2 - 2ax) \sqrt{5x}\).

86. Prob. V. To Multiply and Divide Surds.

Rule. If they are surds of the same rational quantity, add and subtract their exponents.

But if they are surds of different rational quantities, let them be brought to others of the same denomination, by prob. 2. Then, by multiplying or dividing these rational quantities, their product, or quotient, may be set under the common radical sign.

Note. If the surds have any rational coefficients, their product or quotient must be prefixed.

Ex. 1. Required the product of \(\sqrt{a^2}\) and \(\sqrt{a^2}\).

\(\sqrt{a^2} \times \sqrt{a^2} = a^{\frac{2}{2}} \times a^{\frac{2}{2}} = a^{\frac{2+2}{2}} = a^{\frac{4}{2}} = a^2\), Anf.

Ex. 2. Divide \(\sqrt{a^3 - b^3}\) by \(\sqrt{a + b}\).

These surds when reduced to the same denomination are \((a^2 - b^2)^{\frac{1}{2}}\) and \((a + b)^{\frac{1}{2}}\). Hence \(\frac{\sqrt{a^3 - b^3}}{\sqrt{a + b}} = \frac{(a^2 - b^2)^{\frac{1}{2}}}{(a + b)^{\frac{1}{2}}} = \frac{(a + b)(a - b)}{(a + b)^{\frac{1}{2}}} = \frac{(a + b)^{\frac{1}{2}}}{(a + b)^{\frac{1}{2}}} = \sqrt{(a + b)(a - b)}\).

Ex. 3. Required the product of \(\sqrt{8}\) and \(\sqrt{5}\).

\(5 \sqrt{8} \times 3 \sqrt{5} = 5 \times 3 \times \sqrt{8} \times \sqrt{5} = 15 \times \sqrt{40} = 15 \times \sqrt{4 \times 10} = 30 \sqrt{10}\).

Ex. 4. Divide \(8 \sqrt{56}\) by \(4 \sqrt{2}\).

\(\frac{8 \sqrt{56}}{4 \sqrt{2}} = 2 \sqrt{\frac{56}{2}} = 2 \sqrt{28}\). Ex. 5. Required the product of \( x^m \) and \( x^n \); also the quotient arising from the division of \( a^m \) by \( b^n \).

First \( x^m \times x^n = x^{m+n} \), and \( x^{mn} = \sqrt{x^{mn}} \),

And \( \frac{a^m}{b^n} = \left( \frac{a}{b} \right)^{mn} = \frac{\sqrt{a^n}}{\sqrt{b^n}} \).

88. Prob. VI. To Involve and Evolve Surds.

Surds are involved or evolved in the same manner as any other quantities, namely, by multiplying or dividing their exponents by the index of the power, or root required. Thus the square of \( 3\sqrt{3} \) is \( 3 \times 3 \times (3)^{\frac{1}{2}} = 9\sqrt{9} \). The \( n \)th power of \( x^m \) is \( x^{mn} \).

The cube root of \( \frac{1}{8}\sqrt{2} \) is \( \frac{1}{2}(2)^{\frac{1}{6}} = \frac{\sqrt[6]{2}}{2} \) and the \( n \)th root of \( x^m \) is \( x^{\frac{m}{n}} \).

89. If a compound quantity involve one or more surds, its powers may be found by multiplication. Thus the square of \( 3 + \sqrt{5} \) is found as follows:

\[ \begin{align*} 3 + \sqrt{5} \\ 3 + \sqrt{5} \\ 9 + 3\sqrt{5} + 5 \\ 9 + 6\sqrt{5} + 5 = 14 + 6\sqrt{5} \end{align*} \]

the square required.

90. The square root of a binomial, or residual surd \( A+B \), or \( A-B \), may be found thus. Take \( \sqrt{A^2 - B^2} \);

then \( \sqrt{A+B} = \sqrt{\frac{A+D}{2}} + \sqrt{\frac{A-D}{2}} \),

and \( \sqrt{A-B} = \sqrt{\frac{A+D}{2}} - \sqrt{\frac{A-D}{2}} \).

Thus the square root of \( 8 + 2\sqrt{7} \) is \( 1 + \sqrt{7} \); and the square root of \( 3 - \sqrt{8} \) is \( \sqrt{2} - 1 \). With respect to the extraction of the cube or any higher root, no general rule can be given.

Sect. V. Of Proportion.

91. In comparing together any two quantities of the same kind in respect of magnitude, we may consider how much the one is greater than the other, or else how many times the one contains either the whole, or some part of the other; or, which is the same thing, we may consider either what is the difference between the quantities, or what is the quotient arising from the division of the one quantity by the other; the former of these is called their arithmetical ratio, and the latter their geometrical ratio. These denominations, however, have been assumed arbitrarily, and have little or no connexion with the relations they are intended to express.

I. Of Arithmetical Progression.

92. When of four quantities the difference between the first and second is equal to the difference between the third and fourth, the quantities are called arithmetical proportions. Such, for example, are the numerical proportions 2, 5, 9, 12; and, in general, the quantities \( a+d, b+d, c+d \). If the two middle terms are equal, the quantities constitute what are called three arithmetical proportions.

93. The most material property of four arithmetical proportions is the following: If four quantities be arithmetically proportional, the sum of the extreme terms is equal to the sum of the means. Let the quantities be \( a, a+d, b, b+d \), where \( d \) is the difference between the first and second, and also between the third and fourth, the sum of the extremes is \( a+b+d \), and that of the means \( a+d+b \); so that the truth of the proportion is evident. Hence it follows, that if any three quantities be arithmetically proportional, the sum of the two extremes is double the mean.

94. If any three terms of four arithmetical proportions be given, the fourth may be found from the preceding proposition. Let \( a, b, c \) be the first, second, and fourth terms, and let \( x \), the third term, be required; because \( a+c=b+x \); therefore \( x=a+c-b \). In like manner any two of three arithmetical proportions being supposed given, the remaining term may be readily found.

95. If a series of quantities be such, that the difference between any two adjacent terms is always the same, these terms form a continued arithmetical proportion. Thus the numbers 2, 4, 6, 8, 10, &c., form a series in continued arithmetical proportion, and, in general, such a series may be represented thus:

\[ a, a+d, a+2d, a+3d, a+4d, a+5d, \ldots \]

where \( a \) denotes the first term and \( d \) the common difference.

By a little attention to this series, we readily discover that it has the following properties:

1. The last term of the series is equal to the first term, together with the common difference taken as often as there are terms after the first. Thus, when the number of terms is \( n \), the last term is \( a+(n-1)d \); and so on. Hence if \( x \) denote the last term, \( n \) the number of terms, and \( a \) and \( d \) express the first term and common difference, we have \( x=a+(n-1)d \).

2. The sum of the first and last term is equal to the sum of any two terms at the same distance from them. Thus suppose the number of terms to be 7, then the last term is \( a+6d \), and the sum of the first and last, \( 2a+6d \); but the same is also the sum of the second and last but one, of the third and last but two, and so on till we come to the middle term, which, because it is equally distant from the extremes, must be added to itself.

96. From this last mentioned property we derive a rule for finding the sum of all the terms of the series. For if the sum of the first and last be taken, as also the sum of the second and last but one, of the third and last but two, and so on along the series till we come to the sum of the last and first terms, it is evident that we shall have as many sums as there are terms, and each equal to the sum of the first and last terms; but the aggregate of those sums is equal to all the terms of the series taken twice, therefore the sum of the first and last term, taken as often as there are terms, is equal to twice the sum of all the terms, so that if \( s \) denote that sum, we have \( 2s=n(a+x) \), and \( s=\frac{n}{2}(a+x) \). Hence the sum of the odd numbers 1, 3, 5, 7, 9, &c., continued to \( n \) terms, is equal to the square of the number of terms. For in this case \( a = 1, d = 2, z = 1 + (n-1)d = 2n - 1 \), therefore \( s = \frac{n}{2} \times 2n = n^2 \).

II. Of Geometrical Proportion.

97. When four quantities, the quotient arising from the division of the first by the second is equal to that arising from the division of the third by the fourth, these quantities are said to be in geometrical proportion, or are called simply proportionals. Thus 1, 2, 4, 8, 16, 32, are four numbers in geometrical proportion; and, in general, \( na, a, nb, b \) may express any four proportionals, for \( \frac{na}{a} = n \), and also \( \frac{nb}{b} = n \).

98. To denote that any four quantities \( a, b, c, d \) are proportional, it is common to place them thus, \( a : b :: c : d \); or thus \( a : b :: c : d \), which notation, when expressed in words, is read thus, \( a \) is to \( b \) as \( c \) to \( d \), or the ratio of \( a \) to \( b \) is equal to the ratio of \( c \) to \( d \).

The first and third terms of a proportion are called the antecedents, and the second and fourth the consequents.

99. When the two middle terms of a proportion are the same, the remaining terms, and that quantity, constitute three geometrical proportionals; such are 4, 6, 9, and in general \( na, a, \frac{a}{n} \). In this case the middle quantity is called a mean proportional between the other two.

100. The principal properties of four proportionals are the following:

1. If four quantities be proportionals, the product of the extremes is equal to the product of the means. Let \( a, b, c, d \) be four quantities, such that \( a : b :: c : d \); then from the nature of proportionals \( \frac{a}{b} = \frac{c}{d} \); let these equal quotients be multiplied by \( bd \), and we have \( \frac{abd}{b} = \frac{cbd}{d} \), or \( ad = bc \). Hence it follows, that when three quantities are proportional, the product of the extremes is equal to the square of the middle term. It also appears, that if any three of four proportionals be given, the remaining one may be found. Thus let \( a, b, c \), the three first be given, and let it be required to find \( x \) the fourth term; because \( a : b :: c : x \), \( ax = bc \), and dividing by \( a \), \( x = \frac{bc}{a} \). This conclusion may be considered as a demonstration of what is called the rule of three in arithmetic.

2. If four quantities be such that the product of two of them is equal to the product of the other two, these quantities are proportionals.

Let \( a, b, c, d \) be the quantities, which are such that \( ad = bc \), if these equals \( bc \) divided by \( bd \), we get \( \frac{ad}{bd} = \frac{bc}{bd} \) or \( \frac{a}{b} = \frac{c}{d} \); hence it follows, from the definition of proportionals (\$ 97.), that \( a : b :: c : d \). From this property of proportionals it appears, that if three quantities be such that the square of one of them be equal to the product of the other two, these quantities are three proportionals.

101. If four quantities are proportional, that is, if \( a : b :: c : d \), then will each of the following combinations or arrangements of the quantities be also four proportionals.

1st. By inversion \( b : a :: d : c \) 2nd. By alternation \( a : c :: b : d \) 3rd. By composition \( a + b : a :: c + d : c \) or \( a + b : b :: c + d : d \) 4th. By division \( a - b : a :: c - d : c \) or \( a - b : b :: c - d : d \) 5th. By mixing \( a + b : a - b :: c + d : c - d \) 6th. By taking any equimultiples of the antecedents, and also any equimultiples of the consequents \( na : pb :: nc : pd \). 7th. Or by taking any parts of the antecedents and consequents \( \frac{a}{n} : \frac{b}{p} :: \frac{c}{n} : \frac{d}{p} \).

That the preceding combinations of the quantities \( a, b, c, d \) are proportionals, may be readily proved, by taking the products of the extremes and means; for from each of them we derive this conclusion, that \( ad = bc \), which is known to be true, from the original assumption of the quantities.

102. If four quantities be proportional, and also other four, the product of the corresponding terms will be proportional.

Let \( a : b :: c : d \), and \( e : f :: g : h \), Then \( ae : bf :: cg : dh \).

For \( ad = bc \), and \( ef = gh \) (\$ 100.), therefore, multiplying together these equal quantities \( adeh = bcfg \), or \( ae \times dh = bf \times cg \), therefore by the second property (\$ 100.), \( ae : bf :: cg : ah \).

103. Hence it follows, that if there be any number of proportions whatever, the products of the corresponding terms will still be proportional.

104. If a series of quantities be so related to each other, that the quotient arising from the division of any term by that which follows it is always the same quantity, these quantities are said to be in continued geometrical proportion, such are the numbers 2, 4, 8, 16, 32, &c., also \( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \frac{1}{32}, \) &c., and in general a series of such quantities may be represented thus, \( a, ar, ar^2, ar^3, ar^4, ar^5, \) &c. Here \( a \) is the first term, and \( r \) the quotient of any two adjoining terms, which is also called the common ratio.

105. By inspecting this series we find that it has the following properties:

1. The last term is equal to the first, multiplied by the common ratio raised to a power, the index of which is one less than the number of terms. Therefore, if \( z \) denote the last term, and \( n \) the number of terms, \( z = ar^{n-1} \).

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* The quantities in this case must be all of the same kind, that is, if \( a \) and \( b \) denote surfaces, then \( c \) and \( d \) must also denote surfaces, but they cannot represent lines, &c. 2. The product of the first and last term is equal to the product of any two terms equally distant from them; thus, supposing \( ar^5 \) the last term, it is evident that \( a \times ar^5 = ar \times ar^4 = ar^3 \times ar^1 \), &c.

106. The sum of all the terms may be found thus: let \( s \) represent that sum, then, supposing the number of terms to be fixed, \( s = a + ar + ar^2 + ar^3 + ar^4 + ar^5 \), and multiplying these equals by \( r \), \( sr = ar + ar^2 + ar^3 + ar^4 + ar^5 + ar^6 \). If from the lower line, or \( sr = ar + ar^2 + \ldots + ar^5 \), we subtract the upper line, or \( s = a + ar + \ldots + ar^5 \), the remainders will evidently be equal; but on the one side of the sign \( = \) we have \( sr - r \), and on the other \( ar^6 - a \); therefore, \( sr - r = ar^6 - a \), and dividing by \( r - 1 \), \( s = \frac{ar^6 - a}{r - 1} \).

Let us now, instead of 6, substitute \( n \) (for the number of terms put down was 6), and we have the following general rule for finding the sum of a series of quantities in continued geometrical proportion, \( s = \frac{ar^n - a}{r - 1} \), or \( s = \frac{a(r^n - 1)}{r - 1} \).

Sect. VI. Of the Reduction of Equations involving one unknown quantity.

107. The general object of algebraic investigation is to discover certain unknown quantities, by comparing them with other quantities which are given, or supposed to be known. The relation between the known and unknown quantities is either that of equality, or else such as may be reduced to equality; and a proposition which affirms that certain combinations of quantities are equal to one another is called an equation. Such are the following:

\[ \frac{x}{2} + \frac{x}{3} = \frac{24}{x}, \quad 2x + 3y = xy; \]

the first of these equations expresses the relation between an unknown quantity \( x \), and certain known numbers; and the second expresses the relation which the two indefinite quantities \( x \) and \( y \) have to each other.

108. When a quantity stands alone on one side of an equation, the terms on the other side are said to be a value of that quantity. Thus in the equation \( x = ay + b - c \), the quantity \( x \) stands alone on one side, and \( ay + b - c \) is its value.

109. The conditions of a problem may be such as to require several equations and symbols of unknown quantities for their complete expression; these, however, by rules hereafter to be explained, may be reduced to one equation, involving only one unknown quantity and its powers, besides the known quantities; and the method of expressing that quantity, by means of the known quantities, constitutes the theory of equations, one of the most important, as well as most intricate branches of algebraic analysis.

110. An equation is said to be resolved when the unknown quantity is made to stand alone on one side, and only known quantities on the other side; and the value of the unknown quantity is called a root of the equation.

111. Equations containing only one unknown quantity and its powers, are divided into different orders, according to the highest power of that quantity contained in any one of its terms. The equation, however, is supposed to be reduced to such a form, that the unknown quantity is found only in the numerators of the terms, and that the exponents of its powers are expressed by positive integers.

112. If an equation contains only the first power of the unknown quantity, it is called a simple equation, or an equation of the first order. Such is \( ax + b = c \), where \( x \) denotes an unknown, and \( a, b, c \) known quantities.

113. If the equation contains the second power of the unknown quantity, it is said to be of the second degree, or is called a quadratic equation; such is \( 4x^2 + 3x = 12 \), and in general \( ax^2 + bx = c \). If it contains the third power of the unknown quantity, it is of the third degree, or is a cubic equation. Such are \( x^3 - 2x^2 + 4x = 12 \), and \( ax^3 + bx^2 + cx = d \), and so on, with respect to equations of the higher orders. A simple equation is sometimes said to be linear, or to be of one dimension. In like manner, quadratic equations are said to be equations of two dimensions, and cubic equations to be of three dimensions.

114. When in the course of an algebraic investigation we arrive at an equation involving only one unknown quantity, that quantity will often be so entangled in the different terms, as to render several previous reductions necessary before the equation can be expressed under its characteristic form, so as to be resolved by the rules which belong to that form.

These reductions depend upon the operations which have been explained in the former part of this treatise, and the application of a few self-evident principles, namely, that if equal quantities be added to, or subtracted from equal quantities, the sums or remainders will be equal; if equal quantities be multiplied, or divided by the same quantity, the products or quotients will be equal; and, lastly, if equal quantities be raised to the same power, or have the same root extracted out of each, the results will still be equal.

From these considerations are derived the following rules, which apply alike to equations of all orders, and are alone sufficient for the solution of simple equations.

115. Rule 1. Any quantity may be transferred from one side of an equation to the other, by changing its signs.

Thus, if \( x - 3 = 5 \)

Then \( x = 5 + 3 \)

Or \( x = 8 \)

And if \( 3x - 10 = 2x + 5 \)

Then \( 3x - 2x = 5 + 10 \)

Or \( x = 15 \)

Again, if \( ax + b = cx - dx + e \)

Then \( ax - cx + dx = e - b \)

Or \( (a - c + d)x = e - b \)

The reason of this rule is evident, for the transposing a quantity from one side of an equation to the other is nothing more than adding the same quantity to each side of the equation, if the sign of the quantity transposed was \( - \); or subtracting it, if the sign was \( + \).

From this rule we may infer, that if any quantity be found on each side of the equation with the same sign, it may be left out of both. Also, that the signs of all the terms of an equation may be changed into Reduction the contrary without affecting the truth of the equation.

Thus, if \(a + x = b + a - c\) Then \(x = b + c\) And if \(a - x = b - d\) Then \(x = a - d - b\).

116. Rule 2. If the unknown quantity in an equation be multiplied by any quantity, that quantity may be taken away, by dividing all the other terms of the equation by it.

If \(3x = 24\) Then \(x = \frac{24}{3} = 8\) If \(ax = b - c\) Then \(x = \frac{b - c}{a} = \frac{b}{c} - \frac{c}{a}\).

Here equal quantities are divided by the same quantity, and therefore the quotients are equal.

117. Rule 3. If any term of an equation be a fraction, its denominator may be taken away by multiplying all the other terms of the equation by that denominator.

If \(\frac{x}{5} = 7\) Then \(x = 35\) If \(\frac{x}{a} = b - c + d\) Then \(x = ab - ac + ad\) If \(a - x = c\), \(ax - b = cx\).

In these examples, equal quantities are multiplied by the same quantity, and therefore the products are equal.

118. The denominators may be taken away from several terms of an equation by one operation, if we multiply all the terms by any number which is a multiple of each of these denominators.

Thus, if \(\frac{x}{2} + \frac{x}{3} + \frac{x}{4} = 26\). Let all the terms be multiplied by 12, which is a multiple of 2, 3, and 4, and we have

\[ \frac{12x}{2} + \frac{12x}{3} + \frac{12x}{4} = 312, \]

Or \(6x + 4x + 3x = 312\); Hence \(12x = 312\).

Universally, if \(\frac{x}{a} = \frac{x}{b} + \frac{x}{c} = d - e\).

To take away the denominators \(a, b, c\), let the whole equation be multiplied by \(abc\), their product, and we have

\[ bcx - acx + abx = abc(d - e) \]

Or \((bc - ac + ab)x = abc(d - e)\).

119. From the two last rules it appears that if all the terms of an equation be either multiplied or divided by the same quantity, that quantity may be left out of all the terms.

If \(ax = ab - ac\) Then \(x = b - c\) And if \(\frac{x}{a} = \frac{b}{a} + \frac{c}{a}\) Then \(x = b + c\).

120. Rule. If the unknown quantity is found in any term which is a surd, let that surd be made to stand alone on one side of the equation, and the remaining terms on the opposite side; then involve each side to a power denoted by the index of the surd, and thus the unknown quantity shall be freed from the surd expression.

If \(\sqrt{x} + 6 = 10\) Then by transposition \(\sqrt{x} = 10 - 6 = 4\) And squaring both sides \(\sqrt{x} \times \sqrt{x} = 4 \times 4\)

Or \(x = 16\).

Also, if \(\sqrt{a^2 + x^2} = b + x\) By transf. \(\sqrt{a^2 + x^2} = b + x\) And squaring, \(a^2 + x^2 = (b + x)^2 = b^2 + 2bx + x^2\) Hence \(a^2 = b^2 + 2bx\). And if \(\sqrt{a^2x - b^2x} = a\), Then \(a^2x - b^2x = a^2\).

121. Rule. If the side of the equation, which contains the unknown quantity, be a perfect power, the equation may be reduced to another of a lower order, by extracting the root of that power out of each side of the equation.

Thus if \(x^3 = 64a^3\), Then, by extracting the cube root, \(x = 8a\); And if \((a + x)^2 = b^2 - a^2\), Then \(a + x = \sqrt{b^2 - a^2}\).

122. The use of the preceding rules will be further illustrated by the following examples:

Ex. 1. Let \(20 - 3x = 8 - 7x\) By rule 1. \(7x - 3x = 60 + 8 - 20\) Or \(4x = 48\) Therefore by rule 2. \(x = 12\).

Ex. 2. Let \(ax - bx = cx + d\) By rule 1. \(ax - cx = b + d\) Or \((a - c)x = b + d\) And by rule 2. \(x = \frac{b + d}{a - c}\).

Ex. 3. Let \(\frac{x + 1}{2} + \frac{x + 2}{3} = 16 - \frac{x + 3}{4}\) By rule 3. \[ \begin{align*} x + 1 + \frac{2x + 4}{3} &= 32 - \frac{2x + 6}{4} \\ 3x + 3 + 2x + 4 &= 96 - 6x + 18 \\ 12x + 12 + 8x + 16 + 384 - 6x - 18 \end{align*} \] Or \(20x + 28 = 366 - 6x\) Hence, by rule 1. \(26x = 338\) And by rule 2. \(x = 13\).

In this example, instead of taking away the denominators one after another, they might have been all taken away at once, by multiplying the given equation by... Reduction by 12, which is divisible by the numbers 2, 3, and 4; thus we should have got $6x^3 + 6 + 4x + 8 = 192 - 3x^2$, and hence, as before, $x = 13$.

Ex. 4. Let $6x^3 - 2x^2 = 16x^2 + 2x^3$.

Then dividing by $2x^2$, $3x - 10 = 8 + x$

And transposing, $3x - x = 8 + 10$

Or $2x = 18$

And therefore $2x = 9$.

Ex. 5. Let $\frac{b^2}{x} = c$

Then $ax - b^2 = cx$

And $ax - cx = b^2$

Whence $x = \frac{b^2}{a - c}$.

Ex. 6. Let $x - 6 = \frac{x^2}{x + 24}$.

Then $(x - 6)(x + 24) = x^2$

That is, $x^2 + 18x - 144 = x^2$

Therefore $18x = 144$

And $x = 8$.

Ex. 7. Let $ax + b^2 = \frac{ax^2 + ac^2}{a + x}$.

Then $(a + x)(ax + b^2) = ax^2 + ac^2$

Or $a^2x + ab^2 + ax^2 + b^2x = ax^2 + ac^2$

Hence $a^2x + b^2x = ac^2 - ab^2$

And $x = \frac{ac^2 - ab^2}{a^2 + b^2}$.

Ex. 8. Let $\frac{1}{1 + a} = a$.

Then $\frac{1}{1 - a} = a$

And $-x = ax = a - 1$

Or changing the signs, $x + ax = 1 - a$

Hence, $x = \frac{1 - a}{1 + a}$.

Ex. 9. Let $\sqrt{12 + x} = 2 + \sqrt{x}$.

Then by rule 4, $12 + x = 4 + 4\sqrt{x} + x$

And by transposition $8 = 4\sqrt{x}$

And by division $2 = \sqrt{x}$

And again by rule 4, $4 = x$.

Ex. 10. Let $x + \sqrt{a^2 + x^2} = \frac{2a^2}{\sqrt{a^2 + x^2}}$.

Then, by rule 3, $x \sqrt{a^2 + x^2} + a^2 + x^2 = 2a^2$

And by transposition, $x \sqrt{a^2 + x^2} = a^2 - x^2$

Therefore, by rule 4, $x^2 + x^2 = a^2 - 2a^2x^2 + x^4$

Whence $3a^2x^2 = a^4$

And $x^2 = \frac{a^4}{3}$, therefore, rule 5, $x = \frac{a}{\sqrt{3}}$.

Ex. 11. Let $\frac{1 - \sqrt{1 - x^2}}{1 + \sqrt{1 - x^2}} = a$.

Then $1 - \sqrt{1 - x^2} = a + a\sqrt{1 - x^2}$

And $1 - a = a\sqrt{1 - x^2} + \sqrt{1 - x^2} = (1 + a)\sqrt{1 - x^2}$

Whence $\frac{1 - a}{1 + a} = \sqrt{1 - x^2}$

And, taking the square of both sides, $\frac{(1 - a)^2}{(1 + a)^2} = 1 - x^2$

Therefore, by transposition, $x^2 = 1 - \frac{(1 - a)^2}{(1 + a)^2}$

That is, $x^2 = \frac{(1 + a)^2 - (1 - a)^2}{(1 + a)^2} = \frac{4a}{(1 + a)^2}$

Therefore $x = \frac{2\sqrt{a}}{1 + a}$.

Ex. 12. Let $a + x = \sqrt{a^2 + x\sqrt{b^2 + x^2}}$

Then $(a + x)^2 = a^2 + x\sqrt{b^2 + x^2}$

That is, $a^2 + 2ax + x^2 = a^2 + x\sqrt{b^2 + x^2}$

Therefore $2ax + x^2 = x\sqrt{b^2 + x^2}$

And dividing by $x$, $2a + x = \sqrt{b^2 + x^2}$

Again taking the squares of both sides, $4a^2 + 4ax + x^2 = b^2 + x^2$

Whence $4a^2 + 4ax = b^2$

And $4ax = b^2 - 4a^2$; so that $x = \frac{b^2 - 4a^2}{4a}$.

123. In all these examples we have been able to determine the value of the unknown quantity by the rules already delivered, because in every case the first, or at most the second power of that quantity, has been made to stand alone on one side of the equation, while the other consisted only of known quantities; but the same methods of reduction serve to bring equations of all degrees to a proper form for solution. Thus if

$$\frac{1 - p + q + r}{x + 1} = 1 - p - x + \frac{r}{x};$$

by proper reduction, we have $x^3 + px^2 + qx + r$, a cubic equation, which may be resolved by rules to be afterwards explained.

Sect. VII. Of the Reduction of Equations involving more than one unknown quantity.

124. Having shown in the last section in what manner an equation involving one unknown quantity may be resolved, or at least fitted for a final solution, we are next to explain the methods by which two or more equations, involving as many unknown quantities, may at last be reduced to one equation, and one unknown quantity.

As the unknown quantities may be combined together in very different ways, so as to constitute an equation, the methods most proper for their extermination must therefore be various. The three following, however, are of general application, and the last of them may be used with advantage, not only when the unknown quantity to be exterminated arises to the same power in all the equations, but also when the equations contain different powers of that quantity.

125. Method 1. Observe which of the unknown quantities is the least involved, and let its value be found from each equation by the rules of last section.

Let the values thus found be put equal to each other, and hence new equations will arise, from which Reduction that quantity is wholly excluded. Let the same operation be now repeated with the same equations, and the unknown quantities exterminated one by one, till at last an equation be found, which contains only one unknown quantity.

Ex. Let it be required to determine \( x \) and \( y \) from these two equations,

\[ \begin{align*} 2x + 3y &= 23 \\ 5x - 2y &= 10 \end{align*} \]

From the first equation

\[ 2x = 23 - 3y \]

And

\[ x = \frac{23 - 3y}{2} \]

From the second equation

\[ 5x = 10 + 2y \]

And

\[ x = \frac{10 + 2y}{5} \]

Let these values of \( x \) be now put equal to each other,

And we have

\[ \frac{10 + 2y}{5} = \frac{23 - 3y}{2} \]

Or

\[ 2(10 + 2y) = 5(23 - 3y) \]

Therefore

\[ 19y = 95 \]

And

\[ y = 5 \]

And since \( x = \frac{23 - 3y}{2} \), or \( x = \frac{10 + 2y}{5} \), from either of these values we find \( x = 4 \).

126. Method 2. Let the value of the unknown quantity, which is to be exterminated, be found from that equation wherein it is least involved. Let this value, and its powers, be substituted for that quantity, and its respective powers in the other equations; and with the new equations thus arising, let the operation be repeated, till there remain only one equation, and one unknown quantity.

Ex. Let the given equations, as in last method, be

\[ \begin{align*} 2x + 3y &= 23 \\ 5x - 2y &= 10 \end{align*} \]

From the first equation

\[ x = \frac{23 - 3y}{2} \]

And this value of \( x \) being substituted in the second equation, we have

\[ 5 \left( \frac{23 - 3y}{2} \right) - 2y = 10 \]

Or

\[ 115 - 15y - 4y = 20 \]

Therefore

\[ 95 = 19y \]

And

\[ y = 5 \]

And hence \( x = \frac{23 - 3y}{2} = 4 \), as before.

127. Method 3. Let the given equations be multiplied or divided by such numbers or quantities, whether known or unknown, that the term which involved the highest power of the unknown quantity may be the same in each equation.

Then by adding or subtracting the equations, as occasion may require, that term will vanish, and a new equation emerge, wherein the number of dimensions of the unknown quantity in some cases, and in others the number of unknown quantities, will be diminished; and by a repetition of the same, or similar operations, a final equation may be at last obtained, involving only one unknown quantity.

Ex. Let the same example be taken, as in the illustration of the two former methods, namely,

\[ \begin{align*} 2x + 3y &= 23 \\ 5x - 2y &= 10 \end{align*} \]

and from these two equations we are to determine \( x \) and \( y \). To exterminate \( x \), let the first equation be multiplied by 8, and the second by 2, thus we have

\[ \begin{align*} 10x + 15y &= 115 \\ 10x - 4y &= 20 \end{align*} \]

Here the term involving \( x \) is the same in both equations, and it is obvious that by subtracting the one from the other, the resulting equation will contain only \( y \), and known numbers, for by such subtraction we find

\[ 19y = 95, \]

and therefore \( y = 5 \).

Having got the value of \( y \), it is easy to see how \( x \) may be found, from either of the given equations; but it may also be found in the same manner as we found \( y \). For let the first of the given equations be multiplied by 2, and the second by 3, and we have

\[ \begin{align*} 4x + 6y &= 46 \\ 15x - 6y &= 30 \end{align*} \]

By adding these equations, we find

\[ 19x = 76 \]

and therefore \( x = 4 \).

128. The following examples will serve farther to illustrate these different methods of exterminating the unknown quantities from equations.

Ex. 1. Given

\[ \begin{align*} \frac{x}{2} + \frac{y}{3} &= 16 \\ \frac{x}{5} - \frac{y}{9} &= 2 \end{align*} \]

Required \( x \) and \( y \).

By Method 1.

From the first equation we find

\[ x = 32 - \frac{2y}{3} \]

And from the second

\[ x = 10 + \frac{5y}{9} \]

Therefore

\[ 10 + \frac{5y}{9} = 32 - \frac{2y}{3} \]

Or

\[ 90 + 5y = 288 - 6y \]

Hence

\[ 11y = 198 \]

And

\[ y = 18 \]

The value of \( y \) being substituted in either of the values of \( x \), namely, \( 32 - \frac{2y}{3} \) or \( 10 + \frac{5y}{9} \), we find \( x = 20 \).

By Method 2.

Having found from the first given equation \( x = 32 - \frac{2y}{3} \), let this value of \( x \) be substituted in the second, thus we have

\[ \frac{1}{5} \left( 32 - \frac{2y}{3} \right) - \frac{y}{9} = 2 \]

Or

\[ \frac{32}{5} - \frac{2y}{15} - \frac{y}{9} = 2 \]

Hence

\[ 198 = 11y \]

And

\[ 18 = y \] The value of \( y \) being now substituted in either of the given equations, we thence find \( x = 20 \) as before.

By Method 3.

The denominators of the two given equations being taken away by rule 3. of last section, we have

\[ \begin{align*} 3x + 2y &= 96 \\ 9x - 5y &= 90 \end{align*} \]

From three times the first of these equations, or \( 9x + 6y = 288 \), let the second be subtracted, and there remains

\[ 11y = 198 \]

And hence \( y = 18 \).

The value of \( y \) being now substituted in either of the equations \( 3x + 2y = 96 \), \( 9x - 5y = 90 \), we readily find \( x = 20 \).

129. Having now shewn in what manner the different methods of exterminating the unknown quantities may be applied, we shall, in the remaining examples of this section, chiefly make use of the last method, because it is the most easy and expeditions in practice.

Ex. 2. Given

\[ \begin{align*} \frac{x}{2} - \frac{y}{4} &= 8 \\ \frac{x}{5} + \frac{y}{3} &= 8 \end{align*} \]

It is required to determine \( x \) and \( y \).

From the first equation we have \( 4x - 96 = 2y + 64 \). And from the second, \( 12x + 12y + 20x - 480 = 30y - 15x + 1620 \).

These two equations, when abridged, become

\[ \begin{align*} 4x - 2y &= 160 \\ 47x - 18y &= 2100 \end{align*} \]

To exterminate \( y \); from this last equation let 9 times the one preceding it be subtracted.

Thus we find \( 11x = 660 \) And \( x = 60 \)

And because \( 2y = 4x - 160 = 80 \) Therefore \( y = 40 \).

Ex. 3. Given

\[ \begin{align*} ax + by &= c \\ dx + fy &= g \end{align*} \]

To determine \( x \) and \( y \).

To exterminate \( y \), let the first equation be multiplied by \( f \), and the second by \( b \), and we have

\[ afx + bfy = cf \\ bdx + bf = bg \]

Taking now the difference between these equations, we find

\[ afx - bdx = cf - bg \]

Or \((af - bd)x = cf - bg\)

And therefore \( x = \frac{cf - bg}{af - bd} \).

In the same manner may \( y \) be determined, by multiplying the first of the given equations by \( d \), and the second by \( a \); for we find

\[ adx + bdy = cd \\ adx + afy = ag \]

and taking the difference as before, we get:

\[ bdy - afy = cd - ag \]

And therefore \( y = \frac{cd - ag}{bd - af} \).

This last example may be considered as a general solution of the following problem. Two equations expressing the relation between the first powers of two unknown quantities being given, to determine those quantities. For whatever be the number of terms in each equation, it will readily appear, as in example 2d, that by proper reduction, they may be brought to the same form as those given in the 3d example.

130. Let us next consider such equations as involve three unknown quantities.

Ex. 4. Given

\[ \begin{align*} \frac{x}{2} + \frac{y}{3} + \frac{z}{4} &= 29 \\ \frac{x}{2} + \frac{y}{3} + \frac{z}{4} &= 62 \\ \frac{x}{2} + \frac{y}{3} + \frac{z}{4} &= 10 \end{align*} \]

To find \( x \), \( y \), and \( z \).

We shall in this example proceed according to the rules of the first method for exterminating the unknown quantities.

From the first equation \( x = 29 - y - z \) From the second \( x = 62 - 2y - 3z \) From the third \( x = 20 - \frac{2y}{3} - \frac{z}{2} \)

Let these values of \( x \) be put equal to each other, thus we get the two following equations,

\[ \begin{align*} 29 - y - z &= 62 - 2y - 3z \\ 29 - y - z &= 20 - \frac{2y}{3} - \frac{z}{2} \end{align*} \]

Again, from these two equations, by transposition, &c., we find

\[ \begin{align*} y &= 33 - 2z \\ y &= 27 - \frac{3z}{2} \end{align*} \]

Therefore \( 33 - 2z = 27 - \frac{3z}{2} \)

And hence, by reduction, \( z = 12 \)

Whence also \( y = 33 - 2z = 9 \) And \( x = 29 - y - z = 8 \).

Ex. 5. Given

\[ \begin{align*} \frac{x}{2} + \frac{y}{3} + \frac{z}{4} &= 62 \\ \frac{x}{2} + \frac{y}{3} + \frac{z}{4} &= 47 \\ \frac{x}{2} + \frac{y}{3} + \frac{z}{4} &= 38 \end{align*} \]

To find \( x \), \( y \), and \( z \).

Here the given equations, when cleared from fractions, become

\[ \begin{align*} 12x + 8y + 6z &= 1488 \\ 20x + 15y + 12z &= 2820 \\ 30x + 24y + 20z &= 4560 \end{align*} \]

To exterminate \( z \) by the third method, let the first equation be multiplied by 10, the second by 5, and the third by 3, the results will be these:

\[ \begin{align*} 120x + 80y + 60z &= 14880 \\ 100x + 75y + 60z &= 14100 \\ 90x + 72y + 60z &= 13680 \end{align*} \] Reduction Let the second equation be now subtracted from the first, and the third from the second, and we have

\[20x + 5y = 780\] \[10x + 3y = 420\]

Next to exterminate \(y\), let the first of these equations be multiplied by 3, and the second by 5, hence

\[60x + 15y = 2340\] \[50x + 15y = 2100\]

Subtracting now the latter equation from the former,

\[10x = 240\]

Hence

\[x = 24\]

Therefore

\[y = \frac{420 - 10x}{3} = 60\]

And

\[z = \frac{1448 - 12x - 8y}{6} = 120\]

131. From the preceding examples, it is manifest in what manner any number of unknown quantities may be determined, by an equal number of equations, which contain only the first power of those quantities, in the numerators of the terms. Such are the following:

\[ax + by + cz = n\] \[dx + cy + fz = p\] \[gx + hy + kz = q,\]

where \(a, b, c, \&c.\) represent known, and \(x, y, z\) unknown quantities; and in every case of this kind, the unknown quantities may be directly found, for they will be always expressed by whole numbers, or rational fractions, provided that the known quantities, \(a, b, c, \&c.\) are also rational.

132. We shall now add a few examples, in which the equations that result from the extermination of an unknown quantity arise to some of the higher degrees; and therefore their final solution must be referred to the sections which treat of those degrees.

Ex. 6. Let \(x - y = 2\), and \(xy + 5x - 6y = 120\); it is required to exterminate \(x\).

From the first equation \(x = y + 2\); which value being substituted in the other equation according to the second general method (\$126.) it becomes

\[(y + 2)y + 5(y + 2) - 6y = 120\]

That is \(y^2 + 2y + 5y + 10 - 6y = 120\)

therefore the equation required is \(y^2 + y = 110\).

Ex. 7. There is given \(x + y = a\), and \(x^2 + y^2 + b\); to exterminate \(x\).

From the first equation \(x = a - y\), and \(x^2 = (a - y)^2\).

And from the second \(x^2 = b - y^2\).

Therefore \((a - y)^2 = b - y^2\)

That is \(a^2 - 2ay + y^2 = b - y^2\).

Hence \(2y^2 - 2ay + b = a^2\); an equation involving only \(y\).

Ex. 8. Given \(\begin{cases} ax + bx + cy = d \\ fxy + gx + hy = k \end{cases}\) To exterminate \(y\).

From the first equation we find \(y = \frac{d - bx}{ax + c}\)

And from the second \(y = \frac{k - gx}{fx + b}\)

Therefore \(\frac{d - bx}{ax + c} = \frac{k - gx}{fx + b}\), an equation in which the unknown quantity is not found.

Ex. 9. Given \(\begin{cases} y^2 - 3xy + ay = x^2 \\ y^2 + 2ax - by = 4x^2 - b^2 \end{cases}\) To exterminate \(y\).

As the coefficient of \(y^2\) is unity in both equations, if their difference be taken, the highest power of \(y\) will vanish; but to give a general solution, let the terms of the equations be brought all to one side and made equal to 0, thus,

\[y^2 - (3x - a)y - x^2 = 0\] \[y^2 - by + 2ax - 4x^2 + b^2 = 0\]

Let us in the first equation put \(1 = A, -(3x - a) = B, -x^2 = C\); and in the second, \(1 = D, -b = E, 2ax - 4x^2 + b^2 = F\), and the two equations become

\[Ay^2 + By + C = 0\] \[Dy^2 + Ey + F = 0\]

To exterminate \(y^2\), let the first equation be multiplied by \(D\), and the second by \(A\), and we have

\[ADy^2 + BDy + CD = 0\] \[ADy^2 + AEy + AF = 0\]

Therefore, taking the difference of these equations,

\[(BD - AE)y + CD - AF = 0\]

And

\[y = \frac{CD - AF}{BD - AE}\]

Again, to find another value of \(y\), multiply the first equation by \(F\), and the second by \(C\), then

\[AFy^2 + BFy + CF = 0\] \[CDy^2 + CEy + CF = 0\]

Therefore, subtracting as before, we get

\[(AF - CD)y^2 + (BF - CE)y = 0\]

And dividing by \(y\), \((AF - CD)y + BF - CE = 0\),

Therefore, \(y = \frac{CF - BF}{AF - CD}\).

Let this value of \(y\) be put equal to the former value, thus we have \(\frac{AF - CD}{BD - AE} = \frac{CE - BF}{AF - CD}\),

And therefore \((AF - CD)^2 = (BD - AE)(CE - BF)\).

Now as \(y\) does not enter this equation, if we restore the values of \(A, B, C, \&c.\) we have the following equation which involves only \(x\), and known quantities.

\[(b^2 + 2ax - 3x^2)^2 = (a + b - 3x)(2ax - 4x^2 + b^2)\]; this equation when properly reduced will be of the fourth order, and therefore its final resolution belongs not to this place.

Sect. VIII. Questions producing Simple Equations.

133. When a problem is proposed to be resolved by the algebraic method of analysis, its true meaning ought in the first place to be perfectly understood, so that, if necessary, it may be freed from all superfluous and ambiguous expressions; and its conditions exhibited in the clearest point of view possible. The several quantities concerned in the problem are next to be denoted by proper symbols, and their relation to one another expressed agreeably to the algebraic notation. Thus Thus we shall obtain a series of equations, which, if the question be properly limited, will enable us to determine all the unknown quantities required by the rules already delivered in the two preceding sections.

134. In reducing the conditions of a problem to equations, the following rule will be of service. Suppose that the quantities to be determined are actually found, and then consider by what operations the truth of the solution may be verified; then, let the same operations be performed upon the quantities, whether known or unknown, and thus all the conditions of the problem will be reduced to a series of equations, such as is required. For example; suppose that it is required to find two numbers, such, that their sum is 20, and the quotient arising from the division of their difference by the lesser 3; then if we denote the greater of the two numbers by \( x \), and the lesser by \( y \), and proceed as if to prove the truth of the solution, we shall have \( x+y=20 \) for the sum of the numbers, and \( x-y \) for their difference. Now as the former must be equal to 20, and the latter divided by \( y \) equal to 3; the first condition of the problem will be expressed by this equation \( x+y=20 \), and the second by \( \frac{x-y}{y}=3 \), and from these the values of \( x \) and \( y \) may easily be found.

135. When the conditions of a problem have been expressed by equations, or as it were translated from the common language into that of algebra, we must next consider, whether the problem be properly limited; for in some cases, the conditions may be such as to admit of innumerable solutions; and in others, they may involve an absurdity; and thus render the problem altogether impossible.

136. Now by considering the examples of last section, it will readily appear, that to determine any number of unknown quantities, there must be given as many equations as there are unknown quantities. These equations, however, must be such as cannot be derived from each other; and they must not involve any contradiction; for, in the one case, the problem would admit of an unlimited number of answers; and in the other case, it would be impossible. For example, if it were required to determine \( x \) and \( y \) from these two equations, \( 2x-3y=13 \), \( 4x-6y=26 \); as the latter equation is a consequence of the former (for each term of the one is the half of the corresponding term of the other) it is evident, that innumerable values of \( x \) and \( y \) might be found to satisfy both equations. Again, if \( x \) and \( y \) were to be determined from these equations, \( x+2y=8 \), \( 3x+6y=26 \), it will quickly appear, that it is impossible to find such values of \( x \) and \( y \), as will satisfy both equations: for, from the first of them, we find \( 3x=24-6y \); and from the second, \( 3x=26-6y \); and therefore \( 24-6y=26-6y \), or \( 24=26 \), which is absurd; and so also must have been the conditions from which this conclusion is drawn.

137. But there is yet another case in which a problem may be impossible; and that is, when there are more equations than unknown quantities; for it appears, that in this case, by the rules of last section, we would at last find two equations, each involving the same unknown quantity. Now unless these equations happened to agree, the problem would admit of no solution. Upon the whole, therefore, it appears, that a problem is limited, when the conditions afford just as many independent equations as there are unknown quantities to be determined; if there be fewer equations the problem is indeterminate; but if there be more, the problem in general admits of no solution whatever.

138. In expressing the conditions of a problem by equations, it will, in general, be convenient to introduce as few symbols of unknown quantities as possible. Therefore, if two quantities be sought and their sum be given, suppose it \( =x \), then if the one quantity be represented by \( x \), the other may be denoted by \( s-x \). If again their difference be given \( =d \), the quantities may be denoted by \( x \), and \( d+x \), or by \( x \), and \( x-d \). If their product be given \( =p \), the quantities are \( x \), and \( \frac{p}{x} \); and so on.

139. We shall now apply the preceding observations to some examples, which are so chosen as to admit of being resolved by simple equations.

Ex. 1. What is that number, to which if there be added its half, its third, and its fourth part, the sum will be 50.

Let \( x \) denote the number sought. Then its half will be \( \frac{x}{2} \), its third \( \frac{x}{3} \), and its fourth \( \frac{x}{4} \).

Therefore \( x + \frac{x}{2} + \frac{x}{3} + \frac{x}{4} = 50 \).

Hence we find \( 24x + 12x + 8x + 6x = 1200 \).

Or \( 50x = 1200 \).

Therefore \( x = 24 \).

Thus it appears, that the number sought is 24, which upon trial will be found to answer the conditions of the question.

Ex. 2. A post is \( \frac{1}{4} \) of its length in the mud, \( \frac{1}{3} \) in the water, and 10 feet above the water, what is its whole length?

Let its length be \( x \) feet, then the part in the mud is \( \frac{x}{4} \), and that in the water \( \frac{x}{3} \); therefore, from the nature of the question,

\[ \frac{x}{4} + \frac{x}{3} + 10 = x. \]

From which equation we find \( 7x + 120 = 12x \), and \( x = 24 \).

Ex. 3. Two travellers set out at the same time from London and York, whose distance is 150 miles; one of them goes 8 miles a-day, and the other 7; in what time will they meet?

Suppose that they meet after \( x \) days.

Then the one traveller has gone 8\( x \) miles, and the other 7\( x \) miles; now the sum of the distances they travel is, by the question, equal to the distance from London to York.

Therefore \( 8x + 7x = 150 \)

That is \( 15x = 150 \), and \( x = 10 \) days.

Ex. 4. A labourer engaged to serve for 40 days, upon these conditions; that for every day he worked he was to receive 2d. but for every day he played, or was absent, he was to forfeit 8d.; now at the end Let \( x \) be the number of days he worked.

Then will \( 40 - x \) be the number of days he was idle.

Also \( 20x = 20x \) is the sum he earned in pence.

And \( 8 \times (40 - x) = 320 - 8x \) is the sum he forfeited.

Now the difference of these two was \( 1l. 1s. 8d. \) or \( 380d. \)

Therefore \( 20x - (320 - 8x) = 380 \),

That is \( 28x = 700 \).

Hence \( x = 25 \) is the number of days he worked,

And \( 40 - x = 15 \) is the number of days he was idle.

Ex. 5. A market-woman bought a certain number of eggs at 2 a-penny, and as many at 3 a-penny; and sold them all out again at 5 for 2d.; but, instead of getting her own money for them, as she expected, she lost 4d.: what number of eggs did she buy?

Let \( x \) be the number of eggs of each sort.

Then will \( \frac{x}{2} \) be the price of the first sort.

And \( \frac{x}{3} \) is the price of the second sort.

Now the whole number being \( 2x \), we have

\[ 5 : 2x :: 2 : \frac{4x}{5} \]

price of both sorts at \( \frac{5}{2} \) for 2d.

Therefore \( \frac{x}{2} + \frac{x}{3} - \frac{4x}{5} = 4 \), by the question.

Hence \( 15x + 10x - 24x = 120 \),

And \( x = 120 \), the number of each sort.

Ex. 6. A bill of 120l. was paid in guineas and moidores: the number of pieces of both sorts that were used was 100; how many were there of each?

Let the number of guineas be \( x \).

Then the number of moidores will be \( 100 - x \).

Also the value of the guineas, reckoned in shillings, will be \( 21x \); and that of the moidores \( 27(100 - x) = 2700 - 27x \).

Therefore by the question, \( 21x + 2700 - 27x = 2400 \).

Hence we find \( 6x = 300 \), and \( x = 50 \).

So that the number of pieces of each sort was 50.

Ex. 7. A footman agreed to serve his master for 8l. a-year, and livery; but was turned away at the end of 7 months, and received only 2l. 13s. 4d. and his livery; what was its value?

Suppose \( x \) the value of the livery, in pence.

Then his wages for a year were to be \( x + 1920 \) pence.

But for 7 months he received \( x + 640 \) pence.

Now he was paid in proportion to the time he served.

\[ \frac{m}{m} \]

Therefore \( 12 : 7 :: x + 1920 : x + 640 \).

And taking the product of the extremes and means,

\[ 12x + 7680 = 7x + 13440 \]

Hence \( 5x = 5760d. \) and \( x = 1152d. = 4l. 16s. \)

Ex. 8. A person at play lost \( \frac{1}{4} \) of his money, and then won 3s.; after which he lost \( \frac{1}{4} \) of what he then had, and then won 2s.; lastly, he lost \( \frac{1}{4} \) of what he then had; and, this done, found he had only 12s. left: what had he at first?

Suppose he began to play with \( x \) shillings.

He lost \( \frac{1}{4} \) of his money, or \( \frac{x}{4} \), and had left \( x - \frac{x}{4} \)

\[ = \frac{3x}{4} \]

He won 3s. and had then \( \frac{3x}{4} + 3 = \frac{3x + 12}{4} \).

He lost \( \frac{1}{4} \) of \( \frac{3x + 12}{4} \), or \( \frac{x + 4}{4} \), and had left \( \frac{3x + 12}{4} - \frac{x + 4}{4} = \frac{2x + 8}{4} \).

He won 2s. and had then \( \frac{2x + 8}{4} + 2 = \frac{2x + 16}{4} \).

He lost \( \frac{1}{4} \) of \( \frac{2x + 16}{4} \) or \( \frac{2x + 16}{28} \), and had left \( \frac{2x + 16}{4} - \frac{2x + 16}{28} = \frac{12x + 96}{28} \).

And because he had now 12s. left, we have this equation \( \frac{12x + 96}{28} = 12 \).

Hence \( 12x = 240 \), and \( x = 20 \).

Ex. 9. Two tradesmen, A and B, are employed upon a piece of work; A can perform it alone in 15 hours, and B in 10 hours: in what time will they do it when working together.

Suppose that they can do it in \( x \) hours, and let the whole work be denoted by 1.

Then \( 15 : x :: 1 : \frac{x}{15} \) is the part of the work done by A.

And \( 10 : x :: 1 : \frac{x}{10} \) is the part done by B.

Now, by the question, they are to perform the whole work between them;

Therefore, \( \frac{x}{15} + \frac{x}{10} = 1 \).

Hence \( 25x = 150 \), and \( x = 6 \) hours.

Ex. 10. The sum of any two quantities being given \( = s \), and their difference \( = d \), it is required to find each of the quantities.

Let \( x \) denote the greater of the two quantities, and \( y \) the lesser.

Then \( x + y = s \), and \( x - y = d \).

Taking the sum of the equations we get \( 2x = s + d \);

And subtracting the second from the first, \( 2y = s - d \);

Therefore \( x = \frac{s + d}{2} \), and \( y = \frac{s - d}{2} \).

Ex. 11. A gentleman distributing money among some poor people, found he wanted 10s. to be able to give each 5s.; therefore he gave only 4s. to each, and had 5s. left. Required the number of shillings and poor people.

Let the number of shillings be \( x \), and that of the poor people \( y \); then, from the nature of the question, we have these two equations,

\[ 5y = x + 10 \]

From the first equation, \( x = 5y - 10 \),

And from the second, \( x = 4y + 5 \);

Therefore \( 5y - 10 = 4y + 5 \).

Hence \( y = 15 \), and \( x = 4y + 5 = 65 \). Ex. 12. A farmer kept a servant for every 40 acres of ground he rented, and on taking a lease of 104 more acres, he engaged 5 additional servants, after which he had a servant for every 36 acres. Required the number of servants and acres.

Suppose that he had at first \( x \) servants, and \( y \) acres.

From the first condition of the question \( x = \frac{y}{40} \).

And from the second \( x + 5 = \frac{y + 104}{36} \).

By comparing the values of \( x \), as found from these equations, we have \( \frac{y + 104}{36} - 5 = \frac{y}{40} \).

Hence \( 40y + 4160 - 7200 = 36y \), so that \( 4y = 3040 \).

Therefore \( y = 760 \), and \( x = \frac{y}{40} = 19 \).

Ex. 13. Two persons, A and B, were talking of their ages; says A to B, seven years ago I was just three times as old as you were then, and seven years hence I shall be just twice as old as you will be. What is their present ages?

Let the ages of A and B be \( x \) and \( y \) respectively. Their ages seven years ago were \( x - 7 \) and \( y - 7 \), and seven years hence they will be \( x + 7 \) and \( y + 7 \).

Therefore by the question

\[ x - 7 = 3(y - 7) \quad \text{and} \quad x + 7 = 2(y + 7). \]

From the first equation, \( x = 3y - 14 \).

And from the second \( x = 2y + 7 \).

Therefore \( 3y - 14 = 2y + 7 \); hence \( y = 21 \).

And because \( x = 2y + 7 \), therefore \( x = 49 \).

Ex. 14. A hare is 50 leaps before a greyhound, and takes 4 leaps to the greyhound's 3, but 2 of the greyhound's leaps are as much as 3 of the hare's. How many leaps must the greyhound take to catch the hare?

In this example there is only one quantity required, it will, however, be convenient to make use of two letters; therefore let \( x \) denote the number of leaps of the greyhound, and \( y \) those of the hare; then, by considering the proportion between the number of leaps each takes in the same time, we have

\[ 3 : 4 :: x : y, \quad \text{hence} \quad 3y = 4x. \]

Again, by considering the proportion between the number of leaps each must take to run the same distance, we find \( x : 50 + y :: 2 : 3 \), hence \( 100 + 2y = 3x \).

From the first equation we find \( 6y = 8x \).

And from the second \( 6y = 9x - 300 \).

Hence \( 9x - 300 = 8x \), and \( x = 300 \).

Ex. 15. To divide the number 90 into 4 such parts, that if the first be increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2; the sum, difference, product, and quotient, shall be all equal to each other.

In this question there are four quantities to be determined; but instead of introducing several letters, having put \( x \) to denote the first of them, we may find an expression for each of the remaining ones, as follows:

Because \( x + 2 = \) second quantity — 2,

Therefore \( x + 4 = \) the second quantity.

And because \( x + 2 = \) third \( \times 2 \),

Therefore \( \frac{x + 2}{2} = \) the third quantity.

And in like manner \( 2(x + 2) = \) the fourth quantity.

Now by the question, the sum of all the four \( = 90 \);

Therefore \( x + x + 4 + \frac{x + 2}{2} + 2(x + 2) = 90 \);

Hence \( 9x = 162 \), and \( x = 18 \).

Therefore the numbers required are 18, 22, 10, and 40.

Ex. 16. A and B together can perform a piece of work in 12 hours, A and C in 20, and B and C in 15 hours; in what time will each be able to perform it when working separately?

That we may have a general solution, let us suppose A and B can perform the work in \( a \) hours, A and C in \( b \) hours, and B and C in \( c \) hours. Let \( x \), \( y \), and \( z \), denote the times in which A, B, and C, could perform it respectively, if each wrought alone; and let the whole work be represented by 1.

Then \( x : a :: 1 : \frac{a}{x} = \) the part done by A in \( a \) hours

\( y : a :: 1 : \frac{a}{y} = \) the part done by B in \( a \) hours

Also \( x : b :: 1 : \frac{b}{x} = \) the part done by A in \( b \) hours

\( z : b :: 1 : \frac{b}{z} = \) the part done by C in \( b \) hours

And \( y : c :: 1 : \frac{c}{y} = \) the part done by B in \( c \) hours

\( z : c :: 1 : \frac{c}{z} = \) the part done by C in \( c \) hours.

Now by the question we have the three following equations.

\[ \frac{a}{x} + \frac{a}{y} = 1, \quad \frac{b}{x} + \frac{b}{z} = 1, \quad \frac{c}{y} + \frac{c}{z} = 1. \]

Let the first equation be divided by \( a \), the second by \( b \), and the third by \( c \), thus we have

\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{a}, \quad \frac{1}{x} + \frac{1}{z} = \frac{1}{b}, \quad \frac{1}{y} + \frac{1}{z} = \frac{1}{c}. \]

If these be added together, and their sum divided by 2, we find

\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c}. \]

From this equation let each of the three last be subtracted in its turn; thus we get

\[ \frac{1}{z} = \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c} = \frac{abc + ac + bc}{2abc}, \]

\[ \frac{1}{y} = \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c} = \frac{abc - ac + bc}{2abc}, \]

\[ \frac{1}{x} = \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c} = \frac{abc - ab + ac + bc}{2abc}. \]

Hence \( z = \frac{2abc}{abc + ac + bc} = \frac{7200}{120} = 60 \)

\( y = \frac{2abc}{abc - ac + bc} = \frac{7200}{360} = 20 \)

\( x = \frac{2abc}{abc - ab + ac + bc} = \frac{7200}{240} = 30 \). Sect. IX. Of Quadratic Equations.

140. We are next to explain the manner of resolving equations of the second degree, or quadratic equations. These involve the second power of the unknown quantity, as has been already observed (§ 113.), and may be divided into two kinds, pure and affected.

141. I. Pure quadratic equations are such as after proper reduction have the square of the unknown quantity in one term, while the remaining terms contain only known quantities. Thus, \(x^2 = 64\), and \(ax^2 + bx = c\) are examples of pure quadratics.

142. II. Affected quadratic equations, contain the square of the unknown quantity in one term, and its first or simple power in another, and the remaining terms consist entirely of known quantities. Such are the following, \(x^2 + 3x = 28\), \(2x^2 - 33 - 5x\), \(ax^2 + bx = c = d\).

143. The manner of resolving a pure quadratic equation is sufficiently evident; if the unknown quantity be made to stand alone on one side, with unity as a coefficient, while the other side consists entirely of known quantities, and if the square root of each side be taken, we shall immediately obtain the value of the simple power of the unknown quantity as already directed by Rule 5th of Sect. VI.

144. In extracting the square root of any quantity, however, it is necessary to observe, that the sign of the root may be either \(+\) or \(-\). This is an evident consequence of the rule for the signs in multiplication; for since by that rule any quantity, whether positive or negative, if multiplied by itself, will produce a positive quantity, and therefore the square of \(+a\), as well as that of \(-a\), is \(+a^2\); so, on the contrary, the square root of \(-a^2\) is to be considered either as \(+a\) or as \(-a\), and may accordingly be expressed thus \(\pm a\).

145. Having remarked that the square of any quantity, whatever be its sign, is always positive; it evidently follows, that no real quantity whatever, when multiplied by itself, can produce a negative quantity; and therefore if the square root of a negative quantity be required, no such root can be assigned. Hence it also follows, that if a problem requires for its solution the extraction of the square root of a negative quantity, some contradiction must necessarily be involved, either in the condition of the problem, or in the process of reasoning by which that solution has been obtained.

146. When an affected quadratic equation is to be resolved, it may always, by proper reduction, be brought to one or other of the three following forms.

1. \(x^2 + px = q\) 2. \(x^2 - px = q\) 3. \(x^2 - px = -q\)

But as the manner of resolving each of the three forms is the very same, it will be sufficient if we consider any one of them.

147. Referring therefore the first equation, or \(x^2 + px = q\); let us compare the side of it which involves the unknown quantity \(x\) with the square of a binomial \(x + a\); that is, let us compare \(x^2 + px\) with \(x^2 + 2ax + a^2 = (x + a)^2\); and it will presently appear, that if we suppose \(p = 2a\), or \(\frac{p}{2} = a\), the quantities \(x^2 + px\) and \(x^2 + 2ax\) will be equal; and as \(x^2 + 2ax\) is rendered a complete square, by adding to it \(a^2\), so also may \(x^2 + px\) be completed into a square, by adding to it \(\frac{p^2}{4}\), which is equal to \(a^2\); therefore, let \(\frac{p^2}{4}\) be added to both sides of the equation \(x^2 + px = q\), and we have

\[x^2 + px + \frac{p^2}{4} = \frac{p^2}{4} + q,\]

or \((x + \frac{p}{2})^2 = \frac{p^2}{4} + q;\)

and extracting the square root of each side, \(x + \frac{p}{2} = \pm \sqrt{\frac{p^2}{4} + q};\) hence \(x = -\frac{p}{2} \pm \sqrt{\frac{p^2}{4} + q}.\)

148. From these observations we derive the following general rule for resolving affected quadratic equations.

1. Transpose all the terms involving the unknown quantity to one side, and the known quantities to the other side, and so that the term involving the square of the unknown quantity may be positive.

2. If the square of the unknown quantity be multiplied by a coefficient, let the other terms be divided by it, so that the coefficient of the square of the unknown quantity may be 1.

3. Add to both sides the square of half the coefficient of the unknown quantity itself, and the side of the equation involving the unknown quantity will now be a complete square.

4. Extract the square root of both sides of the equation, by which it becomes simple with respect to the unknown quantity; and by transposition, that quantity may be made to stand alone on one side of the equation, while the other side consists of known quantities; and therefore the equation is resolved.

Note. The square root of the first side of the equation is always equal to the sum, or difference of the unknown quantity, and half the coefficient of the second term. If the line of that term be \(+\), it is equal to the sum, but if it be \(-\), then it is equal to the difference.

Ex. 1. Given \(x^2 + 2x = 35\), to determine \(x\).

Here the coefficient of the second term is 2, therefore adding the square of its half to each side, we have

\[x^2 + 2x + 1 = 35 + 1 = 36.\]

And extracting the square root \(x + 1 = \sqrt{36} = 6.\)

Hence \(x = 6 - 1\), that is \(x = +5\), or \(x = -7\), and either of these numbers will be found to satisfy the equation, for \(5 \times 5 + 2 \times 5 = 35\), also \(-7 \times -7 + 2 \times -7 = 35\).

Ex. 2. Given \(\frac{x^2}{6} - 12 = x\), to find \(x\).

This equation, when reduced, becomes \(x^2 - 6x = 72\).

And by completing the square, \(x^2 - 6x + 9 = 72 + 9 = 81.\)

Hence by extracting the square root, \(x - 3 = \pm 9.\)

And \(x = 9 + 3\), therefore \(x = +12\), or \(x = -6\), and upon trial we find that each of these values satisfies then it would be possible to resolve \( x^2 + px - q \) into as many factors, \( x - c, x - d, \ldots \), but the product of Equations more than two factors must necessarily contain the third or higher powers of \( x \); and as \( x^2 + px - q \) contains no higher power than the second; therefore no such solution can take place.

152. Since it appears that \( x^2 + px - q \) may be considered as the product of two factors \( x - a \) and \( x + b \), let us examine the nature of these factors; accordingly, taking their product by actual multiplication, we find it \( (x-a)(x+b) = ab \); and since this quantity must be equal to \( x^2 + px - q \), it follows, that \( b = -a \) and \( ab = q \), or changing the signs of the terms of both equations, \( a - b = p \), \( ab = -q \). Now if we consider that \( +a \) and \( -b \), are the roots of the equation \( x^2 + px - q \); it is evident that \( a - b \) is the sum of the roots, and \( -ab \) their product. So that from the equations \( a - b = p \), and \( -ab = q \), we derive the following proposition relating to the roots of any quadratic equation. The sum of the roots of any quadratic equation \( x^2 + px - q \) is equal to \( -p \), that is, to the coefficient of the second term, having its sign changed; and their product is equal to \( -q \), or to the latter side of the equation, having its sign also changed.

153. This proposition enables us to resolve several important questions concerning the roots of a quadratic equation, without actually resolving that equation. Thus we learn from it, that if \( q \), the term which does not involve the unknown quantity, (called sometimes the absolute number) be positive, the equation has one of its roots positive, and the other negative; but if that term be negative, the roots are either both positive or both negative. It also follows, that in the former case the root which is denoted by the least number will have the same sign with the second term, and in the latter case, the common sign of the roots will be the contrary to that of the second term.

154. From this property of the roots we may also derive a general solution to any quadratic equation \( x^2 + px = q \); for we have only to determine two quantities whose sum is \( -p \), and product \( -q \), and whose quantities shall be the two values of \( x \), or the two roots of the equation.

Without considering the signs of the roots, let us call them \( v \) and \( z \), then

\[ v + z = -p, \quad vz = -q. \]

From the square of each side of the first equation let four times the second be subtracted, and we have

\[ v^2 - 2vz + z^2 = p^2 + 4q, \quad \text{or} \quad (v-z)^2 = p^2 + 4q, \]

therefore by extracting the square root, \( v - z = \pm \sqrt{p^2 + 4q} \); from this equation, and from the equation \( v + z = -p \), we readily obtain \( v = -\frac{p}{2} \pm \frac{\sqrt{p^2 + 4q}}{2} \).

Then \( z = -\frac{p}{2} \mp \frac{\sqrt{p^2 + 4q}}{2} \), that is, if \( v = -\frac{p}{2} + \frac{\sqrt{p^2 + 4q}}{2} \), then \( z = -\frac{p}{2} - \frac{\sqrt{p^2 + 4q}}{2} \), and if \( v = -\frac{p}{2} - \frac{\sqrt{p^2 + 4q}}{2} \), then \( z = -\frac{p}{2} + \frac{\sqrt{p^2 + 4q}}{2} \).

But the value of \( v \), upon the one supposition, is the same as the value of \( z \) upon the other supposition, and vice versa; therefore, in reality, the only two distinct values Ex. 1. It is required to divide the number 10 quadratic into two such parts, that the sum of their squares may be 58.

Let \( x \) be the one number. Then, since their sum is 10, we have \( 10 - x \) for the other.

And by the question \( x^2 + (10-x)^2 = 58 \) That is \( x^2 + 100 - 20x + x^2 = 58 \) Or \( 2x^2 - 20x = 58 - 100 = -42 \) Hence \( x^2 - 10x = -21 \)

And completing the square \( x^2 - 10x + 25 = 25 - 21 = 4 \) Hence, by extracting the root, \( x - 5 = \pm \sqrt{4} = \pm 2 \). And \( x = 7 \) or \( x = 3 \).

If we take the greatest value of \( x \), viz. 7, then the other number \( 10 - x \) will be 3; and if we take the least value of \( x \), viz. 3, then the other number is 7. Thus it appears, that the greatest value of the one number corresponds to the least value of the other; and indeed this must necessarily be the case, seeing that both numbers are alike concerned in the question. Hence, upon the whole, the only numbers that will answer the conditions of the question are 7 and 3.

Ex. 2. What two numbers are those whose product is 28; and such, that twice the greater, together with three the lesser, is equal to 26.

Let \( x \) be the greatest and \( y \) the least number, then, from the nature of the question, we have these two equations,

\[ xy = 28, \quad 2x + 3y = 26. \]

From the first equation we have \( y = \frac{28}{x} \), And from the second \( y = \frac{26 - 2x}{3} \).

Hence, \( \frac{26 - 2x}{3} = \frac{28}{x} \).

And, reducing, \( 26x - 2x^2 = 84 \) Or \( 2x^2 - 26x = -84 \) Hence \( x^2 - 13x = -42 \) And comp. the sq. \( x^2 - 13x + \frac{169}{4} = \frac{169}{4} - 42 \) Hence, by extracting the root \( x - \frac{13}{2} = \pm \sqrt{\frac{169}{4} - 42} \) Therefore \( x = \frac{13}{2} \pm \frac{1}{2} \)

That is \( x = 7 \), or \( x = 6 \).

And since \( y = \frac{28}{x} \), we have \( y = 4 \), or \( y = \frac{7}{3} \).

Thus we have obtained two sets of numbers, which fulfil the conditions required, viz.

\( x = 7, \quad y = 4 \); Or \( x = 6, \quad y = \frac{7}{3} \).

And besides these, there can be no other numbers.

Ex. 3. A company dining together at an inn, find their bill amount to 175 shillings; two of them were not allowed to pay, and the rest found that their shares amounted to 10 shillings a-man more than if they had all paid. How many were in company?

Suppose their number to be \( x \). Then, if all had paid, the share of each would have been \( \frac{175}{x} \). But, because only \( x = 2 \) paid, the share of each was

\[ \frac{175}{x - 2} \]

Therefore, by the question,

\[ \frac{175}{x - 2} = \frac{175}{x} = 10. \]

And by proper reduction \( 175x - 175x + 350 = 10x^2 - 220 \).

That is

\[ 10x^2 - 20x = 350 \]

Or

\[ x^2 - 2x + 1 = 36 \]

Hence, by extracting the root, \( x^2 + 1 = 36 \).

Therefore, \( x = 5 \), or \( x = -7 \). But from the nature of the question, the negative root can be of no use; therefore \( x = 6 \).

Ex. 4. A merchant sold a piece of cloth for 24l. and gained as much per cent. as the cloth cost him; what was the price of the cloth?

Suppose that it cost \( x \) pounds,

Then the gain was \( 24 - x \),

And by the question \( 100 : x :: x : 24 - x \).

Therefore, taking the product of the extremes and means,

\[ 2400 - 100x = x^2, \]

Or

\[ x^2 + 100x = 2400, \]

And comp. the eq. \( x^2 + 100x + 2500 = 4900 \),

Hence, taking the root, \( x + 50 = \pm 70 \),

And

\[ x = 20 \text{ or } -120. \]

Here, as in the last question, the negative root cannot apply; therefore \( x = 20 \) pounds, the price required.

Ex. 5. A grazier bought as many sheep as cost him 60l. out of which he reserved 15, and sold the remainder for 54l. and gained 28s. each upon them. How many sheep did he buy, and what did each cost him?

Suppose that he bought \( x \) sheep,

Then each would cost him \( \frac{1200}{x} \) shillings.

Therefore, after reserving 15, he sold each of the remaining \( x - 15 \) for \( \frac{1200}{x} + 2 \) shillings.

Hence he would receive for them \( (x - 15)(\frac{1200}{x} + 2) \) shillings. And, because 54l. = 1080 shillings, we have by the question \( (x - 15)(\frac{1200}{x} + 2) = 1080 \).

Which by proper reduction becomes \( x^2 + 45x = 9000 \).

Or, completing the square, \( x^2 + 45x + \frac{2025}{4} = \frac{38025}{4} \).

Therefore, extracting the root, \( x = \frac{195}{2} - \frac{45}{2} \).

And taking the positive root, \( x = 75 \), the number of sheep; and consequently \( \frac{1200}{75} = 16 \) shillings, the price of each.

Ex. 6. What number is that, which, when divided by the product of its two digits, the quotient is 3; and if 18 be added to it, the digits are inverted. Let \( x \) and \( y \) denote the digits; then the number itself will be expressed by \( 10x + y \); and that number, in which the digits are inverted, by \( 10y + x \). Thus the conditions of the problem will be expressed by these two equations,

\[ \frac{10x + y}{xy} = 3, \quad 10x + y + 18 = 10y + x. \]

From the first equation we have \( y = \frac{10x}{3x - 1} \),

And from the second \( y = x + 2 \);

Therefore \( x + 2 = \frac{10x}{3x - 1} \),

And \( 3x^2 - 5x - 2 = 10x \).

Hence \( x^2 - \frac{5}{3}x = \frac{2}{3} \),

And comp. the eq. \( x^2 - \frac{5}{3}x + \frac{25}{36} = \frac{25}{36} + \frac{4}{9} \);

Therefore, taking the root \( x = \frac{-5 \pm \sqrt{49}}{6} \),

So that \( x = 2 \), or \( x = -\frac{7}{2} \).

Here it is evident that the negative root is useless; hence we have \( y = x + 2 = 4 \), and 24 for the number required.

Ex. 7. It is required to find two numbers whose product is 100; and the difference of their square roots 3.

Let \( x \) be the one number; then \( \frac{100}{x} \) must denote the other.

Now by the question \( \frac{10}{\sqrt{x}} - \sqrt{x} = 3 \);

Hence we have \( 10 - x = 3\sqrt{x} = 3x \),

Or

\[ x + 3x = 10, \]

And comp. the eq. \( x + 3x + \frac{9}{4} = 10 + \frac{9}{4} = \frac{49}{4} \),

and taking the root \( x = \frac{-3 \pm \sqrt{49}}{2} \),

So that \( x = 5 \) or \( x = -2 \),

and therefore \( x = 25 \) or \( x = 4 \).

If \( x = 4 \), the other number is \( \frac{100}{4} = 25 \), and if \( x = 25 \), then the other number is 4; so that, in either case, the two numbers which answer the conditions of the question are 4 and 25.

Ex. 8. It is required to find two numbers, of which the product shall be 6, and the sum of their cubes 35.

Let \( v \) be the one number, then \( \frac{6}{v} \) will be the other.

Therefore, by the question, \( x^3 + \frac{216}{x^3} = 35 \);

Hence \( x^6 + 216 = 35x^3 \),

Or \( x^6 - 35x^3 = -216 \).

This equation, by putting \( x^3 = y \), becomes

\( y^2 - 35y = -216 \);

Hence we find \( y = 27 \), or \( y = 8 \).

And since \( x^3 = y \); therefore \( x = 3 \), or \( x = 2 \).

If \( x = 3 \), then the other number is 2, and if \( x = 2 \), the other number is 3; so that 2 and 3 are the numbers required.

In general, if it be required to find two numbers, which are exactly alike concerned in a question that produces a quadratic equation; the two numbers sought will be the roots of that equation. A similar observation applies to any number of quantities which require for the determination the resolution of an equation of any degree whatever.

Sect. X. Of Equations in General.

160. Before we proceed to the resolution of cubic, and the higher orders of equations, it will be proper Equations to explain some general properties, which belong to equations of every degree; and also certain operations, which must frequently be performed upon equations, before they are fitted for a final solution.

161. In treating of equations in general, we shall suppose all the terms transposed to one side, and put equal to 0; this we have already done in explaining the nature of quadratics, and in like manner an equation of the fourth degree will stand thus:

\[ x^4 + px^3 + qx^2 + rx + s = 0 \]

where \( x \) denotes an unknown quantity, and \( p, q, r, s \), known quantities, either positive or negative. In this equation the coefficient of the highest power of \( x \) is unity, but if it had been any other quantity, that quantity might have been taken away, and the equation reduced to the above form, by rules already explained, Sect. VI.

162. The terms of an equation being thus arranged, if such a quantity be found, as when substituted for \( x \), will render both sides \( = c \); and therefore satisfy the equation, that quantity, whether it be positive or negative or even imaginary, is to be considered as a root of the equation. But we have seen that every quadratic equation has always two roots, real or imaginary, we may therefore suppose that a similar diversity of roots will take place in all equations of a higher degree; and this supposition we shall presently find to be well founded, by means of the following proposition, which is of great importance in the theory of equations.

If a root of any equation, as \( x^4 + px^3 + qx^2 + rx + s = 0 \), be represented by \( a \), the first side of that equation is divisible by \( x - a \).

For since \( x^4 + px^3 + qx^2 + rx + s = 0 \),

And also \( a^4 + pa^3 + qa^2 + ra + s = 0 \);

Therefore, by subtraction, \( x^4 - a^4 + p(x^3 - a^3) + q(x^2 - a^2) + r(x - a) = 0 \).

163. But any quantity of this form \( x^n - a^n \), where \( n \) denotes a whole positive number, is equal to

\[ (x-a)(x^{n-1} + ax^{n-2} + a^2x^{n-3} + \cdots + a^{n-2}x + a^{n-1}), \]

as may be easily proved by multiplication; therefore, putting \( x = 4, 3 \) and 2 successively, we have

\[ \begin{align*} x^4 - a^4 &= (x-a)(x^3 + ax^2 + a^2x + a^3) \\ x^3 - a^3 &= (x-a)(x^2 + ax + a^2) \\ x^2 - a^2 &= (x-a)(x+a) \\ x - a &= (x-a) \end{align*} \]

and by substitution, and collecting into one term the coefficients of the like powers of \( x \), the equation

\[ x^4 - a^4 + p(x^3 - a^3) + q(x^2 - a^2) + r(x - a) = 0 \]

becomes

\[ (x-a)(x^3 + (a+p)x^2 + (a^2+pa+q)x + a^3 + pa^2 + qa + r) = 0, \]

so that putting \( p' = a + p, q' = a^2 + pa + q, r' = a^3 + pa^2 + qa + r \), we have

\[ x^4 + p'x^3 + q'x^2 + rx + s = (x-a)(x^3 + p'x^2 + q'x + r'). \]

Hence, if the proposed equation \( x^4 + px^3 + qx^2 + rx + s = 0 \) be divided by \( x - a \), the quotient will be \( x^3 + p'x^2 + q'x + r' \), an integer quantity; and since the same mode of reasoning will apply to any equation whatever, the truth of the proposition is evident.

164. We have found that \( (x-a)(x^3 + p'x^2 + q'x + r') = 0 \), and as a product becomes \( = 0 \), when any one of its factors \( = 0 \), therefore, the equation will have its conditions fulfilled, not only when \( x-a = 0 \), but Equations also when \( x^3 + p'x^2 + q'x + r' = 0 \).

Let us now suppose that \( b \) is a root of this equation, then by reasoning exactly as in last article, and putting \( p'' = b + p', q'' = b^2 + p'b + q' \), we shall have

\[ x^3 + p''x^2 + q''x + r'' = (x-b)(x^2 + p''x + q'') = 0, \]

and therefore

\[ x^4 + p''x^3 + q''x^2 + rx + s = (x-a)(x-b)(x^2 + p''x + q''). \]

165. By proceeding in the same manner with the quadratic equation \( x^2 + p''x + q'' = 0 \), we shall find that if \( c \) denote one of its roots, then

\[ x^4 + p''x^3 + q''x^2 + rx + s = (x-c)(x+c+p''). \]

So that if we put \( d = -(c+p'') \), we at last find

\[ x^4 + p''x^3 + q''x^2 + rx + s = (x-a)(x-b)(x-c)(x-d); \]

and since each of the factors \( x-a, x-b, x-c, x-d \) may be assumed \( = 0 \), it follows, that there are four different values of \( x \), which will render the equation \( x^4 + p''x^3 + q''x^2 + rx + s = 0 \), namely, \( x = a, x = b, x = c, x = d \).

166. The mode of reasoning which has been just now employed in a particular case, may be applied to an equation of any order whatever; we may therefore conclude, that every equation may be considered as the product of as many simple factors, as the number denoting its order contains unity; and therefore, that the number of roots in any equation is precisely equal to the exponent of the highest power of the unknown quantity contained in that equation.

167. By considering equations of all degrees as formed from the product of factors \( x-a, x-b, x-c, \ldots \), we discover a number of curious relations, which subsist between the roots of any equation whatever, and its coefficients. Thus, if we limit the number of factors to four, and suppose that \( a, b, c, d \) are the roots of this equation of the fourth degree

\[ x^4 + px^3 + qx^2 + rx + s = 0, \]

we shall also have \( (x-a)(x-b)(x-c)(x-d) = 0 \);

and therefore, by actual multiplication,

\[ \begin{array}{cccc} x^4 & +ab & +ac & +ad \\ x^3 & +bd & +bc & +bd \\ x^2 & +cd & +cd & +cd \\ x & +abcd & +abcd & +abcd \\ \end{array} \]

168. If we compare together the coefficients of the same powers of \( x \), we find the following series of equations:

\[ \begin{align*} a + b + c + d &= -p \\ ab + ac + ad + bc + bd + cd &= -q \\ abc + abd + acd + bcd &= -r \\ abcd &= s; \end{align*} \]

and as a similar series of equations will be obtained for every equation whatever, we hence derive the following propositions, which are of the greatest importance in the theory of equations.

1. The coefficient of the second term of any equation taken with a contrary sign, is equal to the sum of all the roots.

2. The coefficient of the third term is equal to the sum of the products of the roots multiplied together two and two.

3. The coefficient of the fourth term, taken with a contrary Equations contrary sign, is equal to the sum of the roots multiplied together three and three, and so on in the remaining coefficients, till we come to the last term of the equation, which is equal to the product of all the roots, having their signs changed.

169. Instead of supposing an equation to be produced by multiplying together simple equations, we may consider it as formed by the product of equations of any degree, provided that the sum of their dimensions is equal to that of the proposed equation. Thus, an equation of the fourth degree may be formed either from a simple and cubic equation, or from two quadratic equations.

170. If \( n \) denote the degree of an equation, we have shown, that by considering it as the product of simple factors, that equation will have \( n \) divisors of the first degree; but if we suppose the simple factors to be combined two and two, they will form quantities of the second degree, which are also factors of the equation; and since there may be formed \( \frac{n(n-1)}{2} \) such combinations, any equation will admit of \( \frac{n(n-1)}{2} \) divisors of the second degree.

171. For example, the equation \( x^4 + px^3 + qx^2 + rx + s = 0 \), which we have considered as equal to

\[ (x-a)(x-b)(x-c)(x-d) = 0, \]

may be formed of the product of two factors of the second degree, in these five different ways.

By the product of \((x-a)(x-b)\) and \((x-c)(x-d)\)

\[ \begin{align*} (x-a)(x-c) & (x-b)(x-d) \\ (x-a)(x-d) & (x-b)(x-c) \\ (x-b)(x-c) & (x-a)(x-d) \\ (x-b)(x-d) & (x-a)(x-c) \\ (x-c)(x-d) & (x-a)(x-b) \end{align*} \]

Thus an equation of the fourth degree may have \( \frac{4 \times 3}{2} = 6 \) quadratic divisors.

172. By combining the simple factors three and three, we shall have divisors of the third degree, of which the number for an equation of the \( n \)th order will be \( \frac{n(n-1)(n-2)}{6} \); and so on.

173. When the roots of an equation are all positive, its simple factors will have this form \( x-a, x-b, x-c, \ldots \); and if for the sake of brevity we take only these three, the cubic equation which results from their product will have this form,

\[ x^3 - px^2 + qx - r = 0, \]

where \( p = a + b + c, q = ab + ac + bc, r = abc \),

and here it appears that the signs of the terms are \( + \) and \( - \) alternately.

Hence we infer, that when the roots of an equation are all positive, the signs of its terms are positive and negative alternately.

174. If again the roots of the equation be all negative, and therefore its factors \( x+a, x+b, x+c \), then \( p, q, \) and \( r \) being as before, the resulting equation will stand thus:

\[ x^3 + px^2 + qx + r = 0. \]

And hence we conclude, that when the roots are all negative, there is no change whatever in the signs.

175. In general, if the roots of an equation be all real, that equation will have as many positive roots as in general, there are changes of the signs from \( + \) to \( - \), or from \( - \) to \( + \); and the remaining roots are negative. This rule, however, does not apply when the equation has imaginary roots, unless such roots be considered as either positive or negative.

176. That the rule is true when applied to quadratic equations will be evident from Sect. IX. With respect to cubic equations, the rule also applies when the roots are either all positive, or all negative, as we have just now shown.

When a cubic equation has one positive root, and the other two negatives, its factors will be \( x-a, x+b, x+c \), and the equation itself

\[ \begin{align*} x^3 - a & -ab \\ + b & x^2 - ac \\ + c & + bc \end{align*} \]

Here there must always be one change of the signs, since the first term is positive, and the last negative; and there can be no more than one; for if the second term is negative, or \( b+c \) less than \( a \), then \((b+c)^2\) will be less than \((b+c)a\); but \((b+c)^2\) is always greater than \( bc \), therefore \( bc \) will be much less than \((b+c)a\) or \( ab+ac \), so that the third term must also be negative, and therefore in this case only one change of the signs.

If again the second term be positive, then because the sign of the last term is negative, whatever be the sign of the third term, there can still be no more than one change of the signs.

When the equation has two positive roots and one negative, its factors are \( x-a, x-b, x-c \), and the equation

\[ \begin{align*} x^3 - a & + ab \\ - b & x^2 - ac \\ + c & - bc \end{align*} \]

Here there must always be two changes of the signs; for if \( a+b \) be greater than \( c \), the second term is negative, and the last term being always positive, there must be two changes, whether the sign of the third term be positive or negative. If again \( a+b \) be less than \( c \), and therefore the second term positive; it may be thrown as before, that \( ab \) is much less than \( ac+bc \); and hence the third term will be negative; so that in either case there must be two changes of the signs.

We may conclude therefore, upon the whole, that in cubic equations there are always as many positive roots, as changes of the signs from \( + \) to \( - \), or from \( - \) to \( + \); and by the same method of reasoning, the rule will be found to extend to all equations whatever.

177. It appears from the manner in which the coefficients of an equation are formed from its roots, that when the roots are all real, the coefficients must consist entirely of real quantities. But it does not follow, on the contrary, that when the coefficients are real, the roots are also real; for we have already found, that in a quadratic equation, \( x^2 + px + q = 0 \), where \( p \) and \( q \) denote real quantities, the roots are sometimes both imaginary.

178. When the roots of a quadratic equation are imaginary, they have always this form, \( a+\sqrt{-b^2}, a-\sqrt{-b^2} \), which quantities may also be expressed thus, Equations \(a + b \sqrt{-1}, a - b \sqrt{-1}\), so that we have these two factors in general, \(x - a - b \sqrt{-1}, x - a + b \sqrt{-1}\), and taking their product,

\[ x^2 - 2ax + a^2 + b^2 = 0. \]

Thus we see that two imaginary factors may be of such a form as to admit of their product being expressed by a real quantity; and hence the origin of imaginary roots in quadratic equations.

179. It appears by induction, that no real equation can be formed from imaginary factors, unless those factors be taken in pairs, and each pair have the form \(x - a - b \sqrt{-1}, x - a + b \sqrt{-1}\); for the product of three, or any odd number of imaginary factors, whatever be their form, is still an imaginary quantity. Thus, if we take the product of any three of these four imaginary expressions, \(x + a + b \sqrt{-1}, x + a - b \sqrt{-1}, x + c + d \sqrt{-1}, x + c - d \sqrt{-1}\), we may form four different equations, each of which will involve imaginary quantities. If, however, each equation be multiplied by the remaining factor, which had not previously entered into its composition, the product will be found to be rational, and the same for all the four.

180. Hence we may deduce the three following inferences respecting the roots of equations:

1. If an equation have imaginary roots, it must have two, or four, or some even number of such roots. 2. If the degree of an equation be denoted by an odd number, that equation must have at least one real root. 3. If the degree of an equation be denoted by an even number, and that equation have one real root, it will also have another real root.

181. We shall now explain some transformations which are frequently necessary to prepare the higher orders of equations for a solution.

Any equation may have its positive roots changed into negative roots of the same value, and its negative roots into such as are positive, by changing the signs of the terms alternately, beginning with the second. The truth of this remark will be evident, if we take two equations,

\[ (x-a)(x-b)(x+c) = 0, \] \[ (x+a)(x+b)(x-c) = 0, \]

(which are such, that the positive roots of the one have the same values as the negative roots of the other), and multiply together their respective factors, for these equations will stand thus:

\[ \begin{align*} x^3 &- ax^2 + abx - acx + abc = 0 \\ &+ bx^2 - bcx + acx - abc = 0 \end{align*} \]

where it appears that the signs of the first and third terms are the same in each, but the signs of the second and fourth are just the opposite of each other. And this will be found to hold true, not only of cubic equations, but of all equations, to whatever order they belong.

182. It will sometimes be useful to transform an equation into another, that shall have each of its roots greater or less than the corresponding roots of the other equation, by some given quantity.

Let \((x-a)(x-b)(x+c) = 0\) be any proposed equation, which is to be transformed into another, having its roots greater or less than those of the proposed equation by the given quantity \(n\); then, because the roots of the transformed equation are to be \(+a+n, +b+n\) and \(-c-n\), the equation itself will be

\[ (y-n-a)(y-n-b)(y-n+c) = 0. \]

Hence the reason of the following rule is evident.

If the new equation is to have its roots greater than those of the proposed equation, instead of \(x\) and its powers, substitute \(y-n\) and its powers; but if the roots are to be less, then instead of \(x\) substitute \(y+n\); and in either case, a new equation will be produced, the roots of which shall have the property required.

183. By means of the preceding rule, an equation may be changed into another, which has its roots either all positive, or all negative; but it is chiefly used in preparing cubic and biquadratic equations for a solution, by transforming them into others of the same degrees, but which want their second term.

Let \(x^3 + px^2 + qx + r = 0\) be any cubic equation; if we substitute \(y+n\) for \(x\), the equation is changed into the following:

\[ \begin{align*} y^3 &+ 3ny^2 + 3pn^2 + pn^3 \\ &+ qy^2 + 2pnq + pn^2q + pn^3q \\ &+ ry + pnq + pn^2q + pn^3q + pn^4q + pn^5q = 0. \end{align*} \]

Now, that this equation may want its second term, it is evident, that we have only to suppose \(3n+p=0\), or \(n=-\frac{p}{3}\), for this assumption being made, and the value of \(n\) substituted in the remaining terms, the equation becomes

\[ y^3 + \left(q-\frac{p^2}{3}\right)y + \frac{2p^3}{27} - \frac{pq}{3} + r = 0. \]

or, putting \(-\frac{p^2}{3} + q = q'\), and \(\frac{2p^3}{27} - \frac{pq}{3} + r = r'\), the same equation may also stand thus,

\[ y^3 + q'y + r' = 0. \]

184. In general, any equation whatever may be transformed into another, which shall want its second term by the following rule.

Divide the coefficient of the second term of the proposed equation by the exponent of the first term, and add the quotient, with its sign changed, to a new unknown quantity; the sum being substituted for the unknown quantity in the proposed equation, a new equation will be produced, which will want the second term, as required.

185. By this rule, any affected quadratic equation may be readily resolved; for by transforming it into another equation, which wants the second term, we thus reduce its solution to that of a pure quadratic. Thus if the quadratic equation \(x^2 - 5x + 6 = 0\) be proposed; by substituting \(y+\frac{5}{2}\) for \(x\), we find

\[ y^2 + \frac{5}{2}y + \frac{25}{4} - \frac{5}{2}y + \frac{25}{4} + 6 = 0, \]

or \(y^2 + \frac{25}{4} = 0\).

Hence \(y = \pm \frac{5}{2}\), and since \(x = y + \frac{5}{2}\), therefore \(x = \pm \frac{5}{2} + \frac{5}{2} = \pm 3\) or \(\pm 2\).

186. It has been shewn (\$169.) that in any equation, the coefficient of the second term, having its sign changed, is equal to the sum of all the roots, or subtracting. subtracting from their signs, it is equal to the difference between the sum of the positive, and the sum of the negative roots. Therefore, if the second term be wanting, the sum of the positive roots in the equation must necessarily be equal to that of the negative roots.

187. Instead of taking away the second term from an equation, any other term may be made to vanish, by an assumption similar to that which has been employed to take away the second term. Thus if in § 183, we assume $3n^2 + 2pn + q = 0$, by resolving this quadratic equation, a value of $n$ will be found, which, when substituted in the equation, will cause the third term to vanish; and by the resolution of a cubic equation the third term might be taken away; and so on.

188. Another species of transformation, of use in the resolution of equations, is that by which an equation, having the coefficients of some of its terms expressed by fractional quantities, is changed into another, the coefficients of which are all integers.

Let $\frac{p}{a}x^3 + \frac{q}{b}x^2 + \frac{r}{c}x + \frac{s}{d} = 0$ denote an equation to be so transformed; and let us assume $y = abcx$; and therefore $x = \frac{y}{abc}$; then, by substitution, our equation becomes

$$\frac{y^3}{a^3b^3c^3} + \frac{p}{a^3b^3c^3}y^2 + \frac{q}{ab^3c^3}y + \frac{r}{abc} = 0,$$

and multiplying the whole equation by $a^3b^3c^3$, we have

$$y^3 + pby^2 + qcy + r = 0.$$

Thus we have an equation free from fractions; while at the same time the coefficient of the highest power of the unknown quantity is unity, as before.

189. This transformation may always be performed by the following rule. Instead of the unknown quantity substitute a new unknown quantity divided by the product of all the denominators; then, by proper reduction, the equation will be found to have the form required.

190. If, however, the equation have this form,

$$x^3 + \frac{p}{a}x^2 + \frac{q}{a}x + \frac{r}{a} = 0,$$

it will be sufficient to assume $y = ax$, and therefore $x = \frac{y}{a}$; for then we have

$$\frac{y^3}{a^3} + \frac{p}{a^3}y^2 + \frac{q}{a^3}y + \frac{r}{a} = 0,$$

And $y^3 + py^2 + qy + r = 0$,

which last equation has the form required.

Sect. XI. Of Cubic Equations.

191. Cubic equations, as well as equations of every higher degree, are, like quadratics, divided into two classes; they are said to be pure, when they contain only one power of the unknown quantity; and affected, when they contain two or more powers of that quantity.

192. Pure cubic equations are therefore of this form, $x^3 = 125$, or $x^3 = -27$, or, in general, $x^3 = r$; and hence it appears, that the value of the simple power of the unknown quantity may always be found, without difficulty, by extracting the cube root of each side of the equation; thus from the first of the three preceding examples we find $x = 5$, from the second $x = -3$, and from the third $x = \sqrt[3]{-7}$.

193. It would seem at first sight, that the only value which $x$ can have in the cubic equation $x^3 = r$, or putting $r = c^3$, $x^3 - c^3 = 0$, is this one, $x = c$; but since $x^3 - c^3$ may be resolved into these two factors $x - c$ and $x^2 + cx + c^2$, it follows, that besides the value of $x$ already found, which results from making the factor $x - c = 0$, it has yet other two values, which may be found by making the other factor $x^2 + cx + c^2 = 0$; and accordingly by resolving the quadratic equation

$$x^2 + cx + c^2 = 0,$$

we find these values to be

$$-\frac{c + \sqrt{-3c^2}}{2}, \quad \text{and} \quad -\frac{c - \sqrt{-3c^2}}{2}, \quad \text{or} \quad -\frac{1 + \sqrt{-3}}{2}c \quad \text{and} \quad -\frac{1 - \sqrt{-3}}{2}c.$$

Thus it appears, that any cubic equation of this form, $x^3 = c^3$, or $x^3 - c^3 = 0$, has these three roots,

$$x = c, \quad x = -\frac{1 + \sqrt{-3}}{2}c, \quad x = -\frac{1 - \sqrt{-3}}{2}c,$$

the first of which is real, but the two last are imaginary. If, however, each of the imaginary values of $x$ be raised to the third power, the same results will be obtained as from the real value of $x$; the original equation $x^3 - c^3 = 0$ may also be reproduced, by multiplying together the three factors $x - c$, $x - \frac{1 + \sqrt{-3}}{2}c$, and $x - \frac{1 - \sqrt{-3}}{2}c$.

194. Let us now consider such cubic equations as have all their terms, and which are therefore of this form,

$$x^3 + Ax^2 + Bx + C = 0,$$

where $A$, $B$, and $C$ denote known quantities, either positive or negative.

It has been shewn (§ 184,) how an equation having all its terms may be transformed into another, which wants the second term; let us therefore assume $x = y - \frac{A}{3}$, as directed in that article; then, by proper substitution, the above equation will be changed into another of this form,

$$y^3 + qy + r = 0,$$

where $q$ and $r$ denote known quantities, whether positive or negative; now the roots of this equation being once found, it is evident that those of the former may also be readily obtained by means of the assumed equation $x = y - \frac{A}{3}$.

195. Refining, therefore, the equation $y^3 + qy + r = 0$, let us suppose $y = v + z$, and it becomes

$$v^3 + 3v^2z + 3vz^2 + z^3 + qv + qz + r = 0.$$

Thus we have got a new equation, which, as it involves two unknown quantities, $v$ and $z$, may be resolved into any two other equations, which will simplify the determination of those quantities.

Now it appears, that the only way in which we can divide Cubic Equations.

divide that equation into two others, so as to simplify the question, is the following

\[3v^2z + 3vz^2 + vz + qz = 0\]

The first of these equations may also be expressed thus,

\[(3vz + q)(v + z) = 0\]

Hence we must either suppose that \(v + z = 0\), or that \(3vz + q = 0\); but the former supposition cannot be admitted without supposing also that \(y = 0\), which does not agree with the hypothesis of the equation \(y^3 + qy + r = 0\); therefore we must adopt the latter. So that to determine \(v\) and \(z\) we have these two equations,

\[3vz + q = 0, \quad v^3 + z^3 + r = 0.\]

From the first, we find \(vz = -\frac{q}{3}\), and \(v^3z^3 = -\frac{q^3}{27}\);

and from the second \(v^3 + z^3 = -r\), so that to determine the quantities \(v^3\) and \(z^3\), we have given their sum, and product; now this is a problem which we have already resolved when treating of quadratic equations, § 155; and by proceeding in the same manner, in the present case we shall find

\[v^3 = -\frac{r}{2} + \sqrt{\frac{r^2}{4} + \frac{q^3}{27}}, \quad z^3 = -\frac{r}{2} - \sqrt{\frac{r^2}{4} + \frac{q^3}{27}}.\]

Thus we have at last obtained a value of the unknown quantity \(y\), in terms of the known quantities \(q\) and \(r\); therefore the equation is resolved.

166. But this is only one of three values which \(y\) may have; let us, for the sake of brevity, put

\[A = -\frac{r}{2} + \sqrt{\frac{r^2}{4} + \frac{q^3}{27}}, \quad B = -\frac{r}{2} - \sqrt{\frac{r^2}{4} + \frac{q^3}{27}},\]

and denote the imaginary expressions

\[-\frac{1}{2} + \frac{i\sqrt{3}}{2}, \quad -\frac{1}{2} - \frac{i\sqrt{3}}{2}\]

by \(\alpha\) and \(\beta\). Then, from what has been shewn (§ 193), it is evident that \(v\) and \(z\) have each these three values,

\[v = \frac{3}{\sqrt{A}}, \quad v = \alpha \frac{3}{\sqrt{B}}, \quad v = \beta \frac{3}{\sqrt{B}},\]

\[z = \frac{3}{\sqrt{B}}, \quad z = \alpha \frac{3}{\sqrt{B}}, \quad z = \beta \frac{3}{\sqrt{B}}.\]

To determine the corresponding values of \(v\) and \(z\), we must consider that \(vz = -\frac{q}{3} = \sqrt{AB}\); now if we observe that \(\alpha \beta = 1\), it will immediately appear that \(v + z\) has these three values,

\[v + z = \frac{3}{\sqrt{A}} + \frac{3}{\sqrt{B}},\]

\[v + z = \alpha \frac{3}{\sqrt{A}} + \beta \frac{3}{\sqrt{B}},\]

\[v + z = \beta \frac{3}{\sqrt{A}} + \alpha \frac{3}{\sqrt{B}}.\]

Hence the three values of \(y\) are also these,

\[y = \frac{3}{\sqrt{A}} + \frac{3}{\sqrt{B}},\]

\[y = \alpha \frac{3}{\sqrt{A}} + \beta \frac{3}{\sqrt{B}},\]

\[y = \beta \frac{3}{\sqrt{A}} + \alpha \frac{3}{\sqrt{B}}.\]

The first of these formulae is commonly known by the name of Cardan's rule; but it is well known that Cardan was not the inventor, and that it ought to be attributed to Nicholas Tartalea, and Scipio Ferreus, who discovered it much about the same time, and independently of each other (see the Introduction).

197. The formulae given in last article for the roots of a cubic equation may be put under a different form, and perhaps better adapted to the purposes of arithmetical calculation as follows. Because \(vz = -\frac{q}{3}\),

therefore \(z = -\frac{q}{3v} = -\frac{q}{3} \times \frac{1}{\sqrt{A}}\), hence \(v + z = \sqrt{A} - \frac{q}{3\sqrt{A}}\);

thus it appears that the three values of \(y\) may also be expressed thus,

\[y = \frac{3}{\sqrt{A}} - \frac{q}{3\sqrt{A}},\]

\[y = \alpha \frac{3}{\sqrt{A}} - \frac{q}{3\sqrt{A}},\]

\[y = \beta \frac{3}{\sqrt{A}} - \frac{q}{3\sqrt{A}}.\]

198. To show the manner of applying these formulae, let it be required to determine \(x\) from the cubic equation

\[x^3 + 3x^2 + 9x - 13 = 0.\]

And as this equation has all its terms, the first step towards its resolution is to transform it into another which shall want the second term, by substituting \(y = 1\) for \(x\) as directed (§ 184). The operation will stand thus,

\[\begin{align*} &x^3 = y^3 - 3y^2 + 3y - 1 \\ &+ 3x^2 = + 3y^2 - 6y + 3 \\ &+ 9x = + 9y - 9 \\ &- 13 = - 13 \end{align*}\]

The transformed equation is \(y^3 + 6y - 20 = 0\)

which being compared with the general equation

\[y^3 + qy + r = 0\]

gives \(q = 6, r = -20\); hence

\[A = \sqrt{-\frac{r}{2} + \sqrt{\frac{r^2}{4} + \frac{q^3}{27}}} = \sqrt{10 + \sqrt{108}}.\]

Therefore, the first formula of last article gives \(y = \frac{3}{\sqrt{10 + \sqrt{108}}} - \frac{2}{3\sqrt{10 + \sqrt{108}}}\); but as this expression involves a radical quantity, let the square root of 108 be taken and added to 10, and the cube root of the sum found; thus we have \(\frac{3}{\sqrt{10 + \sqrt{108}}} = 2.732\), nearly, and and therefore \( \frac{2}{\sqrt{10} + \sqrt{108}} = \frac{2}{2.732} = .732 \); hence we at last find one of the values of \( y \) to be \( 2.732 - .732 = 2 \).

In finding the cube root of the radical quantity \( \sqrt{10} + \sqrt{108} \) we have taken only its approximate value, so as to have the expression for the root under a rational form, and in this way we can always find, as near as we please, the cube root of any surd of the form \( a + \sqrt{b} \) where \( b \) is a positive number. But it will sometimes happen that the cube root of such a surd can be expressed exactly by another surd of the same form; and accordingly, in the present case, it appears that the cube root of \( 10 + \sqrt{108} \) is \( 1 + \sqrt{3} \), as may be proved by actually raising \( 1 + \sqrt{3} \) to the third power. Hence we find

\[ \frac{2}{\sqrt{10} + \sqrt{108}} = \frac{2}{1 + \sqrt{3}} (1 - \sqrt{3}) = -(1 - \sqrt{3}); \]

so that we have \( y = 1 + \sqrt{3} + 1 - \sqrt{3} = 2 \), as before.

The other two values of \( y \) will be had by substituting \( 1 + \sqrt{3} \) and \( 1 - \sqrt{3} \) for \( \sqrt{A} \) and \( \frac{\sqrt{3}}{\sqrt{A}} \) in the second and third formulae of last article, also restoring the values of \( \alpha \) and \( \beta \). We thus have

\[ y = \frac{-1 + \sqrt{3}}{2} \times (1 + \sqrt{3}) + \frac{-1 - \sqrt{3}}{2} \times (1 - \sqrt{3}) = -1 + \sqrt{9}; \]

\[ y = \frac{-1 - \sqrt{3}}{2} \times (1 + \sqrt{3}) + \frac{-1 + \sqrt{3}}{2} \times (1 - \sqrt{3}) = -1 - \sqrt{9}. \]

So that the three values of \( y \) are

\[ +2, -1 + \sqrt{9}, -1 - \sqrt{9}, \]

and since \( x = y + 1 \), the corresponding values of \( x \) are

\[ +1, -2 + \sqrt{9}, -2 - \sqrt{9}. \]

Thus it appears that one of the roots of the proposed equation is real and the other two imaginary.

The two imaginary roots might have been found otherwise, by considering that since one root of the equation is 1, the equation must be divisible by \( x - 1 \) (\$165.). Accordingly the division being actually performed, and the quotient put \( = 0 \), we have this quadratic equation,

\[ x^2 + 4x + 13 = 0; \]

which, when resolved by the rule for quadratics, gives \( x = -2 \pm \sqrt{9} \), the same imaginary value as before.

199. In the application of the preceding formulae (\$196 and 197) to the resolution of the equation \( y^3 + gy + r = 0 \), it is necessary to find the square root of \( \frac{1}{3}g^2 + \frac{1}{2}r^2 \); now when that quantity is positive, as in the equation \( y^3 + 6y - 20 = 0 \), which was resolved in last article, no difficulty occurs, for its root may be found, either exactly, or to as great a degree of accuracy as we please.

As, however, the coefficients \( g \) and \( r \) are independent of each other, it is evident that \( q \) may be negative, and such that \( \frac{1}{3}g^2 \) is greater than \( \frac{1}{2}r^2 \); in this case the expression \( \frac{1}{3}g^2 + \frac{1}{2}r^2 \) will be negative, and therefore its square root an imaginary quantity. Let us take as an example this equation \( y^3 - 6y + 4 = 0 \); here \( q = -6, r = 4, \frac{1}{3}g^2 = 8, \frac{1}{2}r^2 = 4, \sqrt{\frac{1}{3}g^2 + \frac{1}{2}r^2} = \sqrt{-4} = 2\sqrt{-1} \), hence by recurring to the formula (\$166.), we have \( A = 2 + 2\sqrt{-1}, B = 2 - 2\sqrt{-1} \), and therefore the three roots of the equation expressed thus,

\[ y = \frac{3}{\sqrt{2 + 2\sqrt{-1}}} + \frac{3}{\sqrt{2 - 2\sqrt{-1}}}, \]

\[ y = \frac{3}{\sqrt{2 + 2\sqrt{-1}}} + \frac{3}{\sqrt{2 - 2\sqrt{-1}}}, \]

\[ y = \frac{3}{\sqrt{2 + 2\sqrt{-1}}} + \frac{3}{\sqrt{2 - 2\sqrt{-1}}}. \]

Here all the roots appear under an imaginary form; but we are certain from the theory of equations, as explained in Sect. X., that every cubic equation must have at least one real root. The truth is, as we shall show immediately, that in this case, so far from any of the roots being imaginary (as in the former example), they are all real; for it appears by actual involution that the imaginary expression \( 2 + 2\sqrt{-1} \) is the cube of this other imaginary expression \( -1 + \sqrt{-1} \), and in like manner, that \( 2 - 2\sqrt{-1} \) is the cube of \( -1 - \sqrt{-1} \), so that we have

\[ y = \frac{3}{\sqrt{2 + 2\sqrt{-1}}} + \frac{3}{\sqrt{2 - 2\sqrt{-1}}} = -1 + \sqrt{-1} = 2; \]

\[ y = \frac{-1 + \sqrt{-3}}{2} \times (-1 + \sqrt{-1}) + \frac{-1 - \sqrt{-3}}{2} \times (-1 - \sqrt{-1}) = 1 + \sqrt{3}; \]

\[ y = \frac{-1 - \sqrt{-3}}{2} \times (-1 + \sqrt{-1}) + \frac{-1 + \sqrt{-3}}{2} \times (-1 - \sqrt{-1}) = 1 - \sqrt{3}. \]

200. We now proceed to prove in general, that as often as the roots of the equation \( x^3 + qx + r = 0 \) are real, \( q \) is negative, and \( \frac{1}{3}g^2 \) greater than \( \frac{1}{2}r^2 \); and, on the contrary, that if \( \frac{1}{3}g^2 \) be greater than \( \frac{1}{2}r^2 \) the roots are all real.

Let us suppose \( a \) to be a real root of the proposed equation,

Then \( x^3 + qx + r = 0 \),

And \( a^3 + qa + r = 0 \).

And therefore by subtraction \( x^3 - a^3 + q(x-a) = 0 \); hence, dividing \( x^3 - a^3 \), also \( q(x-a) \) by \( x-a \), we have

\[ x^2 + ax + a^2 + q = 0. \]

This quadratic equation is formed from the two remaining roots of the proposed equation, and by resolving it we find

\[ x = \frac{-a \pm \sqrt{-\frac{1}{3}a^2 - q}}{2}. \]

And as, by hypothesis, all the roots are real, it is evident that \( q \) must necessarily be negative, and greater than \( \frac{1}{3}a^2 \); for otherwise the expression \( \sqrt{-\frac{1}{3}a^2 - q} \) would be imaginary. Let us change the sign of \( q \), and put put \( q = \frac{1}{3}a^2 + d \); thus the roots of the equation \( x^3 + qx + r = 0 \) will be

\[ a_1 = \frac{1}{3}a + \sqrt{d}, \quad a_2 = \frac{1}{3}a - \sqrt{d}, \]

and here \( d \) is a positive quantity.

To find an expression for \( r \) in terms of \( a \) and \( d \), let \( \frac{1}{3}a^2 + d \) be substituted for \( q \) in the equation \( a^3 - qa + r = 0 \); we thence find \( r = \frac{1}{3}a^3 + ad \); so that to compare together the quantities \( q \) and \( r \) we have these equations,

\[ q = \frac{1}{3}a^2 + d \\ r = \frac{1}{3}a^3 + ad. \]

In order to make this comparison, let the cube of \( \frac{1}{3}q \) be taken, also the square of \( \frac{1}{3}r \); the results are

\[ \frac{1}{27}q^3 = \frac{1}{9}a^6 + \frac{1}{3}a^4d + \frac{1}{3}a^2d^2 + \frac{1}{27}d^3 \\ \frac{1}{27}r^2 = \frac{1}{9}a^6 - \frac{1}{3}a^4d + \frac{1}{3}a^2d^2; \]

and therefore by subtraction,

\[ \frac{1}{27}q^3 - \frac{1}{27}r^2 = \frac{1}{9}a^6 - \frac{1}{3}a^4d - \frac{1}{3}a^2d^2 + \frac{1}{27}d^3 \\ = 3d(\frac{1}{9}a^4 - \frac{1}{3}a^2d + \frac{1}{3}d^2) \\ = 3d(\frac{1}{9}a^2 - \frac{1}{3}d)^2. \]

Now the square of any real quantity being always positive, it follows, that \( 3d(\frac{1}{9}a^2 - \frac{1}{3}d)^2 \) will be positive when \( d \) is positive; hence it is evident that in this case \( \frac{1}{3}q^3 \) must be greater than \( \frac{1}{27}r^2 \); and that the contrary cannot be true unless \( d \) be negative, that is, unless that \( \frac{1}{3}a + \sqrt{d}, \frac{1}{3}a - \sqrt{d} \), the two other roots of the equation, are imaginary. If we suppose \( d = 0 \), then \( \frac{1}{3}q^3 = \frac{1}{27}r^2 \); and the roots of the equations, which in this case are also real, are \( a_1 = \frac{1}{3}a, a_2 = \frac{1}{3}a \).

Upon the whole, therefore, we infer, that since a cubic equation has always one real root, its roots will be all real as often as \( q \) is negative, and \( \frac{1}{3}q^3 \) greater than \( \frac{1}{27}r^2 \); and consequently, that in this case the formulae for the roots must express real quantities notwithstanding their imaginary form.

201. Let \( y^3 - qy + r = 0 \) denote any equation of the form which has been considered in last article, namely, that which has its roots all real; then, if we put \( a = -\frac{1}{3}r, b = \frac{1}{3}q^3 - \frac{1}{27}r^2 \), one of the roots, as expressed by the first formula, § 196, will be,

\[ y = \sqrt[3]{a + b\sqrt{-1}} + \sqrt[3]{a - b\sqrt{-1}}. \]

This expression, although under an imaginary form, must (as we have shown in last article) represent a real quantity. It will sometimes happen, as in last example, § 199, that the two roots which compose the root are perfect cubes of the form \( (\Lambda + B\sqrt{-1})^3 \) and \( (\Lambda - B\sqrt{-1})^3 \); and then the value of \( y \) becomes

\[ A + B\sqrt{-1} + A - B\sqrt{-1} = 2A. \]

But the rules for determining when this is the case depend upon trials, and are, besides, troublesome in the application; and if we attempt by a direct process to investigate the numerical values of \( A \) and \( B \), we are brought to a cubic equation of the very same form as that whose root is required.

202. This imaginary expression for a real quantity has greatly perplexed mathematicians; and much pains has been taken to obtain the root under another form, but without success. Accordingly, the case of cubic equations, in which the roots are all real, is now called the irreducible case.

203. It is remarkable that the expression

\[ \sqrt{a + b\sqrt{-1}} + \sqrt{a - b\sqrt{-1}}, \]

and in general,

\[ \sqrt[n]{a + b\sqrt{-1}} + \sqrt[n]{a - b\sqrt{-1}}, \]

where \( n \) is any power of 2, admits of being reduced to another form in which no impossible quantity is found.

Thus \( \sqrt{a + b\sqrt{-1}} + \sqrt{a - b\sqrt{-1}} = \sqrt{2a + 2\sqrt{a^2 + b^2}} \)

\[ \sqrt[4]{a + b\sqrt{-1}} + \sqrt[4]{a - b\sqrt{-1}} = \sqrt{\left(2a + 2\sqrt{a^2 + b^2}\right)^2 + \left(\frac{a^2 + b^2}{2}\right)^2}, \]

as is easily proved by first squaring the imaginary formulae and then taking the square root of each. But when \( n \) is 3, it does not seem that such reduction can possibly take place.

204. If each of the roots be expanded into an infinite series and their sum be taken, the imaginary quantity \( \sqrt{-1} \) will vanish; and thus the root may be found by a direct process. There are, however, other methods which seem preferable; and the following, which is derived from the application of algebra to geometry, seems to be the best.

205. It will be demonstrated in Sect. XXV., that if \( a \) denote an arch of a circle, the relation between the cosine of the arch and the cosine of \( \frac{a}{3} \), one-third of that arch, is expressed by the following cubic equation.

\[ \text{Cof. } \frac{a}{3} = \frac{3}{4} \text{ cof. } \frac{a}{3} \text{ cof. } a. \]

Let us assume \( \frac{a}{3} = \frac{y}{n} \), then, by substitution, the equation is transformed into the following:

\[ \frac{y^3}{n^3} - \frac{3y}{4n} = \frac{1}{4} \text{ cof. } a. \]

Or \( y^3 - \frac{3n^2}{4} y = n^3 \times \frac{1}{4} \text{ cof. } a \),

and in this cubic equation one of the roots is evidently \( y = n \times \text{ cof. } \frac{a}{3} \). Now from the arithmetic of lines it appears that \( \text{ cof. } a, \text{ cof. } (360^\circ - a), \text{ and } \text{ cof. } (360^\circ + a) \), are all expressed by the same quantity; therefore the equation must have for a root not only \( n \times \text{ cof. } \frac{a}{3} \), but also \( n \times \text{ cof. } \frac{360^\circ - a}{3} \), and \( n \times \text{ cof. } \frac{360^\circ + a}{3} \). But from the arithmetic of lines, \( \text{ cof. } \frac{360^\circ - a}{3} = \text{ fin. } \frac{90^\circ - a}{3} \), and \( \text{ cof. } \frac{360^\circ + a}{3} = \text{ fin. } \frac{90^\circ + a}{3} \). Therefore the roots of the equation are

\[ n \times \text{ cof. } \frac{a}{3}, -n \times \text{ fin. } \frac{90^\circ - a}{3}, -n \times \text{ fin. } \frac{90^\circ + a}{3}. \]

Let us next suppose that \( y^3 - qy + r = 0 \) is a cubic equa- Cubic Equations whose roots are required, and let us compare it with the former equation \( y^3 - \frac{3n^2}{4} y = n^3 \times \text{cof. } \frac{1}{4} a \); then it is evident that if we assume the quantities \( n \) and \( a \), such that

\[ \frac{3n^2}{4} = q, \quad n^3 \times \text{cof. } \frac{1}{4} a = r, \]

the two equations will become identical, and thus their roots will be expressed by the very same quantities. But from these two assumed equations we find

\[ n = \sqrt{\frac{4q}{3}}, \quad \text{cof. } a = \frac{4r}{n^3} = \frac{\sqrt{27r^3}}{4q^3} = \frac{3r\sqrt{3}}{2q\sqrt{q}}, \]

and since the cofine of an arch cannot exceed unity, therefore, \( \frac{27r^3}{4q^3} \) must be a proper fraction, that is, \( 4q^3 \) must exceed \( 27r^3 \), or \( \frac{1}{4} q^3 \) must exceed \( \frac{1}{3} r^3 \); if we now recollect that \( q \) is a negative quantity, it will immediately appear that the proposed equation must necessarily belong to the irreducible case.

206. The rule, therefore, which we derive from the preceding analysis for resolving that case is as follows.

Let \( y^3 - qy = r \) be the proposed equation.

Find in the trigonometrical tables an arch \( a \), whose natural cofine \( = \frac{3r\sqrt{3}}{2q\sqrt{q}} \).

The roots of the equation are

\[ y = 2\sqrt{\frac{q}{3}} \times \text{cof. } \frac{a}{3}, \]

\[ y = -2\sqrt{\frac{q}{3}} \times \text{fin. } \frac{90^\circ - a}{3}, \]

\[ y = -2\sqrt{\frac{q}{3}} \times \text{fin. } \frac{90^\circ + a}{3}. \]

These formulae will apply, whether \( r \) be positive or negative, by proper attention to the signs: If, however, \( r \) be negative, or the equation have this form, \( y^3 - qy = -r \), the following will be more convenient:

Find in the tables an arch \( a \), whose fine \( = \frac{3r\sqrt{3}}{2q\sqrt{q}} \).

Then the roots of the equation are

\[ y = 2\sqrt{\frac{q}{3}} \times \text{fin. } \frac{a}{3}, \]

\[ y = 2\sqrt{\frac{q}{3}} \times \text{cof. } \frac{90^\circ + a}{3}, \]

\[ y = -2\sqrt{\frac{q}{3}} \times \text{cof. } \frac{90^\circ - a}{3}. \]

The last formulae are derived from the equation

\[ \sin \frac{3a}{3} = \frac{3}{4} \times \text{fin. } \frac{a}{3} = \text{fin. } a \]

in the same manner as the former were found from the first equation of last article.

Ex. 1. It is required to find the roots of the equation \( x^3 - 3x = 1 \).

Here \( \frac{3r\sqrt{3}}{2q\sqrt{q}} = \frac{3\sqrt{3}}{6\sqrt{3}} = \text{cof. } 60^\circ = \text{cof. } a \).

Hence

\[ \begin{align*} x &= 2 \text{ cof. } \frac{60^\circ}{3} = 2 \text{ cof. } 20^\circ = 1.8793852 \\ x &= -2 \text{ fin. } \frac{150^\circ}{3} = -2 \text{ fin. } 50^\circ = -1.5320888 \\ x &= -2 \text{ fin. } \frac{350^\circ}{3} = -2 \text{ fin. } 10^\circ = .3472964. \end{align*} \]

Ex. 2. It is required to find the roots of the equation \( x^3 - 3x = 1 \).

Here \( \frac{3r\sqrt{3}}{2q\sqrt{q}} = \frac{3\sqrt{3}}{6\sqrt{3}} = \frac{1}{2} \text{ sin. } 30^\circ = \text{fin. } a \).

\[ \begin{align*} x &= 2 \text{ fin. } \frac{30^\circ}{3} = 2 \text{ fin. } 10^\circ = .3472964 \\ x &= 2 \text{ cof. } \frac{120^\circ}{3} = 2 \text{ cof. } 40^\circ = 1.5320888 \\ x &= -2 \text{ cof. } \frac{60^\circ}{3} = -2 \text{ cof. } 20^\circ = -1.8793852. \end{align*} \]

Sect. XII. Of Biquadratic Equations.

207. When a biquadratic equation contains all its terms, it hath this form,

\[ x^4 + Ax^3 + Bx^2 + Cx + D = 0, \]

where \( A, B, C, D \), denote any known quantities whatever.

208. We shall first consider pure biquadratics, or such as contain only the first and last terms, and therefore are of this form, \( x^4 = b^4 \). In this case it is evident that \( x \) may be readily had by two extractions of the square root; by the first we find \( x^2 = b^2 \), and by the second \( x = b \). This, however, is only one of the values which \( x \) may have; for since \( x^2 = b^2 \), therefore \( x^2 - b^2 = 0 \); but \( x^2 - b^2 \) may be resolved into two factors \( x^2 - b^2 = (x - b)(x + b) \) each of which admit of a similar resolution; for \( x^2 - b^2 = (x - b)(x + b) \) and \( x^2 + b^2 = (x - b\sqrt{-1})(x + b\sqrt{-1}) \). Hence it appears that the equation \( x^4 = b^4 = 0 \) may also be expressed thus:

\[ (x - b)(x + b)(x - b\sqrt{-1})(x + b\sqrt{-1}) = 0, \]

so that \( x \) may have these four values,

\[ \pm b, \pm b\sqrt{-1}, \pm b\sqrt{-1}, \]

two of which are real and the others imaginary.

209. Next to pure biquadratic equations, in respect of coefficients of resolution, are such as want the second and fourth terms, and therefore have this form,

\[ x^4 + qx^2 + s = 0. \]

These may be resolved in the manner of quadratic equations; for if we put \( y^2 = x^2 \) we have

\[ y^2 + qy + s = 0, \]

from which we find \( y = \frac{-q \pm \sqrt{q^2 - 4s}}{2} \), and therefore

\[ x = \pm \sqrt{-q \pm \sqrt{q^2 - 4s}}. \]

210. When a biquadratic equation has all its terms, the manner of resolving it is not so obvious as in the two former cases, but its resolution may be always reduced to that of a cubic equation. There are various methods by which such a reduction may be effected; the following, which we select as one of the most ingenious, was first given by Euler in the Peterburgh Commentaries, and Biquadratic and afterwards explained more fully in his Elements of Equations. Algebra.

We have already explained, § 184, in what manner an equation which is complete in its terms may be transformed into another equation of the same degree, but which wants the second term; therefore, any proposed biquadratic equation may be reduced to this form,

\[ y^4 + py^2 + qy + r = 0, \]

where the second term is wanting, and where \( p, q, r \) denote any known quantities whatever.

211. That we may form any equation similar to the above, let us assume \( y = \sqrt{a} + \sqrt{b} + \sqrt{c} \), and let us also suppose that the letters \( a, b, c \) denote the roots of the cubic equation

\[ x^3 + Px^2 + Qx - R = 0; \]

then from the theory of equations we have

\[ a + b + c = -P, \quad ab + ac + bc = Qabc = R. \]

Let us now square the assumed formula

\[ y = \sqrt{a} + \sqrt{b} + \sqrt{c}, \]

and we obtain

\[ y^2 = a + b + c + 2(\sqrt{ab} + \sqrt{ac} + \sqrt{bc}) \]

or substituting \(-P\) for \(a + b + c\), and transposing,

\[ y^2 + P = 2(\sqrt{ab} + \sqrt{ac} + \sqrt{bc}). \]

Let this equation be also squared, and we have

\[ y^4 + 2Py^2 + P^2 = 4(ab + ac + bc) + 8(\sqrt{a^2bc} + \sqrt{ab^2c} + \sqrt{abc^2}), \]

and since \(ab + ac + bc = Q\),

\[ y^4 + 2Py^2 + P^2 = 4Q + 8\sqrt{Ry}. \]

Thus we have obtained the biquadratic equation

\[ y^4 + 2Py^2 - 8\sqrt{Ry} + P^2 - 4Q = 0, \]

one of the roots of which \(y = \sqrt{a} + \sqrt{b} + \sqrt{c}\), and in which \(a, b, c\) are the roots of the cubic equation

\[ x^3 + Px^2 + Qx - R = 0. \]

212. That we may apply this resolution to the proposed equation \( y^4 + py^2 + qy + r = 0 \), we must express the assumed coefficients \(P, Q, R\) by means of \(p, q, r\) the coefficients of that equation. For this purpose let us compare together the equations

\[ y^4 + py^2 + qy + r = 0, \] \[ y^4 + 2Py^2 - 8\sqrt{Ry} + P^2 - 4Q = 0, \]

and it immediately appears that \(2P = p, -8\sqrt{R} = q, P^2 - 4Q = r\); and from these three equations we find

\[ P = \frac{p}{2}, \quad Q = \frac{P^2 - 4r}{16}, \quad R = \frac{q^2}{64}. \]

Hence it follows, that the roots of the proposed equation are generally expressed by the formula \(y = \sqrt{a} + \sqrt{b} + \sqrt{c}\); where \(a, b, c\) denote the roots of this cubic equation

\[ x^3 + Px^2 + Qx - R = 0. \]

213. But to find each particular root, we must consider, that as the square root of a number may be either positive or negative, to each of the quantities, \(\sqrt{a}, \sqrt{b}, \sqrt{c}\), may have either the sign \(+\) or \(-\) prefixed to it; and hence our formula will give eight different expressions for the root. It is, however, to be observed, that as the product of the three quantities \(\sqrt{a}, \sqrt{b}, \sqrt{c}\), must be equal to \(\sqrt{R}\) or to \(-\frac{q}{8}\), therefore when \(q\) is positive, their product must be a negative quantity; and this can only be effected by making either one or three of them negative; again, when \(q\) is negative, their product must be a positive quantity, so that in this case they must either be all positive, or two of them must be negative. These considerations enable us to determine, that four of the eight expressions for the root belong to the case in which \(q\) is positive, and the other four to that in which it is negative.

214. We shall now give the result of the preceding investigation, in the form of a practical rule, for resolving biquadratic equations; and as the coefficients of the cubic equation which has been found, § 212, involve fractions, we shall transform it into another, in which the coefficients are integers, by supposing

\[ x = \frac{v}{4}. \]

Thus the equation \(x^3 + Px^2 + \frac{P^2 - 4r}{16}x - \frac{q^2}{64} = 0\) becomes, after reduction, \(v^3 + 2pv^2 + (p^2 - 4r)v - q^2 = 0\); it also follows, that since the roots of the former equation are \(a, b, c\), the roots of the latter are \(\frac{a}{4}, \frac{b}{4}, \frac{c}{4}\), so that our rule may now be expressed thus:

Let \(y^4 + py^2 + qy + r = 0\) be any biquadratic equation wanting its second term. Form this cubic equation

\[ v^3 + 2pv^2 + (p^2 - 4r)v - q^2 = 0, \]

and find its roots, which let us denote by \(a, b, c\).

Then the roots of the proposed biquadratic equation are

\[ y = \frac{1}{2}(\sqrt{a} + \sqrt{b} + \sqrt{c}), \quad y = \frac{1}{2}(-\sqrt{a} + \sqrt{b} + \sqrt{c}), \] \[ y = \frac{1}{2}(-\sqrt{a} + \sqrt{b} + \sqrt{c}), \quad y = \frac{1}{2}(-\sqrt{a} + \sqrt{b} + \sqrt{c}). \]

215. This resolution of biquadratic equations suggests the following general remarks upon the nature of their roots.

1. It is evident from the form of the roots, that if the cubic equation

\[ v^3 + 2pv^2 + (p^2 - 4r)v - q^2 = 0 \]

have all its roots real, and positive, those of the biquadratic equation shall be all real.

2. Since the last term of the cubic equation is negative, when its three roots are real, they must either be all positive, or two of them must be negative and one positive; for the last term is equal to the product of all the roots taken with contrary signs, § 169; so that in this last case two of the three quantities \(a, b, c\) must be negative, and therefore all the four roots of the biquadratic equation imaginary. If, however, the two negative roots be equal, they will destroy each other in two of the roots of the biquadratic equation, which will then become real and equal. Let us suppose for example that \(b\) and \(c\) are negative, and equal; the two first values of \(y\) in each column become then imaginary, By resolving the equation \( v^3 + Av + B = 0 \), we find

\[ v = -\frac{A}{2} \pm \sqrt{\frac{A^2}{4} - B}; \]

here, the roots being supposed imaginary, \( \frac{A^2}{4} - B \) must be a negative quantity. That we may simplify the form of the roots, let us put \( \frac{A}{2} = a \), and \( \frac{A^2}{4} - B = -\beta^2 \), then

\[ v = -a \pm \sqrt{-\beta^2} = -a \pm \beta \sqrt{-1}, \]

and \( v = -a + \beta \sqrt{-1}, v = -a - \beta \sqrt{-1} \).

Hence we have

\[ a = a + \beta \sqrt{-1}, b = a - \beta \sqrt{-1}, c = y; \]

so that in two of the four values of \( y \), we have a quantity of this form,

\[ \sqrt{a + \beta \sqrt{-1}} + \sqrt{a - \beta \sqrt{-1}}; \]

but this quantity, although it appears to be imaginary, is indeed real; for if we first square it, and then take its square root, it becomes

\[ \sqrt{2a + 2\sqrt{a^2 - \beta^2}}, \]

which is a real quantity. The two other roots involve this other expression,

\[ \sqrt{a + \beta \sqrt{-1}} - \sqrt{a - \beta \sqrt{-1}}; \]

which, being treated in the same manner as the former, becomes

\[ \sqrt{2a - 2\sqrt{a^2 + \beta^2}}, \]

an imaginary quantity, and therefore the roots into which it enters are imaginary.

4. We may discover from the coefficients of the proposed biquadratic equation in what case the roots of the cubic equation are all real; for this purpose, the latter is to be transformed into another which shall want the second term, by assuming \( v = u - \frac{2p}{3} \); thus it becomes

\[ u^3 - \left( \frac{p^3}{3} + 4r \right) u = \frac{2p^3}{27} + \frac{8rp}{3} - q^2 = 0; \]

and in this equation the three roots will be real when \( \frac{p^3}{3} + 4r \) is greater than \( \frac{2p^3}{27} + \frac{8rp}{3} + q^2 \).

216. As an example of the method of resolving a biquadratic equation, let it be required to determine the roots of the following,

\[ x^4 - 25x^3 + 60x - 36 = 0. \]

Vol. I. Part II.

217. We have now explained the particular rules by which the roots of equations belonging to each of the first four orders may be determined; and this is the greatest length mathematicians have been able to go in the direct resolution of equations; for as to those of the fifth, and all higher degrees, no general method has hitherto been found, either for resolving them directly, or for reducing them to others of an inferior degree.

It even appears that the formulae which express the roots of cubic equations are by no means of universal application; for in one case, that is, when the roots are all real, they become illusory, so that no conclusion can be drawn from them. The same observation will also apply to the formulae for the roots of biquadratic equations, because, before they can be applied, it is always necessary to find the roots of a cubic equation. But in either cubics or biquadratic equations, even when the formulae involve no imaginary quantities, and therefore can be always applied, it is more convenient in practice to employ some other methods which we are hereafter to explain.

Sect. XIII. Of Reciprocal Equations.

218. Although no general resolution has hitherto been given of equations belonging to the fifth, or any higher degree; yet there are particular equations of all orders, which, by reason of certain peculiarities in the nature of their roots, admit of being reduced to others of a lower degree, and thus, in some cases, equations of the higher orders may be resolved by the rules which have been already explained for the resolution of equations belonging to the first four orders.

219. When the coefficients of the terms of an equation form the same numerical series, whether taken in a direct or an inverted order, as in this example,

\[ x^4 + px^3 + qx^2 + rx + s = 0, \]

that equation may always be transformed into another of a degree denoted by half the exponent of the highest power of the unknown quantity, if that exponent be an even number, or by half the exponent diminished by unity, if it be an odd number.

The same observation will also apply to any equation of this form,

\[ x^4 + px^3 + qx^2 + rx + s = 0, \]

where the given quantity \( a \) and the unknown quantity \( x \) are... Reciprocal \( x \) are precisely alike concerned; for by substituting \( ay \) Equations, for \( x \), it becomes

\[ a^4y^4 + pa^3y^3 + qa^2y^2 + ra^2y + sa = 0; \]

and dividing by \( a^4 \),

\[ y^4 + py^3 + qy^2 + ry + s = 0, \]

an equation of the same kind as the former.

220. That we may effect the proposed transformation upon the equation

\[ x^4 + px^3 + qx^2 + rx + s = 0, \]

let every two terms which are equally distant from the extremes be collected into one, and the whole be divided by \( x^2 \), thus we have

\[ x^2 + \frac{1}{x^2} + p(x + \frac{1}{x}) + q = 0. \]

Let us assume \( x + \frac{1}{x} = z \).

Then \( x^2 + \frac{1}{x^2} = z^2 \) and \( x^2 + \frac{1}{x^2} = z^2 - 2 \).

Thus the equation \( x^2 + \frac{1}{x^2} + p(x + \frac{1}{x}) + q = 0 \)

becomes \( z^2 + pz + q - 2 = 0 \).

And since \( z + \frac{1}{z} = 2 \), therefore \( z^2 - 2z + 1 = 0 \).

221. Hence, upon the whole, to determine the roots of the biquadratic equation

\[ x^4 + px^3 + qx^2 + rx + s = 0, \]

we have the following rule.

Form this quadratic equation,

\[ z^2 + pz + q - 2 = 0, \]

and find its roots, which let us suppose denoted by \( z' \) and \( z'' \). Then the four roots of the proposed equation will be found by resolving two quadratic equations

\[ x^2 - z'x + 1 = 0, \quad x^2 - z''x + 1 = 0. \]

222. It may be observed respecting these two quadratic equations, that since the last term of each is unity, if we put \( a, a' \) to denote the roots of the one, and \( b, b' \) those of the other, we have from the theory of equations \( a'a = 1 \), and therefore \( a' = \frac{1}{a} \); also \( b'b = 1 \), and \( b' = \frac{1}{b} \); now \( a, a', b, b' \) are also the roots of the equation

\[ x^4 + px^3 + qx^2 + rx + s = 0. \]

Hence it appears that the proposed equation has this peculiar property, that the one half of its roots are the reciprocals of the other half; and to that circumstance we are indebted for the simplicity of its resolution.

223. The following equation

\[ x^6 + px^5 + qx^4 + rx^3 + sx^2 + tx + u = 0, \]

which is of the sixth order, admits of a resolution in all respects similar to the former; for by putting it under this form

\[ x^3 + \frac{1}{x^3} + p(x^2 + \frac{1}{x^2}) + q(x + \frac{1}{x}) + r = 0, \]

and putting also \( x + \frac{1}{x} = z \), so that \( x^2 - zx + 1 = 0 \), we have \( x^2 + \frac{1}{x^2} = z^2 - 2 \).

\[ x^3 + \frac{1}{x^3} = z^3 - 3(z + \frac{1}{z}) = z^3 - 3z. \]

Hence, by substitution, the proposed equation is transformed into the following cubic equation

\[ z^3 + pz^2 + (q - 3)z + r - 2p = 0. \]

Therefore, putting \( z, z', z'' \), to denote its roots, the six roots of the proposed equation will be had by resolving these three quadratics

\[ x^2 - zx + 1 = 0, \quad x^2 - z'x + 1 = 0, \quad x^2 - z''x + 1 = 0, \]

and here it is evident, as in the former case, that the roots of each quadratic equation are the reciprocals of each other, so that the one half of the roots of the proposed equation are the reciprocals of the other half.

224. The method of resolution we have employed in the two preceding examples is general for all equations whatever, in which the terms placed at equal distances from the first and last have the same coefficients, and which are called reciprocal equations, because any such equation has the same form when you substitute for \( x \) its reciprocal \( \frac{1}{x} \).

225. If the greatest exponent of the unknown quantity in a reciprocal equation is an odd number, as in this example

\[ x^5 + px^4 + qx^3 + rx^2 + sx + t = 0, \]

the equation will always be satisfied by substituting \( -1 \) for \( x \); hence \( -1 \) must be a root of the equation, and therefore the equation must be divisible by \( x + 1 \).

Accordingly, if the division be actually performed, we shall have in the present case

\[ x^4 + (p - 1)x^3 + (q - p)x^2 + (r - q)x + (s - r) = 0, \]

another reciprocal equation, in which the greatest exponent of \( x \) is an even number, and therefore resolvable in the manner we have already explained.

Sect. XIV. Of Equations which have Equal Roots.

226. When an equation has two or more of its roots equal to one another, those roots may always be discovered, and the equation reduced to another of an inferior degree, by a method of resolution which is peculiar to this class of equations; and which we now proceed to explain.

227. Although the method of resolution we are to employ will apply alike to equations having equal roots, of every degree, yet, for the sake of brevity, we shall take a biquadratic equation

\[ x^4 + px^3 + qx^2 + rx + s = 0, \]

the roots of which may be generally denoted by \( a, b, c, d \). Thus we have, from the theory of equations,

\[ (x-a)(x-b)(x-c)(x-d) = x^4 + px^3 + qx^2 + rx + s. \]

Let us put

\[ A = (x-a)(x-b)(x-c)(x-d), \quad A' = (x-a)(x-b)(x-d), \quad A'' = (x-a)(x-c)(x-d), \quad A''' = (x-b)(x-c)(x-d). \]

Then, Then, by actual multiplication, we have

\[ A = x^3 - a \] \[ A' = x^3 - b \] \[ A'' = x^3 - c \] \[ A''' = x^3 - d \]

and taking the sum of these four equations

\[ A + A' + A'' + A''' = 4x^3 - 3a - 3b - 3c - 3d \]

But since \( a, b, c, d \) are the roots of the equation

\[ x^4 + px^3 + qx^2 + rx + s = 0, \]

we have

\[ -(a + b + c + d) = 3p \] \[ 2(ab + ac + ad + bc + bd + cd) = 2q \] \[ -(abc + abd + acd + bcd) = r \]

Therefore, by substitution,

\[ A + A' + A'' + A''' = 4x^3 + 3px^2 + 2qx + r. \]

228. Let us now suppose that the proposed biquadratic equation has two equal roots, or \( a = b \), then \( x = a = b \), and since one or other of these equal factors enters each of the four products \( A, A', A'', A''' \), it is evident that \( A + A' + A'' + A''' \), or \( 4x^3 + 3px^2 + 2qx + r \) must be divisible by \( x - a \) or \( x - b \). Thus it appears that if the proposed equation

\[ x^4 + px^3 + qx^2 + rx + s = 0 \]

has two equal roots, each of them must also be a root of this equation

\[ 4x^3 + 3px^2 + 2qx + r = 0; \]

for when the first of these equations is divisible by \( (x - a)^2 \), the latter is necessarily divisible by \( x - a \).

229. Let us next suppose that the proposed equation has three equal roots or \( a = b = c \), then at least of the three equal factors \( x - a, x - b, x - c \), must enter each of the four products \( A, A', A'', A''' \); so that in this case \( A + A' + A'' + A''' \), or \( 4x^3 + 3px^2 + 2qx + r \) must be twice divisible by \( x - a \). Hence it follows, that as often as the proposed equation has three equal roots, two of them must also be equal roots of the equation

\[ 4x^3 + 3px^2 + 2qx + r = 0. \]

230. Proceeding in the same manner, it may be shown that whatever number of equal roots are in the proposed equation

\[ x^4 + px^3 + qx^2 + rx + s = 0 \]

they will all remain except one, in this equation

\[ 4x^3 + 3px^2 + 2qx + r = 0, \]

which is evidently derived from the former, by multiplying each of its terms by the exponent of \( x \) in that with equal term, and then diminishing the exponent by unity.

231. If we suppose that the proposed equation has two equal roots or \( a = b \), and also two other equal roots, or \( c = d \), then, by reasoning as before, it will appear that the equation derived from it must have one root equal to \( a \) or \( b \), and another equal to \( c \) or \( d \); so that when the former is divisible both by \( (x - a)^2 \) and \( (x - c)^2 \), the latter will be divisible by \( (x - a)(x - c) \).

232. The same mode of reasoning may be extended to all equations whatever; so that if we suppose

\[ x^n + Px^{n-1} + Qx^{n-2} + \ldots + Sx + T + U = 0, \]

an equation of the \( m \)th degree, to have a divisor of this form,

\[ (x - a)^n (x - d)^p (x - f) \ldots &c. \]

The equation

\[ mx^{m-1} + (m - 1) Px^{m-2} + (m - 2) Qx^{m-3} + \ldots + 2Sx + T = 0, \]

which is of the next lower degree, will have for a divisor

\[ (x - a)^{n-1} (x - d)^{p-1} (x - f)^{r-1} \ldots &c. \]

and as this last product must be a divisor of both equations, it may always be discovered by the rule which has been given (\$ 49.) for finding the greatest common divisor of two algebraic quantities.

233. Again, as this last equation must, in the case of equal roots, have the same properties as the original equation; therefore, if we multiply each of its terms by the exponent of \( x \), and diminish that exponent by unity, as before, we have

\[ m(m - 1)x^{m-2} + (m - 1)(m - 2) Px^{m-3} + (m - 2)(m - 3) Qx^{m-4} + \ldots + 2Sx + T = 0, \]

a new equation, which have for a divisor

\[ (x - a)^{n-1} (x - d)^{p-1} (x - f)^{r-1} \ldots &c. \]

where the exponent of the factors are one less than those of the equation from which it was derived; and as this last divisor is also a divisor of the original equation, it may be discovered in the same manner as the former, namely, by finding the greatest common measure of both equations; and so on we may proceed as far as we please.

234. As a particular example, let us take this equation

\[ x^5 - 13x^4 + 67x^3 - 171x^2 + 216x - 108 = 0, \]

and apply to it the method we have explained, in order to discover whether it has equal roots, and if so, what they are. We must therefore seek the greatest common measure of the proposed equation and this other equation, which is formed agreeably to what has been shown § 228,

\[ 5x^4 - 52x^3 + 201x^2 - 342x + 216 = 0, \]

and the operation being performed, we find that they have a common divisor \( x^3 - 8x^2 + 21x - 18 \), which is of the third degree, and consequently may have several factors. Let us therefore try whether the last equation and the following,

\[ 20x^3 - 156x^2 + 402x - 342 = 0, \]

which is derived from it, as directed in § 228, have any common divisor; and, by proceeding as before, we find Equations find that they admit of this divisor \( x - 3 \), which is also a factor of the last divisor \( x^3 - 8x^2 + 21x - 18 \), and therefore the product of remaining factors is immediately found by division to be \( x^2 - 5x + 6 \), which is evidently resolvable into \( x - 2 \) and \( x - 3 \).

Thus it appears, upon the whole, that the common divisor of the original equation, and that which is immediately derived from it, is \( (x - 2)(x - 3)^2 \); and that the common divisor of the second and third equations is \( x - 3 \). Hence it follows that the proposed equation has \( (x - 2)^2 \) for one factor, and \( (x - 3)^3 \) for another factor; so that the equation itself may be expressed thus, \( (x - 2)^2(x - 3)^3 = 0 \), and the truth of this conclusion may be easily verified by multiplication.

**Sect. XV. Resolution of Equations whose Roots are Rational.**

235. It has been shown in § 169 that the last term of any equation is always the product of its roots taken with contrary signs: Hence it follows that when the roots are rational they may be discovered by the following rule.

Bring all the terms of the equation to one side; find all the divisors of the last term, and substitute them successively for the unknown quantity in the equation. Then each divisor, which produces a result equal to 0, is a root of the proposed equation.

**Ex. 1.** Let \( x^3 - 4x^2 - 7x + 10 = 0 \) be the proposed equation.

Then, the divisors of 10 the last term are 1, 2, 5, 10, each of which may be taken either positively, or negatively, and these being substituted successively for \( x \), we obtain the following results.

| \( x \) | \( x^3 - 4x^2 - 7x + 10 \) | |-------|--------------------------| | 1 | 0 | | 2 | 0 | | 5 | 0 | | 10 | 0 |

Here the divisors which produce results equal to 0 are \( \pm 1, \pm 2, \pm 5, \pm 10 \), and therefore these numbers are the three roots of the proposed equation.

236. When the number of divisors to be tried happens to be considerable, it will be convenient to transform the proposed equation into another, in which the last term has fewer divisors. This may, in general, be done by forming an equation, the roots of which are greater or less than those of the proposed equation by some determinate quantity, as in the following example:

**Ex. 2.** Let \( y^4 - 4y^3 - 8y + 32 = 0 \) be proposed.

Here the divisors to be tried are 1, 2, 4, 8, 16, 32, each taken either positively or negatively; but to prevent the trouble of too many substitutions, let us transform the equation, by putting \( x + 1 \) for \( y \).

Then \( y^4 = x^4 + 4x^3 + 6x^2 + 4x + 1 \)

\[ \begin{align*} -4y^3 &= -4x^3 - 12x^2 - 12x - 4 \\ -8y &= -8x - 8 \\ +32 &= +32 \end{align*} \]

Therefore \( x^4 - 6x^3 - 16x + 21 = 0 \)

This is the transformed equation, and the divisors of the last term are \( \pm 1, \pm 1, \pm 3, \pm 7, \pm 7 \). These being put successively for \( x \), we get \( \pm 1 \) and \( \pm 3 \) for two roots of the equation; and as to the two remaining roots, it is easy to see that they must be imaginary. They may, however, be readily exhibited by considering that the equation \( x^4 - 6x^3 - 16x + 21 = 0 \) is divisible by the product of the two factors \( x - 1 \) and \( x - 3 \), and therefore may be reduced to a quadratic. Accordingly, by performing the division and putting the quotient equal 0, we have this equation,

\[ x^2 + 4x + 7 + 0, \]

the roots of which are the imaginary quantities \( -2 + \sqrt{-3} \) and \( -2 - \sqrt{-3} \); so that since \( y = x + 1 \), the roots of the equation \( y^4 - 4y^3 - 8y + 32 = 0 \) are these,

\( y = -2, y = -4, y = -1 + \sqrt{-3}, y = -1 - \sqrt{-3} \).

If this literal equation were proposed,

\[ x^3 - (3a + b)x^2 + (2a^2 + 3ab)x - 2a^2b = 0, \]

by proceeding as before, we should find \( x = a, x = 2a, x = b \) for the roots.

237. To avoid the trouble of trying all the divisors of the last term, a rule may be investigated for restricting the number to very narrow limits as follows:

Suppose that the cubic equation \( x^3 + px^2 + qx + r = 0 \) is to be resolved. Let it be transformed into another, the roots of which are less than those of the proposed equation by unity: this may be done by assuming \( y = x - 1 \), and the last term of the transformed equation will be \( 1 + p + q + r \). Again, by assuming \( y = x + 1 \) another equation will be formed whose roots exceed those of the proposed equation by unity, and the last term of this other transformed equation will be \( -1 + p + q + r \). And here it is to be observed, that these two quantities \( 1 + p + q + r \) and \( -1 + p + q + r \) are formed from the proposed equation \( x^3 + px^2 + qx + r \) by substituting in it successively \( +1 \) and \( -1 \) for \( x \).

Now the values of \( x \) are some of the divisors of \( r \), which is the term left in the proposed equation, when \( x \) is supposed \( = 0 \); and the values of the \( y \)'s are some of the divisors of \( 1 + p + q + r \) and \( -1 + p + q + r \) respectively; and these values are in arithmetical progression, increasing by the common difference unity; because \( x = 1, x = 2 \) are in that progression; and it is obvious, that the same reasoning will apply to an equation of any degree whatever. Hence the following rule.

Substitute in place of the unknown quantity, successively, three or more terms of the progression 1, 0, -1, &c., and find all the divisors of the sums that result; then take out all the arithmetical progressions that can be found among these divisors, whose common difference is 1, and the values of \( x \) will be among these terms of the progressions, which are the divisors of the result arising from the substitution of \( x = 0 \). When the series increases, the roots will be positive; and, when it decreases, they will be negative.

**Ex. 1.** Let it be required to find a root of the equation \( x^3 - x^2 - 10x + 6 = 0 \). The operation.

| Substit. | Result | Divisors | Ar. Pro. | |----------|--------|----------|---------| | $x = +1$ | $x^3 - x^2 - 10x + 6 = \{ -4, 1, 2, 4; 4$ | | $x = 0$ | | | | | $x = -1$ | | | |

In this example there is only one progression, 4, 3, 2, the term of which opposite to the supposition of $x = 0$ being 3, and the series decreasing, we try if $-3$ substituted for $x$ makes the equation vanish, and as it succeeds, it follows that $-3$ is one of its roots. To find the remaining roots, if $x^3 - x^2 - 10x + 6$ be divided by $x + 3$, and the quotient $x^2 - 4x + 2$ put $= 0$, they will appear to be $2 + \sqrt{2}$, and $2 - \sqrt{2}$.

Ex. 2. Let the proposed equation be

$$x^3 + x^2 - 29x^2 - 9x + 18 = 0.$$

To find its roots.

Here there are four progressions, two increasing and two decreasing; hence by taking their terms, which are opposite to the supposition of $x = 0$, we have these four numbers to be tried as roots of the equation $+3$, $+4$, $-3$, $-5$, all of which are found to succeed.

238. If any of the coefficients of the proposed equation be a fraction, the equation may be transformed into another, having the coefficient of the highest power unity, and those of the remaining terms integers by §189, and the roots of the transformed equation being found, those of the proposed equation may be easily derived from them.

For example, if the proposed equation be

$$x^3 - \frac{7}{4}x^2 + \frac{3}{4}x - 6 = 0.$$

Let us assume $x = \frac{y}{4}$, thus the equation is transformed to

$$\frac{y^3}{64} - \frac{7y^2}{64} + \frac{3y}{16} - 6 = 0,$$

Or $y^3 - 7y^2 + 14y - 384 = 0$,

one root of which is $y = 3$; hence $x = \frac{y}{4} = \frac{3}{4}$.

The proposed equation being now divided by $x - \frac{3}{4}$ is reduced to this quadratic $x^2 - x + 8 = 0$, the roots of which are both impossible.

239. When the coefficients of an equation are integers, and that of the highest power of the unknown quantity unity, if its roots are not found among the divisors of the last term, we may be certain that, whether the equation be pure or affected, its roots cannot be exactly expressed either by whole numbers or rational fractions. This may be demonstrated by means of the following proposition. If a prime number $P$ be a divisor of the product of two numbers $A$ and $B$, it will also be a divisor of at least one of the numbers.

240. Let us suppose that it does not divide $B$, and that $B$ is greater than $P$; then, putting $q$ for the greatest number of times that $P$ can be had in $B$, and $B'$ for the remainder, we have $\frac{B}{P} = q + \frac{B'}{P}$, and therefore

$$\frac{AB}{P} = qA + \frac{AB'}{P}.$$

Hence it appears, that if $P$ be a divisor of $AB$, it is also a divisor of $AB'$. Now $B'$ is less than $P$, for it is the remainder which is found in dividing $B$ by $P$; therefore, seeing we cannot divide $B'$ by $P$, let $P$ be divided by $B'$, and $q'$ put for the quotient, also $B''$ for the remainder; again, let $P$ be divided by $B''$, and $q''$ put for the quotient, and $B'''$ for the remainder, and so on; and as $P$ is supposed to be a prime number, it is evident that this series of operations may be continued till the remainder be found equal to unity, which will at last be the case; for the divisors are the successive remainders of the divisions, and therefore each is less than the divisor which preceded it. By performing these operations we obtain the following series of equations,

$$P = q'B' + B''$$

$$P = q''B'' + B''', \ldots$$

and therefore

$$B' = \frac{P - B''}{q'},$$

$$B'' = \frac{P - B'''}{q''},$$

$$\ldots$$

Hence we have $AB' = \frac{AP - AB''}{q'}$, and

$$\frac{q'AB'}{P} = \frac{AP - AB''}{P} = A - \frac{AB''}{P}.$$

Now if $AB$ be divisible by $P$, we have shown that $AB'$, and consequently $q'AB'$ is divisible by $P$; therefore, from the last equation, it appears that $AB''$ must also be divisible by $P$.

Again, from the preceding series of equations, we have $AB'' = \frac{AP - AB'''}{q''}$, and therefore

$$\frac{q''AB''}{P} = \frac{AP - AB'''}{P} = A - \frac{AB'''}{P};$$

hence we conclude that $AB'''$ is also divisible by $P$.

Proceeding in this manner, and observing that the series of quantities $B', B'', B''', \ldots$ continually decrease till one of them $= 1$, it is evident that we shall at last come to a product of this form $AX1$, which must Equations must be divisible by \( P \), and hence the truth of the proposition is manifest.

241. It follows from this proposition, that if the prime number \( P \), which we have supposed not to be a divisor of \( B \), is at the same time not a divisor of \( A \), it cannot be a divisor of \( AB \) the product of \( A \) and \( B \).

242. Let \( \frac{b}{a} \) be a fraction in its lowest terms, then the numbers \( a \) and \( b \) have no common divisor; but from what has been just now shewn, it appears, that if a prime number be not a divisor of \( a \) it cannot be a divisor of \( a^2 \) or \( a^3 \); and in like manner, that if a prime number is not a divisor of \( b \), it cannot be a divisor of \( b^2 \) or \( b^3 \); therefore, it is evident that \( a^2 \) and \( b^2 \) have no common divisor, and thus the fraction \( \frac{b^2}{a^2} \) is also in its lowest terms.

Hence it follows that the square of any fractional quantity is still a fraction, and cannot possibly be a whole number; and, on the contrary, that the square root of a whole number cannot possibly be a fraction; so that all such whole numbers as are not perfect squares can neither have their roots expressed by integers nor by fractions.

243. Since that if a prime number is not a divisor of \( a \), it is also not a divisor of \( a^2 \); therefore if it is not a divisor of \( a \), it cannot be a divisor of \( a \times a \) or \( a^2 \), § 241, and by reasoning in this way, it is obvious that if a prime number is not a divisor of \( a \), it cannot be a divisor of \( a^2 \); also, that if it is not a divisor of \( b \), it cannot be a divisor of \( b^2 \), therefore if \( \frac{b}{a} \) is a fraction in its lowest terms, \( \frac{b^2}{a^2} \) is also a fraction in its lowest terms; so that any power whatever of a fraction is also a fraction, and on the contrary, any root of a whole number is also a whole number. Hence it follows that if the root of a whole number is not expressible by an integer, such root cannot be expressed by a fraction, but is therefore irrational or incommensurable.

244. Let us next suppose that

\[ x^n + Px^{n-1} + Qx^{n-2} \ldots + Tx + U = 0 \]

is any equation whatever, in which \( P, Q, \ldots \) denote integer numbers; then if its roots are not integers they cannot possibly be rational fractions. For, if possible, let us suppose \( x = \frac{b}{a} \), a fraction reduced to its lowest terms, then, by substitution,

\[ \frac{a^n}{b^n} + \frac{a^{n-1}}{b^{n-1}} + \frac{Q}{b^{n-2}} \ldots + \frac{T}{b} + U = 0, \]

and, reducing all the terms to a common denominator,

\[ a^n + Pa^{n-1}b + Qa^{n-2}b \ldots + Tab^{n-1} + Ub^n = 0, \]

which equation may also be expressed thus,

\[ a^n + b(Pa^{n-1} + Qa^{n-2}b \ldots + Tab^{n-1} + Ub^n) = 0, \]

where the equation consists of two parts, one of which is divisible by \( b \). But by hypothesis \( a \) and \( b \) have no common measure, therefore \( a^n \) is not divisible by \( b \), § 243; hence it is evident that the two parts of the equation cannot destroy each other as they ought to do; therefore \( x \) cannot possibly be a fraction.

Sect. XVI. Resolution of Equations by Approximation.

245. When the roots of an equation cannot be accurately expressed by rational numbers, it is necessary to have recourse to the methods of approximation, and by these we can always determine the numerical values of the roots to as great a degree of accuracy as we please.

246. The application of the methods of approximation is rendered easy by means of the following principles:

If two numbers, either whole or fractional, be found, which, when substituted for the unknown quantity in any equation, produce results with contrary signs; we may conclude that at least one root of the proposed equation is between those numbers, and is consequently real.

Let the proposed equation be

\[ x^3 - 5x^2 + 10x - 15 = 0, \]

which, by collecting the positive terms into one sum, and the negative into another, may also be expressed thus,

\[ x^3 + 10x - (5x^2 + 15) = 0; \]

then to determine a root of the equation, we must find such a number as when substituted for \( x \) will render

\[ x^3 + 10x = 5x^2 + 15. \]

Let us suppose \( x \) to have every degree of magnitude from 0 upwards in the scale of number, then \( x^3 + 10x \) and \( 5x^2 + 15 \) will both continually increase, but with different degrees of quickness, as appears from the following table.

| Successive values of \( x \) | 0, 1, 2, 3, 4, 5, 6, &c. | |-----------------------------|-------------------------| | \( x^3 + 10x \) | 0, 11, 28, 57, 104, 175, 276, &c. | | \( 5x^2 + 15 \) | 15, 35, 60, 95, 140, 195, &c. |

By inspecting this table, it appears that while \( x \) increases from 0 to a certain numerical value, which exceeds 3, the positive part of the equation, or \( x^3 + 10x \), is always less than the negative part, or \( 5x^2 + 15 \); so that the expression

\[ x^3 + 10x - (5x^2 + 15), \]

or \( x^3 - 5x^2 + 10x - 15 \)

must necessarily be negative.

It also appears that when \( x \) has increased beyond that numerical value, and which is evidently less than 4, the positive part of the equation, instead of being less than the negative part, is now greater, and therefore the expression

\[ x^3 - 5x^2 + 10x - 15 \]

is changed from a negative to a positive quantity.

247. Hence we may conclude that there is some real and determinate value of \( x \), which is greater than 3, but less than 4, and which will render the positive and negative parts of the equation equal to one another; therefore that value of \( x \) must be a root of the proposed equation; and as what has been just now shewn in a particular case will readily apply to any equation whatever, the truth of what has been asserted at § 246 is obvious.

248. Two 248. Two limits, between which all the roots of any equation are contained, may be determined by the following proposition.

Let $N$ be the greatest negative coefficient in any equation. Change the signs of the terms taken alternately, beginning with the second, and let $N'$ be the greatest negative coefficient after the signs are so changed. The positive roots of the equation are contained between $0$ and $N + 1$, and the negative roots between $0$ and $-N' - 1$.

Suppose the equation to be

$$x^4 - px^3 + qx^2 - rx - s = 0,$$

which may be also expressed thus:

$$x^4 \left(1 - \frac{p}{x} + \frac{q}{x^2} - \frac{r}{x^3} - \frac{s}{x^4}\right) = 0.$$

Then, whatever be the values of the coefficients $p, q, r, \ldots$, it is evident that $x$ may be taken to great as to render each of the quantities $\frac{p}{x}, \frac{q}{x^2}, \frac{r}{x^3}, \frac{s}{x^4}$ as small as we please, and therefore their sum, or $-\frac{p}{x} + \frac{q}{x^2} - \frac{r}{x^3} - \frac{s}{x^4}$, less than $1$; but in that case the quantity

$$x^4 \left(1 - \frac{p}{x} + \frac{q}{x^2} - \frac{r}{x^3} + \frac{s}{x^4}\right),$$

or $x^4 - px^3 + qx^2 - rx + s$

will be positive, and such, that the first term $x^4$ is greater than the sum of all the remaining terms; therefore also $x^4 - px^3$ the sum of the positive terms will be much greater than $px^3 - rx + s$ the sum of the negative terms alone.

Hence it follows, that if a number be found, which when substituted for $x$, renders the expression $x^4 - px^3 + qx^2 - rx - s$ positive, and which is also such that every greater number has the same property, that number will exceed the greatest positive root of the equation.

Now, if we suppose $N$ to be the greatest negative coefficient, it is evident that the positive part of the equation, or $x^4 + qx^2$, is greater than $px^3 + rx + s$, provided that $x^4$ is greater than $N x^3 + N x^2 + N x + N$, or $N (x^3 + x^2 + x + 1)$; but $x^3 + x^2 + x + 1 = \frac{x^4 - 1}{x - 1}$, therefore a positive result will be obtained, if for $x$ there be substituted a number such that $x^4 > N(x^3 + x^2 + x + 1)$, or $x^4 > Nx^3 + Nx^2 + Nx + N$. Now this last condition will evidently be fulfilled if we take $x^4 > Nx^3$, and from this equation we find $x = N + 1$; but it farther appears that the same condition will also be fulfilled as often as $x^5 > Nx^4$, or $x^4 > N$, that is, $x > N + 1$, therefore $N + 1$ must be a limit to the greatest positive root of the proposed equation, as was to be shown.

249. If $-y$ be substituted for $+x$, the equation $x^4 - px^3 + qx^2 - rx - s = 0$ will be transformed into $y^4 + py^3 + qy^2 + ry - s = 0$; which equation differs from the former only in the signs of the second, fourth, &c. terms; and as the positive roots of this last equation are the same as the negative roots of the proposed equation, it is evident that their limit must be such as has been assigned.

250. From the two preceding propositions it will not be difficult to discover, by means of a few trials, the nearest integers to the roots of any proposed numerical equation, and those being found, we may approximate to the roots continually, as in the following example:

$$x^4 - 4x^3 - 3x + 27 = 0.$$

Here the greatest negative coefficient being $4$, it follows, § 248, that the greatest positive root is less than $5$. If $-y$ be substituted for $x$, the equation is transformed to

$$y^4 + 4y^3 + 3y + 27 = 0,$$

an equation having all its terms positive; therefore, it can have no positive roots, and consequently the proposed equation can have no negative roots; its real roots must therefore be contained between $0$ and $+5$.

251. To determine the limits of each root in particular, let $0, 1, 2, 3, 4$ be substituted successively for $x$; thus we obtain the following corresponding results.

| Substitutions for $x$ | Results | |----------------------|---------| | 0 | +27 | | 1 | +21 | | 2 | +15 | | 3 | +9 | | 4 | +15 |

Hence it appears that the equation has two real roots, one between $2$ and $3$, and another between $3$ and $4$.

252. That we may approximate to the first root, let us suppose $x = 2 + y$, where $y$ is a fraction less than unity, and therefore its second and higher powers but small in comparison to its first power; hence, in finding an approximate value of $y$, they may be rejected. Thus we have

$$x^4 = 16 + 32y, \text{ &c.}$$

$$-4x^3 = -32 - 48y, \text{ &c.}$$

$$-3x = -6 - 3y,$$

$$+27 = +27$$

Hence $0 = 5 - 19y$ nearly,

and $y = \frac{5}{19} = .26$; therefore, for a first approximation, we have $x = 2.26$.

Let us next suppose $x = 2.26 + y'$, then, rejecting as before the second and higher powers of $y'$ on account of their smallness, we have

$$x^4 = 26.087 + 46.172y', \text{ &c.}$$

$$-4x^3 = -46.172 - 61.201y', \text{ &c.}$$

$$-3x = -6.780 - 3y',$$

$$+27 = +27$$

$$0 = .135 - 18.119y' \text{ nearly.}$$

Hence $y' = \frac{.135}{18.119} = .0075$, and $x = 2.26 + y' = 2.2675$.

This value of $x$ is true to the last figure, but a more accurate value may be obtained by supposing $x = 2.675 + y''$, and finding the value of $y''$ in the same manner as we have already found those of $y'$ and $y$; and thus

* The sign $\geq$ denotes that the quantities between which it is placed are unequal. Thus $a \geq b$, signifies that $a$ is greater than $b$, and $a \leq c$, that $a$ is less than $c$. the approximation may be continued till any required degree of accuracy be obtained.

The second root of the equation, which we have already found to be between 3 and 4, may be investigated in the same manner as the first, and will appear to be 3.6797, the approximation being carried on to the fourth figure of the decimal, in determining each root.

253. In the preceding example we have shown how to approximate to the roots of an affected equation, but the same method will also apply to pure equations.

For example, let it be required to determine \( x \) from this equation \( x^2 = 2 \).

Because \( x \) is greater than 1, and less than 2, but nearer to the former number than to the latter, let us assume \( x = 1 + y \), then, rejecting the powers of \( y \) which exceed the first, we have \( x^2 = 1 + 3y \), and therefore \( 2 = 1 + 3y \), and \( y = \frac{1}{3} \cdot 3 \) nearly; hence \( x = 1.3 \) nearly.

Let us next assume \( x = 1.3 + y' \); then, proceeding as before, we find \( 2 = 2.197 + 5.07y' \), hence \( y' = \frac{1.97}{5.07} = -0.39 \), and \( x = 1.3 - 0.39 = 1.26 \) nearly.

To find a still nearer approximation, let us suppose \( x = 1.26 + y'' \), then from this assumption we find \( y'' = -0.00079 \), and therefore \( x = 1.25921 \), which value is true to the last figure.

254. By assuming an equation of any order with literal coefficients, a general formula may be investigated, for approximating to the roots of equations belonging to that particular order.

Let us take for an example the cubic equation,

\[ x^3 + px^2 + qx + r = 0, \]

and suppose that \( x = a + y \), where \( a \) is nearly equal to \( x \), and \( y \) is a small fraction. Then, by substituting \( a + y \) for \( x \) in the proposed equation, and rejecting the powers of \( y \) which exceed the first, on account of their smallness, we have

\[ a^3 + pa^2 + qa + r + (3a^2 + 3pa + q)y = 0. \]

Hence \( y = \frac{-a^3 - pa^2 - qa - r}{3a^2 + 2pa + q} \),

and \( x = a + \frac{a^3 + pa^2 + qa + r}{3a^2 + 2pa + q} = \frac{2a^3 + pa^2 - r}{3a^2 + 2pa + q} \).

255. Let it be required to approximate to a root of the cubic equation \( x^3 + 2x^2 + 3x - 50 = 0 \). Here \( p = 2 \), \( q = 3 \), and \( r = -50 \); and by trials it appears that \( x \) is between 2 and 3, but nearest the latter number; therefore, for the first approximation, \( a \) may be supposed = 3, hence we find

\[ x = \frac{2a^3 + pa^2 - r}{3a^2 + 2pa + q} = \frac{128}{44} = 2.9. \]

By substituting \( \frac{128}{44} \) for \( a \) in the formula, and proceeding as before, a value of \( x \) would be found more exact than the former, and so on we may go as far as we please.

256. The method we have hitherto employed for approximating to the roots of equations, is known by the name of the method of successive substitutions, and was first proposed by Newton. It has been since improved by Lagrange, who has given it a form which has the advantage of shewing the progress made in the approximation by each operation. This improved form we now proceed to explain.

Let \( a \) denote the whole number, next less to the root sought, and \( \frac{1}{y} \) a fraction, which, when added to \( a \), completes the root, then \( x = a + \frac{1}{y} \). If this value of \( x \) be substituted in the proposed equation, a new equation involving \( y \) will be had, which, when cleared of fractions, will necessarily have a root greater than unity.

Let \( b \) be the whole number which is next less than that root, then, for a first approximation, we have \( x = a + \frac{1}{b} \). But \( b \) being only an approximate value of \( y \), in the same manner as \( a \) is an approximate value of \( x \), we may suppose \( y = b + \frac{1}{y'} \), then, by substituting \( b + \frac{1}{y'} \) for \( y \), we shall have a new equation, involving only \( y' \), which must be greater than unity; putting therefore \( b' \) to denote the next whole number less than the root of the equation involving \( y' \), we have \( y = b + \frac{1}{b'} = \frac{bb' + 1}{b'} \), and substituting this value in that of \( x \), the result is

\[ x = a + \frac{b'}{bb' + 1} \]

for a second approximate value of \( x \).

To find a third value we may take \( y' = b' + \frac{1}{y''} \), then if \( b'' \) denote the next whole number less than \( y'' \), we have \( a' = b' + \frac{1}{b''} = \frac{bb'' + b'}{b''} \), whence

\[ y = b + \frac{b''}{bb'' + 1} = \frac{bb'' + b'}{bb'' + 1}, \]

and

\[ x = a + \frac{b''}{bb'' + b'} + \frac{b'}{bb' + 1} \]

and so on to obtain more accurate approximations.

257. We shall apply this method to the following example,

\[ x^3 - 7x + 7 = 0. \]

Here the positive roots must be between 0 and 8, let us therefore substitute successively, 0, 1, 2, ... to 8, and we obtain results as follow:

| Substitutions | Results | |---------------|---------| | 0 | +7 | | 1 | +1 | | 2 | +1 | | 3 | +13 | | 4 | +43 | | 5 | +97 | | 6 | +181 | | 7 | +301 | | 8 | +463 |

but as these results have all the same sign, nothing can be concluded respecting the magnitude of the roots from that circumstance alone. It is, however, observable, that while \( x \) increases from 0 to 1 the results decrease; but that whatever succeeding magnitudes \( x \) has greater than 2, the results increase; we may therefore reasonably conclude that if the equation have any positive roots they must be between 1 and 2. Accordingly, by substituting 1.2, 1.4, 1.6, and 1.8 successively for \( x \), we find these results +.328, -.056, -.104, +.232, and... Approximately, and as there are here two changes of the signs, it follows that the equation has two positive roots, one between 1.2 and 1.4, and another between 1.6 and 1.8.

Hence it appears, that to find either value of \( x \), we may assume \( x = 1 + \frac{1}{y} \); thus, by substitution, we have

\[ y^3 - 4y^2 + 3y + 1 = 0. \]

The limit of the positive root of this last equation is 5, and by substituting 1, 2, 3, 4, successively for \( y \), it will be found to have two, one of which is between 1 and 2, and the other between 2 and 3. Therefore, for a first approximation, we have

\[ x = 1 + \frac{1}{y}, \quad x = 1 + \frac{1}{y}, \quad \text{that is, } x = 2, \quad x = 3. \]

To approach nearer to the first value of \( y \), let us take

\[ y = 1 + \frac{1}{y}, \]

and therefore

\[ y^3 - 2y^2 - y + 1 = 0. \]

This last equation will be found to have only one real root between 2 and 3, from which it appears, that \( y = 1 + \frac{1}{y} = \frac{3}{2} \), and \( x = 1 + \frac{1}{y} = \frac{5}{2} \).

Let us next suppose \( y' = 2 + \frac{1}{y'} \); hence we find

\[ y'^3 - 3y'^2 - 4y' - 1 = 0, \]

and from this equation \( y' \) is found to be between 4 and 5. Taking the least limit, we have

\[ y' = 2 + \frac{1}{y'} = \frac{5}{2}, \quad y = 1 + \frac{1}{y} = \frac{3}{2}, \quad x = 1 + \frac{1}{y} = \frac{5}{2}. \]

It is easy to continue this process by assuming \( y'' = 4 + \frac{1}{y''} \), and so on, as far as may be judged necessary.

We return to the second value of \( x \), which was found \( \frac{3}{2} \) by the first approximation, and which corresponds to \( y = 2 \). Putting \( y = 2 + \frac{1}{y} \), and substituting this value in the equation \( y^3 - 4y^2 + 3y + 1 = 0 \), which was formerly found, we get

\[ y^3 + y^2 - 2y - 1 = 0, \]

this equation, as well as the corresponding equation employed in determining the other value of \( x \), has only one root greater than unity, which root being between 1 and 2, let us take \( y = 1 \), we thence find

\[ y = 3, \quad x = 1 + \frac{1}{y} = \frac{4}{3}. \]

Put \( y = 1 + \frac{1}{y} \), and we thence find by substitution

\[ y'^3 - 3y'^2 - 4y' - 1 = 0, \]

an equation which gives \( y' \) between 4 and 5; hence, as before,

\[ y' = \frac{5}{2}, \quad y = \frac{3}{2}, \quad x = \frac{5}{2}. \]

That we may proceed in the approximation, we have only to suppose \( y'' = 4 + \frac{1}{y''} \), and so on. The equation

\[ x^3 - 7x + 7 \text{ has also a negative root between } -3 \text{ and } \]

Sect. XVII. Of Infinite Series.

259. The resolving of any proposed quantity into a series, is a problem of considerable importance in the application of algebra to the higher branches of the mathematics; and there are various methods by which it may be performed, suited to the particular forms of the quantities which may become the subject of consideration.

260. Any rational fraction may be resolved into a series, by the common operation of algebraic division, as in the following examples:

Ex. 1. To change \( \frac{ax}{a-x} \) into an infinite series.

\[ \begin{align*} \frac{ax}{a-x} &= ax + \frac{x^2}{a} + \frac{x^3}{a^2} + \frac{x^4}{a^3} + \ldots \\ &= x + \frac{x^2}{a} + \frac{x^3}{a^2} + \frac{x^4}{a^3} + \ldots \end{align*} \]

Thus it appears, that

\[ \frac{ax}{a-x} = x + \frac{x^2}{a} + \frac{x^3}{a^2} + \frac{x^4}{a^3} + \ldots \]

Here the law of the series being evident, the terms may be continued at pleasure.

Ex. 2. It is required to convert \( \frac{a^2}{(a+x)^2} \) into an infinite series. Infinite Series

\[a^2 + 2ax + x^2) = \frac{1}{a} - \frac{2x}{a} + \frac{3x^2}{a^2} - \frac{4x^3}{a^3} + \ldots\]

\[= \frac{a^2 + 2ax + x^2}{-2ax - x^2}\]

\[= \frac{-2ax - 4x^2}{a^2 + 3x^2 + \frac{2x^3}{a}}\]

\[+ \frac{3x^3 + \frac{6x^3}{a} + \frac{3x^3}{a^2}}{\frac{4x^3}{a} - \frac{3x^4}{a^2}}\]

Therefore \(\frac{a^2}{(a+x)^2} = \frac{1}{a} - \frac{2x}{a} + \frac{3x^2}{a^2} - \frac{4x^3}{a^3} + \frac{5x^4}{a^4} + \ldots\)

the law of continuation being evident.

261. A second method by which algebraic quantities, whether rational or irrational, may be converted into series, and which is also of very extensive use in the higher parts of the mathematics, consists in assuming a series with indeterminate coefficients, and having its terms proceeding according to the powers of some quantity contained in the proposed expression.

That we may explain this method, let us suppose that the fraction \(\frac{a^2}{a^2 + ax + x^2}\) is to be converted into a series proceeding by the powers of \(x\); we are therefore to assume

\[\frac{a^2}{a^2 + ax + x^2} = A + Bx + Cx^2 + Dx^3 + Ex^4 + \ldots\]

where \(A\) denotes these terms of the series into which \(x\) does not at all enter; \(Bx\) the terms which contain only the first power of \(x\); \(Cx^2\) the terms which contain only the second power, and so on. Let both sides of the equation be multiplied by \(a^2 + ax + x^2\), so as to take away the denominator of the fraction, and let the numerator \(a^2\) be transposed to the other side, so that the whole expression may be \(= 0\), then

\[\begin{align*} a^2A + a^2B &+ a^2C + a^2D + a^2E \\ -a^2 + aA &+ aB + aC + aD + aE \\ &+ A + B + C + D + E = 0 \end{align*}\]

Now the quantities \(A, B, C, D, \ldots\) being supposed to be entirely independent of any particular value of \(x\), it follows that the whole expression can only be \(= 0\), upon the supposition that the terms which multiply the same powers of \(x\) are separately \(= 0\); for if that were not the case, it would follow that \(x\) had a certain determinate relation to the quantities \(A, B, C, \ldots\) which is contrary to what we have all along supposed. To determine the quantities \(A, B, C, \ldots\) therefore, we have this series of equations,

\[\begin{align*} a^2A - a^2 &= 0 &\text{Hence } A = 1 \\ a^2B + aA &= 0 &\text{B} = -\frac{A}{a} = -\frac{1}{a} \\ a^2C + aB + A &= 0 &\text{C} = -\frac{B}{a} = -\frac{1}{a^2} \\ a^2D + aC + B &= 0 &\text{D} = -\frac{C}{a} = -\frac{1}{a^3} \\ a^2E + aD + C &= 0 &\text{E} = -\frac{D}{a} = -\frac{1}{a^4} \\ &\ldots \end{align*}\]

Here the law of relation which takes place among the quantities \(A, B, C, D, \ldots\) is evident, viz. that if \(P, Q, R, \ldots\) denote any three coefficients which immediately follow each other,

\[a^2R + aQ + P = 0;\]

and from this equation, by means of the coefficients already determined, we find \(F = 0, G = \frac{1}{a^6}, H = -\frac{1}{a^7}, K = 0, \ldots\)

Therefore, resuming the assumed equation, and substituting for \(A, B, C, \ldots\) their respective values, we have

\[\frac{a^2}{a^2 + ax + x^2} = \frac{1}{a} - \frac{x}{a} + \frac{x^2}{a^2} - \frac{x^3}{a^3} + \frac{x^4}{a^4} - \frac{x^5}{a^5} + \ldots\]

262. As a second example of the method of indeter-

This method of resolving a quantity into an infinite series will be found more expeditious than any other, as often as the operations of division and evolution are to be performed at the same time, as in these expressions:

\[\frac{1}{\sqrt{a^2 + x^2}}, \text{or } \frac{\sqrt{a^2 - x^2}}{3\sqrt{a^2 + x^3}}.\]

263. The binomial theorem affords a third method of resolving quantities into series; but before we explain this method, it will be proper to shew how the theorem itself may be investigated.

Let \(a + x\) be any binomial quantity, which is to be raised raided to a power denoted by \( \frac{m}{n} \), where \( m \) and \( n \) denote any numbers either positive or negative. Or because \( a+x = a(1+\frac{x}{a}) \), if we put \( \frac{x}{a} = y \), then \( (a+x)^{\frac{m}{n}} = a^{\frac{m}{n}} \times (1+y)^{\frac{m}{n}} \); therefore instead of \( a+x \) we may consider \( 1+y \), which is somewhat more simple in its form.

264. By considering some of the first powers of \( 1+x \), viz.

\[ \begin{align*} (1+x)^1 &= 1 + x \\ (1+x)^2 &= 1 + 2x + x^2 \\ (1+x)^3 &= 1 + 3x + 3x^2 + x^3 \\ (1+x)^4 &= 1 + 4x + 6x^2 + 4x^3 + x^4 \\ &\vdots \end{align*} \]

it appears that the powers of \( 1+x \) have this form

\[ 1 + Ax + Bx^2 + Cx^3 + Dx^4 + \ldots \]

where the coefficients \( A, B, C, D, \ldots \) are numbers which are altogether independent of any particular value of \( x \). It also appears that the series cannot contain any negative power of \( x \); for if any of its terms had this form \( \frac{Q}{x^n} \), then the supposition of \( x=0 \) would render that term indefinitely great, whereas the whole series ought in that case to be reduced to unity.

265. Let us therefore assume

\[ (1+y)^{\frac{m}{n}} = 1 + Ay + By^2 + Cy^3 + Dy^4 + \ldots \]

Then we have also

\[ (1+x)^{\frac{m}{n}} = 1 + Ax + Bx^2 + Cx^3 + Dx^4 + \ldots \]

Let us put \( (1+y)^{\frac{m}{n}} = u \), \( (1+x)^{\frac{m}{n}} = v \), and therefore \( u^m - v^m = A(y-z) + B(y^2-z^2) + C(y^3-z^3) + D(y^4-z^4) + \ldots \)

But from the equation originally assumed we have

\[ \frac{m}{n}(1+y)^{\frac{m}{n}} = \frac{m}{n} + \frac{m}{n}Ay + \frac{m}{n}By^2 + \frac{m}{n}Cy^3 + \frac{m}{n}Dy^4 + \ldots \]

therefore

\[ \frac{m}{n} + \frac{m}{n}Ay + \frac{m}{n}By^2 + \frac{m}{n}Cy^3 + \frac{m}{n}Dy^4 + \ldots \]

And as the coefficients of the terms have no connexion with any particular value of \( y \), it follows, that the coefficient of any power of \( y \) on the one side of the equation must be equal to the coefficient of the same power of \( y \) on the other side. Therefore, to determine \( A, B, C, \ldots \) we have the following series of equations:

\[ \begin{align*} A &= \\ B &= \\ C &= \\ D &= \\ &\vdots \end{align*} \] \[ A = \frac{m}{n} \] Hence \( A = \frac{m}{n} \)

\[ 2B + A = \frac{m}{n} A \]

\[ 3C + 2B = \frac{m}{n} B \]

\[ 4D + 3C = \frac{m}{n} C \]

\[ 5E + 4D = \frac{m}{n} D \]

\[ \text{&c.} \]

Or, substituting for \( A, B, C, \) &c. their values as determined from the preceding equations:

\[ A = \frac{m}{n} \]

\[ B = \frac{m(m-n)}{1 \cdot 2 \cdot n^2} \]

\[ C = \frac{m(m-n)(m-2n)}{1 \cdot 2 \cdot 3 \cdot n^3} \]

\[ D = \frac{m(m-n)(m-2n)(m-3n)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot n^4} \]

\[ E = \frac{m(m-n)(m-2n)(m-3n)(m-4n)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot n^5} \]

\[ \text{&c.} \]

267. Referring now the assumed equation

\[ (1+y)^{\frac{m}{n}} = 1 + Ay + By^2 + Cy^3 + \text{&c.} \]

and observing that \( \frac{x}{y} = y \) and \( (a+x)^{\frac{m}{n}} = a^{\frac{m}{n}} (1 + \frac{mx}{na} + \frac{\Lambda(m-n)x^2}{2n} + \frac{B(m-2n)}{3n} x^3 + \frac{C(m-3n)x^4}{4n} + \text{&c.}) \)

we have

\[ (a+x)^{\frac{m}{n}} = a^{\frac{m}{n}} + \frac{m}{n} a^{\frac{m-1}{n}} x + \frac{m(m-n)}{1 \cdot 2 \cdot n^2} a^{\frac{m-2}{n}} x^2 + \frac{m(m-n)(m-2n)}{1 \cdot 2 \cdot 3 \cdot n^3} a^{\frac{m-3}{n}} x^3 + \frac{m(m-n)(m-2n)(m-3n)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot n^4} a^{\frac{m-4}{n}} x^4 + \text{&c.} \]

where \( A, B, C, \) &c. denote the coefficients of the preceding terms, or

\[ (a+x)^{\frac{m}{n}} = a^{\frac{m}{n}} + \frac{m}{n} a^{\frac{m-1}{n}} x + \frac{m(m-n)}{1 \cdot 2 \cdot n^2} a^{\frac{m-2}{n}} x^2 + \frac{m(m-n)(m-2n)}{1 \cdot 2 \cdot 3 \cdot n^3} a^{\frac{m-3}{n}} x^3 + \frac{m(m-n)(m-2n)(m-3n)}{1 \cdot 2 \cdot 3 \cdot 4 \cdot n^4} a^{\frac{m-4}{n}} x^4 + \text{&c.} \]

and either of these formulae may be considered as a general theorem for raising a binomial quantity \( a+x \) to any power whatever.

268. In determining the value of the expression \( \frac{u^m - v^m}{u^n - v^n} \) when \( u = v \), it has been taken for granted that \( \frac{m}{n} \) is positive, but the same conclusion will be obtained when \( \frac{m}{n} \) is negative. For, changing \( +m \) into \( -m \), and observing that

\[ \frac{u^m - v^m}{u^n - v^n} = \frac{1}{u^m v^n} \left( \frac{v^m - u^m}{u^n - v^n} \right) \]

we have

\[ \frac{u^m - v^m}{u^n - v^n} = \frac{1}{u^m v^n} \left( \frac{v^m - u^m}{u^n - v^n} \right) \]

Now we have already found, that when \( u = v \), the fraction \( \frac{u^m - v^m}{u^n - v^n} \) becomes \( \frac{mu^{m-1}}{nu^{n-1}} \); therefore, in the same case,

\[ \frac{u^m - v^m}{u^n - v^n} = \frac{1}{u^m v^n} \times \frac{mu^{m-1}}{nu^{n-1}} = \frac{u^m - v^m}{nu^{n-1}} \]

and from this last expression we derive the same value for \( u^m \) or \( (1+y)^{\frac{m}{n}} \) as before, regard being had to the change of the sign of the exponent.

269. If we suppose \( m \) to be a positive integer, and \( n = 1 \), the series given in last article for the powers of \( a+x \) will always terminate, as appears also from the operation of involution; but if \( m \) be negative, or \( \frac{m}{n} \) a fraction, the series will consist of an indefinite number of terms. Examples of the application of the theorem have been already given upon the first supposition, when treating of involution; we now proceed to show how it is to be applied to the expansion of algebraic quantities into series upon either of the two last hypotheses.

270. Ex. I. It is required to express \( \frac{r^3}{(r+x)^3} \) by means of a series.

Because \( \frac{r}{r+x} = \frac{1}{1+\frac{x}{r}} \),

Therefore \( \frac{r^3}{(r+x)^3} = \left( \frac{1}{1+\frac{x}{r}} \right)^3 = \left( 1+\frac{x}{r} \right)^{-3} \)

Let \( \left( 1+\frac{x}{r} \right)^{-3} \) be compared with \( (a+x)^{\frac{m}{n}} \) and we have

\[ a = 1, \quad x = \frac{x}{r}, \quad m = -3, \quad n = 1. \]

Hence, by substituting these values of \( a, x, m, n \) in the first general formula of § 267, we have

\[ \begin{align*} \frac{r^3}{(r+x)^3} &= 1 - \frac{3x}{r} + \frac{3 \cdot 4 \cdot x^2}{1 \cdot 2 \cdot r^3} - \frac{3 \cdot 4 \cdot 5 \cdot x^3}{1 \cdot 2 \cdot 3 \cdot r^3} + \text{&c.} \\ &= 1 - \frac{3x}{r} + \frac{6x^2}{1 \cdot 2 \cdot r^2} - \frac{10x^3}{1 \cdot 2 \cdot 3 \cdot r^3} + \text{&c.} \end{align*} \] Ex. 2. It is required to express \( \sqrt[3]{a + b} \) by the form of a series.

Because \( a + b = a\left(1 + \frac{b}{a}\right) \),

Therefore \( \sqrt[3]{a + b} = \sqrt[3]{a} \times \sqrt[3]{1 + \frac{b}{a}} = a^{\frac{1}{3}} \left(1 + \frac{b}{a}\right)^{\frac{1}{3}} \)

By comparing \( \left(1 + \frac{b}{a}\right)^{\frac{1}{3}} \) with \( (a + x)^{\frac{m}{n}} \) we have \( a = 1, b = \frac{b}{a}, m = 1, n = 3 \),

and substitute as in last example

\[ \begin{align*} \sqrt[3]{a + b} &= a^{\frac{1}{3}} \left(1 + \frac{b}{a}\right)^{\frac{1}{3}} \\ &= a^{\frac{1}{3}} \left(1 + \frac{b}{3a} - \frac{b^2}{9a^2} + \frac{5b^3}{81a^3} - \cdots \right) \end{align*} \]

Ex. 3. It is required to resolve \( \frac{r^3}{(r^3 + z^3)^{\frac{2}{3}}} \) into a series.

Because \( \frac{r^3}{(r^3 + z^3)^{\frac{2}{3}}} = r^3 \times (r^3 + z^3)^{-\frac{2}{3}} \), if we raise \( r^3 + z^3 \) to the \(-\frac{2}{3}\) power, and multiply the resulting series by \( r^3 \), we shall have the series required. Or the given quantity may be reduced to a more simple form thus;

because \( r^3 + z^3 = r^3 \left(1 + \frac{z^3}{r^3}\right) \).

Therefore \( (r^3 + z^3)^{\frac{2}{3}} = r^3 \left(1 + \frac{z^3}{r^3}\right)^{\frac{2}{3}} \), and

\[ \frac{r^3}{(r^3 + z^3)^{\frac{2}{3}}} = \frac{1}{\left(1 + \frac{z^3}{r^3}\right)^{\frac{2}{3}}} = \left(1 + \frac{z^3}{r^3}\right)^{-\frac{2}{3}} \]

Hence

\[ \begin{align*} \frac{r^3}{(r^3 + z^3)^{\frac{2}{3}}} &= \left(1 + \frac{z^3}{r^3}\right)^{-\frac{2}{3}} \\ &= 1 - \frac{2z^3}{3r^3} + \frac{2 \cdot 5 \cdot z^6}{3 \cdot 6 \cdot r^6} - \frac{2 \cdot 5 \cdot 8 \cdot 11 \cdot z^9}{3 \cdot 6 \cdot 9 \cdot 12 \cdot r^9} + \cdots \end{align*} \]

Ex. 4. It is required to find a series equal to \( \frac{\sqrt{a^2 + x^2}}{\sqrt{a^2 - x^2}} \).

First by the binomial theorem we have

\[ \sqrt{a^2 + x^2} = (a^2 + x^2)^{\frac{1}{2}} = a + \frac{x^2}{2a} - \frac{x^4}{8a^2} + \frac{x^6}{16a^3} - \cdots \]

\[ \sqrt{a^2 - x^2} = (a^2 - x^2)^{\frac{1}{2}} = a - \frac{x^2}{2a} + \frac{3x^4}{8a^2} - \frac{5x^6}{16a^3} + \cdots \]

Therefore, by taking the product of the two series, and proceeding in the operation only to such terms as involve the 6th power of \( x \), we find

\[ \frac{\sqrt{a^2 + x^2}}{\sqrt{a^2 - x^2}} = 1 + \frac{x^2}{a^2} - \frac{x^4}{2a^4} + \frac{x^6}{2a^6} + \cdots \]

Sect. XVIII. Of the Reversion of Series.

271. The method of indeterminate coefficients, which we have already employed when treating of infinite series, may also be applied to what is called the reverting of series; that is, having any quantity expressed by an infinite series composed of the powers of another quantity, to express, on the contrary, the latter quantity by means of an infinite series composed of the powers of the former.

272. Let \( y = n + ax + bx^2 + cx^3 + dx^4 + \cdots \)

Then to revert the series we must find the value of \( x \) in terms of \( y \). For this purpose we shall transpose \( n \), and put \( z = y - n \), then

\[ z = ax + bx^2 + cx^3 + dx^4 + \cdots \]

Now when \( x = 0 \), it is evident that \( z = 0 \), therefore we may assume for \( x \) a series of this form,

\[ x = A + Bx + Cx^2 + Dx^3 + Ex^4 + \cdots \]

where the coefficients \( A, B, C, D, \ldots \) denote quantities as yet unknown, but which are entirely independent of the quantity \( x \). To determine these quantities let the first, second, third, etc., powers of the series,

\[ Ax + Bx^2 + Cx^3 + Dx^4 + \cdots \] Of Logarithms be found by multiplication, and substituted for \( x, x^2, \) etc., respectively, in the equation

\[ o = -x + ax + bx^2 + cx^3 + \ldots \]

thus we have

\[ \begin{align*} -x &= -x \\ +ax &= aAx + aBx^2 + aCx^3 + aDx^4 + \ldots \\ +bx^2 &= bAx^2 + 2bABx^3 + 2bACx^4 + \ldots \\ +cx^3 &= cAx^3 + 3cABx^4 + \ldots \\ &\vdots \\ +dx^4 &= dAx^4 + \ldots \\ &\vdots \end{align*} \]

and, putting the coefficients of \( x, x^2, x^3, \) etc., each \( = 0, \)

\[ aA - 1 = 0, \quad aB + bA^2 = 0, \quad aC + 2bAB + cA^3 = 0, \] \[ aD + 2bAC + bB^2 + 3cAB^2 + dA^4 = 0, \quad \text{etc.} \]

these equations give

\[ \begin{align*} A &= \frac{1}{a} \\ B &= \frac{b}{a^3} \\ C &= \frac{2b^2 - ac}{a^5} \\ D &= \frac{5b^3 - 5abc + a^2d}{a^7} \\ &\vdots \end{align*} \]

Therefore \( x = \frac{1}{a} - \frac{b}{a^3} x^2 + \frac{2b^2 - ac}{a^5} x^3 - \frac{5b^3 - 5abc + a^2d}{a^7} x^4 + \ldots \)

---

273. As an example of the application of this formula, let it be required to determine \( x \) from the equation

\[ y = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]

In this case we have

\[ z = y, \quad a = 1, \quad b = -\frac{1}{2}, \quad c = \frac{1}{6}, \quad d = -\frac{1}{24}, \quad \text{etc.} \]

Therefore, substituting these values, we have

\[ x = y + \frac{y^2}{2} + \frac{y^3}{6} + \frac{y^4}{24} + \ldots \]

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274. In the equation

\[ ay + by^2 + cy^3 + \ldots \]

in which both sides are expressed by series, and it is required to find \( y \) in terms of \( x, \) we must assume, as before,

\[ y = Ax + Bx^2 + Cx^3 + Dx^4 + \ldots \]

and substitute this series and its powers for \( y \) and its powers in the proposed equation; afterwards, by bringing all the terms to one side, and making the coefficients of each power of \( y, \) \( = 0, \) a series of equations will be had by which the quantities \( A, B, C, D, \) etc., may be determined.

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Sect. XIX. Of Logarithms and Exponential Quantities.

275. All positive numbers may be considered as powers of any one given affirmative number. The powers of 2, for instance, may become equal, either exactly, or nearer than by any assignable difference, to all logarithms, etc., numbers whatever, from 0 upwards. If the exponents be integers, we shall have only the numbers which form the geometrical progression, 1, 2, 4, 8, 16, etc.; but the intermediate numbers may be expressed, at least nearly, by means of fractional exponents. Thus the numbers from 0 to 10 may be expressed by the powers of 2 as follows:

\[ \begin{align*} 2^0 &= 1 \\ 2^1 &= 2 \\ 2^{1.585} &= 3 \\ 2^2 &= 4 \\ 2^{3.322} &= 5 \\ 2^{3.879} &= 6 \\ 2^{4.807} &= 7 \\ 2^{5.322} &= 8 \\ 2^{5.779} &= 9 \\ 2^{6.322} &= 10 \end{align*} \]

In like manner may fractions be expressed by the powers of 2. Thus

\[ \begin{align*} \frac{1}{2^{3.322}} &= 2^{-3.322} \\ \frac{1}{2^{4.807}} &= 2^{-4.807} \\ \frac{1}{2^{5.322}} &= 2^{-5.322} \\ \frac{1}{2^{5.779}} &= 2^{-5.779} \\ &\vdots \end{align*} \]

where it is observable that the exponents are now negative.

In the same manner may all numbers be expressed by the powers of 10. Thus,

\[ \begin{align*} 10^0 &= 1 \\ 10^{-3.322} &= 2^{-3.322} \\ 10^{-4.807} &= 2^{-4.807} \\ 10^{-5.322} &= 2^{-5.322} \\ &\vdots \end{align*} \]

276. Even a fraction might be taken in place of 2, or 10, in the preceding examples; and such exponents might be found as would give its powers equal to all numbers, from 0 upwards. There are therefore no limitations with respect to the magnitude of the number, by the powers of which all other numbers are to be expressed, except that it must neither be equal to unity, nor negative. If it were = 1, then all its powers would also be = 1, and if it were negative, there are numbers to which none of its powers could possibly be equal.

277. If therefore \( y \) denote any number whatever, and \( r \) a given number, a number \( x \) may be found, such, that \( r^x = y, \) and \( x, \) that is, the exponent of \( r \) which gives a number equal to \( y, \) is called the logarithm of \( y. \)

278. The given number \( r, \) by the powers of which all other numbers are expressed, is called the radical number of the logarithms, which are the indices of those powers.

279. From the preceding definition of logarithms their properties are easily deduced, as follows:

1. The sum of two logarithms is equal to the logarithm of their product. Let \( y \) and \( y' \) be two numbers, and \( x \) and \( x' \) their logarithms, so that \( r^x = y, \) and \( r^{x'} = y', \) then \( r^x \times r^{x'} = yy'; \) hence, from the definition, \( x + x' \) is the logarithm of \( yy', \) that is, the sum of the logarithms of \( y \) and \( y' \) is the logarithm of \( yy'. \)

2. The difference of the logarithms of two numbers is equal to the logarithm of their quotient; for if if \( r^x = y \) and \( r^{x'} = y' \), then \( \frac{r^x}{r^{x'}} = \frac{y}{y'} \) or \( r^{x-x'} = \frac{y}{y'} \); therefore, by the definition, \( x - x' \) is the logarithm of \( \frac{y}{y'} \); that is, the difference of the logarithms of \( y \) and \( y' \) is the logarithm of \( \frac{y}{y'} \).

3. Let \( n \) be any number whatever, then, \( \log_N^n = n \times \log_N N \). For \( N^n \) is \( N \) multiplied into itself \( n \) times, therefore the logarithm of \( N^n \) is equal the logarithm of \( N \) added to itself \( n \) times, or to \( n \times \log_N N \).

280. From these properties of logarithms it follows, that if we possess tables by which we can assign the logarithm corresponding to any given number, and also the number corresponding to any given logarithm, the operations of multiplication and division of numbers may be reduced to the addition and subtraction of their logarithms, and the operations of involution and evolution to the more simple operations of multiplication and division. Thus, if two numbers \( x \) and \( y \) are to be multiplied together, by taking the sum of their logarithms, we obtain the logarithm of their product, and, by inspecting the table, the product itself. A similar observation applies to the quotient of two numbers, and also to any power or to any root of a number.

281. The general properties of logarithms are independent of any particular value of the radical number, and hence there may be various systems of logarithms, according to the radical number employed in their construction. Thus if the radical number be 10, we shall have the common system of logarithms, but if it were 27182818 we should have the logarithms first constructed by Lord Napier, which are called hyperbolic logarithms.

282. We have already observed (§ 277), that the relation between any number and its logarithm is expressed by the equation \( r^x = y \), where \( y \) denotes a number, \( x \) its logarithm, and \( r \) the radical number of the system, and any two of these three quantities being given the remaining one may be found. If either \( y \) or \( r \) were the quantity required, the exponent would involve no difficulty; if, however, the exponent \( x \) were considered as the unknown quantity while \( r \) and \( y \) were supposed given, the equation to be resolved would be of a different form than any that we have hitherto considered. Equations of this form are called exponential equations. To resolve such an equation is evidently the same thing as to determine the logarithm of a given number, and this problem we shall now proceed to investigate.

283. We therefore resolve the equation \( r^x = y \), where \( r, x, \) and \( y \) denote, as before, we are to find a value of \( x \) in terms of \( r \) and \( y \). Let us suppose \( r = 1 + a \) and \( y = 1 + v \), then our equation will stand thus:

\[ (1 + a)^x = 1 + v. \]

So that, by raising both sides to the power \( n \), where \( n \) denotes an indefinite number, which is to disappear in the course of the investigation, we have \((1 + a)^n = (1 + v)^n\), and resolving both sides of the equation into series by means of the binomial theorem,

\[ 1 + nx + \frac{n(n-1)}{1 \cdot 2}a^2 + \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3}a^3 + \ldots, \]

and by supposing the factors which constitute the terms of each series to be actually multiplied, and the products arranged according to the power of \( n \), the last equation will have this form,

\[ xa + \frac{x(n-1)}{1 \cdot 2}a^2 + \frac{x(n-1)(n-2)}{1 \cdot 2 \cdot 3}a^3 + \ldots, \]

Here the coefficients of the power \( n \), viz. \( P, P', P'' \), &c., \( Q, Q', \) &c., \( R, R', \) &c., also \( p, p', p'' \), &c., \( q, q', \) &c., \( r, r', \) &c., are expressions which denote certain combinations of the powers of \( x \) in the first series, and certain numbers in the second; but as they are all to vanish in the course of the investigation, it is not necessary that they should be expressed in any other way than by a single letter.

284. Now each side of this last equation may evidently be resolved into two parts, one of which is entirely free from the quantity \( n \), and the other involves that quantity, hence the same equation may also stand thus,

\[ xa + \frac{x}{2}a^2 + \frac{x}{3}a^3 + \ldots, \]

This equation must hold true, whatever be the value of \( n \), which is a quantity entirely arbitrary, and therefore ought to vanish from the equation expressing the relation between \( x \) and \( v \); hence it follows that the terms on each side of the equation, which involve \( n \), ought to destroy each other, and thus there will remain Of Logarithms, &c., that is,

\[ x = \frac{a^3}{2} + \frac{a^4}{3} - \frac{a^5}{4} + \ldots \]

or

\[ (a - \frac{a^3}{2} + \frac{a^4}{3} - \frac{a^5}{4} + \ldots) x = v - \frac{v^2}{2} + \frac{v^3}{3} - \frac{v^4}{4} + \ldots \]

Let us now put \( A \) to denote the constant multiplier

\[ a - \frac{a^3}{2} + \frac{a^4}{4} + \ldots = (r-1) - \frac{(r-1)^3}{2} + \frac{(r-1)^4}{3} + \ldots \]

and substitute for \( y \), its value \( y = \frac{1}{A} \left( \frac{(y-1)^2}{2} + \frac{(y-1)^3}{3} - \frac{(y-1)^4}{4} + \ldots \right) \)

and by this formula the logarithm of any number a little greater than unity may be readily found.

285. If \( y \) be nearly \( = 2 \), the series will, however, converge too slowly to be of use, and if it exceed 2, the series will diverge, and therefore cannot be directly applied to the finding of its logarithm. But a series which shall converge faster, and be applicable to every case, may be investigated as follows:

\[ \log_y \frac{n+x}{n} = \frac{1}{A} \left( \frac{2x}{2n+x} + \frac{1}{3} \frac{2x^3}{(2n+x)^3} + \frac{1}{5} \frac{2x^5}{(2n+x)^5} + \ldots \right) \]

But \( \log_y \frac{n+x}{n} = \log_y (n+x) - \log_y n \), therefore

\[ \log_y (n+x) = \log_y n + \frac{1}{A} \left( \frac{2x}{2n+x} + \frac{1}{3} \frac{2x^3}{(2n+x)^3} + \frac{2x^5}{(2n+x)^5} + \ldots \right) \]

This series gives the logarithm of \( n+x \) by means of the logarithm of \( n \), and converges very fast when \( n \) is considerable.

287. It appears from the series which have been found for \( \log_y \) in § 284 and 285, that the logarithm of a number is always the product of two quantities; one of these is variable, and depends upon the number itself, but the other, viz. \( \frac{1}{A} \) is constant, and depends entirely on the radical number of the system. This quantity has been called by writers on logarithms the modulus of the system.

288. The most simple system of logarithms, in respect to facility of computation, is that in which \( \frac{1}{A} = 1 \) or \( A = 1 \). The logarithms of this system are the same as those first invented by Napier, and are also called hyperbolic logarithms.

The hyperbolic logarithm of any numbers \( y \), is therefore (\$ 284),

\[ y - 1 - \frac{(y-1)^2}{2} + \frac{(y-1)^3}{3} - \ldots \]

and that of \( r \), the radical number of any system, is

Because \( \log_y (1+v) = \frac{1}{A} \left( \frac{v^2}{2} + \frac{v^3}{3} - \frac{v^4}{4} + \ldots \right) \)

By substituting \(-v\) for \(+v\), we have

\[ \log_y (1-v) = \frac{1}{A} \left( -\frac{v^2}{2} - \frac{v^3}{3} - \frac{v^4}{4} + \ldots \right) \]

Now, \( \log_y (1+v) - \log_y (1-v) = \log_y \frac{1+v}{1-v} \),

therefore subtracting the latter series from the former, we have

\[ \log_y \frac{1+v}{1-v} = \frac{1}{A} \left( 2v + \frac{2v^3}{3} + \frac{2v^5}{5} + \ldots \right) \]

Put \( \frac{1+v}{1-v} = y \), then \( v = \frac{y-1}{y+1} \), and the last series becomes

\[ \log_y = \frac{1}{A} \left( \frac{2(y-1)}{y+1} + \frac{2(y-1)^3}{3(y+1)^3} + \frac{2(y-1)^5}{5(y+1)^5} + \ldots \right) \]

This series will always converge whatever be the value of \( y \), and by means of it the logarithms of small numbers may be found with great facility.

286. When a number is composite, its logarithm will most easily be found, by adding together the logarithms of its factors; but if it be a prime number, its logarithm may be derived from that of some convenient composite number, either greater or less, and an infinite series. Let \( n \) be a number of which the logarithm is already found; then substituting \( \frac{n+x}{n} \) for \( y \) in the last formula, we have

\[ \log_y \frac{n+x}{n} = \frac{1}{A} \left( \frac{2x}{2n+x} + \frac{1}{3} \frac{2x^3}{(2n+x)^3} + \frac{2x^5}{(2n+x)^5} + \ldots \right) \]

but this last series is the same as we have denoted by \( A \); hence it follows, that the modulus of any system is the reciprocal of the hyperbolic logarithm of the radical number of that system. Thus it appears, that the logarithms of numbers, according to any proposed system, may be readily found from the hyperbolic logarithm of the same numbers, and the hyperbolic logarithm of the radical number of that system.

289. Let \( L \) denote the hyp. log. of any number, and \( l, l' \) the logarithms of the same number according to two other systems whose moduli are \( m \) and \( m' \); then

\[ l = mL, l' = m'L; \]

therefore \( \frac{l}{m} = \frac{l'}{m'} \) and \( m : m' :: l : l' \).

That is, the logarithms of the same number, according to different systems, are directly proportional to the moduli of these systems, and therefore have a given ratio to one another.

290. We shall now apply the series here investigated to the calculation of the hyperbolic logarithm of 10, the reciprocal of which is the modulus of the common system. Of Logarithms; and also to the calculation of the common logarithm of 2. The hyp. log. of 10 may be obtained by substituting 10 for y in the formula

\[ \text{hyp. log. } y = \frac{2(y-1)}{y+1} + \frac{2(y-1)^3}{3(y+1)} + \frac{2(y-1)^5}{5(y+1)} + \ldots \]

but the resulting series \( \frac{2^9}{11} + \frac{2^9}{3^3} + \frac{2^9}{5^5} + \ldots \) converges too slowly to be of any practical utility, it will therefore be better to derive the logarithm of 10 from those of 2 and 5. By substituting 2 in the formula we have

\[ \text{hyp. log. } 2 = \frac{1}{3} + \frac{1}{3^3} + \frac{1}{5^5} + \frac{1}{7^7} + \ldots \]

This series converges very fast, so that by reducing its terms to decimal fractions, and taking the sum of the first seven terms, we find the hyp. log. of 2 to be 0.6931472.

The hyp. log. of 5 may be found in the same manner, but more easily from the formula given in § 286. For the log. of 2 being given, that of \( 4^2 = 2^4 \) is also given, § 279. Therefore, substituting log. 4 = log. 2 for log. n, and 1 for x, in the series

\[ \text{hyp. log. } (n+x) = \text{hyp. log. } n + 2 \left( \frac{x}{2n+x} + \frac{x^3}{(2n+x)^3} + \ldots \right) \]

we have

\[ \text{hyp. log. } 5 = 2 \text{ hyp. log. } 2 + 2 \left( \frac{1}{9} + \frac{1}{3^3} + \frac{1}{5^5} + \ldots \right) \]

The first three terms of this series are sufficient to give the result true to the seventh decimal, so that we have hyp. log. 5 = 1.6094379, and hyp. log. 10 = hyp. log. 2 + hyp. log. 5 = 2.3025851.

Hence the modulus of the common system of logarithms, or \( \frac{1}{\text{hyp. log. } 10} \), is found = 0.4342945. The same number, because of its great utility in the construction of tables of logarithms, has been calculated to a much greater number of decimals. A celebrated calculator of the last century, Mr A. Sharp, found it to be

\[ 0.4342948190325182765112891891660508229 \] \[ 4397005803666566114454. \]

Having found the hyp. log. of 2 to be 0.6931472, the common logarithm of 2 is had immediately, by multiplying the hyp. log. of 2 by the modulus of the system; thus we find

\[ \text{com. log. } 2 = 4.342945 \times 0.6931472 = 3.010300. \]

291. We have already observed, § 282, that to determine the logarithm of a given number, is the same problem as to determine the value of \( x \) in an equation of this form, \( a^x = b \), where the unknown quantity is an exponent. But in order to resolve such an equation, it is not necessary to have recourse to series; for a table of logarithms being once supposed constructed, the value of \( x \) may be determined thus. It appears, from § 279, that \( x \times \log_a = \log_b \). Hence it follows,

\[ x = \frac{\log_b}{\log_a}. \]

The use of this formula will appear in next section, which treats of computations relative to annuities.

292. The theory of logarithms requires the solution of this other problem. Having given the radical number of a system, and a logarithm, to determine the corresponding number. Or having given the equation \( r^x = y \), where \( r, x, \) and \( y \) denote, as in § 282, to find a series which shall express \( y \) in terms of \( r \) and \( x \).

293. For this purpose, let us suppose \( r = 1 + a \), then our equation becomes \( y = (1 + a)^x \), which may also be expressed thus:

\[ y = [(1 + a)^x]^n, \]

where \( n \) is an indefinite quantity, which is to disappear in the course of the investigation.

By the binomial theorem we have

\[(1 + a)^n = 1 + na + \frac{n(n-1)}{1 \cdot 2}a^2 + \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3}a^3 + \ldots \]

this equation, by multiplying together the factors which compose the terms of the series, and arranging the results according to the powers of \( n \), may also be expressed thus:

\[(1 + a)^n = 1 + An + Bn^2 + Cn^3 + \ldots \]

where it will readily appear that

\[ A = a - \frac{a^2}{2} + \frac{a^3}{3} - \frac{a^4}{4} + \ldots \]

as to the values of \( B, C, \ldots \) it is of no importance to know them, for they will all disappear in the course of the investigation. Hence, by substituting for \((1 + a)^n\) its value, as expressed by this last series, we have

\[ y = (1 + An + Bn^2 + Cn^3 + \ldots)^n \]

and expanding the latter part of this equation by means of the binomial theorem, it becomes

\[ y = 1 + \frac{x}{n}(An + Bn^2 + \ldots) + \frac{x(x-n)}{1 \cdot 2 \cdot n^2}(An + Bn^2 + \ldots)^2 + \ldots \]

But \( An + Bn^2 + \ldots = n(A + Bn + \ldots) \) also \((An + Bn^2 + \ldots)^2 = n^2(A + Bn + \ldots)^2 \), and \((An + Bn^2 + \ldots)^3 = n^3(A + Bn + \ldots)^3 \), &c.

therefore, by leaving out of each term of the series the powers of \( n \), which are common to the numerator and denominator, the equation will stand thus:

\[ y = 1 + \frac{x}{n}(A + Bn + \ldots) + \frac{x(x-n)}{1 \cdot 2 \cdot n^2}(A + Bn + \ldots)^2 + \ldots \]

Now \( n \) is here an arbitrary quantity, and ought, from the nature of the original equation, to disappear from the value of \( y \); the terms of the equation which are multiplied Interest and multiplied by \( n \) ought therefore to destroy each other; and this being the case, the equation is reduced to

\[ r^x = y = 1 + \frac{xA}{1} + \frac{x^2A^2}{1 \cdot 2} + \frac{x^3A^3}{1 \cdot 2 \cdot 3} + \frac{x^4A^4}{1 \cdot 2 \cdot 3 \cdot 4} + \ldots \]

and since we have found

\[ \Lambda = a - \frac{a^2}{2} + \frac{a^3}{3} - \frac{a^4}{4} + \ldots \]

\[ = (r-1) - \frac{(r-1)^2}{2} + \frac{(r-1)^3}{3} - \frac{(r-1)^4}{4} \]

It is evident from § 288, that \( \Lambda \) is the hyperbolic logarithm of the radical number of the system.

294. If, in the equation \( r^x = y \), we suppose \( x = 1 \), the value of \( y \) becomes

\[ r = 1 + \frac{\Lambda}{1} + \frac{\Lambda^2}{1 \cdot 2} + \frac{\Lambda^3}{1 \cdot 2 \cdot 3} + \ldots \]

Here the radical number is expressed by means of its hyperbolic logarithm. Again, if we suppose \( x = \frac{1}{\Lambda} \), then

\[ r^{\frac{1}{\Lambda}} = 1 + \frac{1}{1} + \frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 2 \cdot 3} + \ldots \]

Thus it appears that the quantity \( r^{\frac{1}{\Lambda}} \) is equal to a constant number, which, by taking the sum of a sufficient number of terms of the series, will be found \( = 2.718281828459045 \ldots \). Let us denote this number by \( e \), then \( r^{\frac{1}{\Lambda}} = e \), and hence \( r = e^\Lambda \). Now, if we remark that \( \Lambda \) is the hyp. log. of \( r \), it must be evident (\$ 277. and 278.), that \( e \) is the radical number of the hyperbolic system of logarithms.

Again, since \( r^{\frac{1}{\Lambda}} = e \), therefore \( \frac{1}{\Lambda} \times \log_r e = \log_e e \)

and \( \Lambda = \frac{\log_r r}{\log_r e} \), here \( \log_r r \) and \( \log_r e \) denote logarithms taken according to any system whatever.

295. If we now resume the equation

\[ r^x = y = 1 + \frac{xA}{1} + \frac{x^2A^2}{1 \cdot 2} + \frac{x^3A^3}{1 \cdot 2 \cdot 3} + \ldots \]

and substitute for \( A \) its value \( \frac{\log_r r}{\log_r e} \), we shall have the following general expression for any exponential quantity whatever,

\[ r^x = 1 + \frac{x(\log_r r)}{1} + \frac{x^2(\log_r r)^2}{1 \cdot 2} + \frac{x^3(\log_r r)^3}{1 \cdot 2 \cdot 3} + \ldots \]

which, by supposing \( r = e \), becomes

\[ e^x = 1 + \frac{x}{1} + \frac{x^2}{1 \cdot 2} + \frac{x^3}{1 \cdot 2 \cdot 3} + \ldots \]

Sect. XX. Of Interest and Annuities.

296. The theory of logarithms finds its application in some measure to calculations relating to interest and annuities; these we now proceed to explain. There are two hypotheses according to either of which money compound put out at interest may be supposed to be improved. We may suppose that the interest, which is always proportional to the sum lent, or principal, is also proportional to the time during which the principal is employed; and on this hypothesis the money is said to be improved at simple interest. Or we may suppose that the interest which ought to be paid to the lender at successive stated periods, is added to the principal instead of being actually paid, and thus their amount converted into a new principal. When money is laid out according to this second hypothesis, it is said to be improved at compound interest.

297. In calculations relating to interest, the things to be considered are the principal, or sum lent; the rate of interest, or sum paid for the use of 100l. for one year; the time during which the principal is lent; and the amount, or sum of the principal and interest, at the end of that time.

Let \( p \) denote the principal, 1l. being the unit.

\( r \) the interest of 1l. for one year, at the given rate.

\( t \) the time, one year being the unit.

\( a \) the amount.

We shall now examine the relations which subsist between these quantities, according to each of the two hypotheses of simple and compound interest.

I. Simple Interest.

298. Because the interest of 1l. for one year is \( r \), the interest of 1l. for \( t \) years must be \( rt \), and the interest of \( p \) pounds for the same time \( prt \), hence we have this formula,

\[ p + prt = a, \]

from which we find

\[ p = \frac{a}{1 + rt}, \quad r = \frac{a - p}{pt}, \quad t = \frac{a - p}{pr}. \]

As the manner of applying these formulae to questions relating to simple interest is sufficiently obvious, we proceed to consider compound interest.

II. Compound Interest.

299. In addition to the symbols already assumed, let \( R = 1 + r = \) amount of 1l. in one year; then, from the nature of compound interest, \( R \) is also the principal at the beginning of the second year. Now, interest being always proportional to the principal, we have

\[ 1 : r :: R : rR = \text{the interest of } R \text{ for a year}, \]

and \( R + rR = (1 + r)R = R^2 = \text{amount of } R \text{ in a year}, \)

therefore \( R^2 \) is the amount of 1l. in two years, which sum being assumed as a new principal, we find, as before, its interest for a year to be \( rR^2 \), and its amount \( R^3 + rR^2 = (1 + r)R^2 = R^3 \); so that \( R^3 \) is the amount of 1l. in three years. Proceeding in this manner, we find, in general, that the amount of 1l. in \( t \) years is \( R^t \), and of \( p \) pounds \( prt \); hence we have this formula

\[ pR^t = a, \] Annuities, which from the nature of logarithms may also be expressed thus:

\[ \log_p + t \times \log_R = \log_a. \]

Hence we find

\[ p = \frac{a}{R^t}, \quad R = \sqrt{\frac{a}{p}}, \]

or, by logarithms,

\[ \log_p = \log_a - t \times \log_R, \quad \log_R = \frac{\log_a - \log_p}{t}. \]

300. As an example of the use of these formulae, let it be required to determine what sum improved at 5 per cent. compound interest will amount to £500 in 42 years. In this case we have given \( a = 500, r = 0.05, R = 1.05, t = 42 \), to find \( p \).

From

\[ \log_a = \log_500 = 2.6989700 \]

subtract

\[ t \times \log_R = 42 \times \log_1.05 = 0.8899506 \]

remains \( \log_p = 1.8090194 \)

therefore \( p = 64.421 = 64l. 8s. 5d. \) the sum required.

Ex. 2. In what time will a sum laid out at 4 per cent. compound interest be doubled.

Let any sum be expressed by unity, then we have given \( p = 1, r = 0.04, R = 1.04, a = 2 \), to find \( t \).

From the formula

\[ t = \frac{\log_a - \log_p}{\log_R} = \frac{\log_2}{\log_1.04} \]

find \( t = \frac{3010300}{0170333} = 17.7 \) years nearly.

301. In treating of compound interest, we have supposed the interest to be joined to the principal at the end of the year. But we might have supposed it to be added at the end of every half year or every quarter, or even every instant; and suitable rules might have been found for performing calculations according to each hypothesis. As such suppositions are, however, never made in actual business, we shall not at present say anything more of them.

III. Annuities.

302. An annuity is a payment made annually for a term of years; and the chief problem relating to it is to determine its present worth, that is, the sum a person ought to pay immediately to another, upon condition of receiving from the latter a certain sum annually for a given time. In resolving this problem, it is supposed that the buyer improves his annuity from the time he receives it, and the seller the purchase money, in a certain manner, during the continuance of the annuity, so that at the end of the time the amount of each may be the same. There may be various suppositions as to the way in which the annuity and its purchase money may be improved; but the only one commonly applied to practice is the highest improvement possible of both, viz. by compound interest. As the taking compound interest is, however, prohibited by law, the realizing of this supposed improvement requires punctual payment of interest; and therefore the interest in such calculations is usually made low.

303. Let \( A \) denote the annuity; \( P \) the present worth, or purchase money; \( t \) the time of its continuance; \( r \) and \( R \) denote as before.

The seller, by improving the price \( P \) at compound interest during the time \( t \), has \( PR^t \).

The purchaser is supposed to receive the first annuity \( A \) at the end of one year, which being improved for \( t-1 \) years, amounts to \( AR^{t-1} \). He receives the second year's annuity at the end of the second year, which, being improved for \( t-2 \) years, amounts to \( AR^{t-2} \). In like manner the third year's annuity becomes \( AR^{t-3} \), and so on to the last year's annuity, which is simply \( A \). Therefore, the whole amount of the improved annuities is the geometrical series

\[ A + AR + AR^2 + AR^3 + \cdots + AR^{t-1}, \]

the sum of which, by § 106, is \( A \frac{R^t - 1}{R - 1} = A \frac{R^{t-1}}{r} \);

and since this sum must be equal to the amount of the purchase money, or \( PR^t \), we have

\[ PR^t = A \frac{R^{t-1}}{r}; \]

and from this equation we find

\[ P = \frac{A}{r} \left( 1 - \frac{1}{R^t} \right), \quad A = \frac{PR^t}{R - 1}, \quad t = \frac{\log_A - \log_P}{\log_R} \]

As to \( r \), it can only be found by the resolution of an equation of the \( t \) order.

304. To find the present value of an annuity in reversion, that is, an annuity which is to commence at the end of \( n \) years, and continue during \( t \) years; first find its value for \( n+t \) years, and then for \( n \) years; and subtract the latter from the former, we thus obtain the following formula:

\[ P = \frac{A}{rR^n} \left( 1 - \frac{1}{R^t} \right). \]

305. If the annuity is to commence immediately, and to continue for ever, then, because in this case \( R^t \) is infinitely great, and therefore \( \frac{1}{R^t} = 0 \), the formula

\[ P = \frac{A}{r} \left( 1 + \frac{1}{R^t} \right) \]

becomes simply \( P = \frac{A}{r} \).

And if the annuity is to commence after \( n \) years, and continue for ever, the formula

\[ P = \frac{A}{rR^n} \left( 1 - \frac{1}{R^t} \right) \]

becomes \( P = \frac{A}{rR^n} \).

Sect. XXI. Of Continued Fractions.

306. Every quantity which admits of being expressed by a common fraction may also be expressed in Continued the form of what is called a continued fraction. The nature of such fractions will be easily understood by the following example.

Let the common fraction be $\frac{314159}{100000}$, or, which is the same, $3 + \frac{14159}{100000}$. Since $100000 = 7 \times 14159 + 887$, therefore $\frac{14159}{100000} = \frac{14159}{7 \times 14159 + 887} = \frac{1}{887 + \frac{1}{14159}}$, and

$$\frac{314159}{100000} = 3 + \frac{1}{887 + \frac{1}{14159}}.$$

Now $\frac{887}{14159} = \frac{887}{15 \times 887 + 854} = \frac{1}{854 + \frac{1}{887}}$, and substituting this for $\frac{887}{14159}$ in the value of $\frac{314159}{100000}$, already found, we have

$$\frac{314159}{100000} = 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{854 + \frac{1}{887}}}}.$$

Again, $\frac{854}{887} = \frac{854}{854 + 33} = \frac{1}{1 + \frac{33}{854}}$, which being substituted as before, gives

$$\frac{314159}{100000} = 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1 + \frac{33}{854}}}}.$$

By operations similar to the preceding, we find $\frac{33}{854} = \frac{1}{25 + \frac{29}{33}} = \frac{1}{25 + \frac{1}{33}}$, therefore, by substitution,

$$\frac{314159}{100000} = 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{25 + \frac{1}{7 + \frac{1}{4}}}}}.$$

By an operation, in all respects the same as has been just now performed, may any fraction whatever be reduced to the form

$$a + \frac{1}{b + \frac{1}{c + \frac{1}{d + \ldots}}}$$

and it is then called a continued fraction.

307. It is easy to see in what manner the inverse of the preceding operation is to be performed, or a continued fraction reduced to a common fraction.

Thus if the continued fraction be

$$a + \frac{1}{b + \frac{1}{c + \frac{1}{d}}}$$

it will evidently be reduced to a common fraction by adding the reciprocal of $d$ to $b$, and the reciprocal of that sum to $b$, and again the reciprocal of this last sum to $a$; now the reciprocal of $d$, or $\frac{1}{d}$, added to $c$ is $c + \frac{1}{d} = \frac{cd + 1}{d}$; again, the reciprocal of this sum, or $\frac{d}{cd + 1}$, added to $b$, is $b + \frac{d}{cd + 1} = \frac{bcd + b + d}{cd + 1}$, and the reciprocal of this last quantity, viz., $\frac{bcd + b + d}{cd + 1}$, when added to $a$, gives

$$\frac{abcd + ab + ad + cd + 1}{bcd + b + d} = a + \frac{1}{b + \frac{1}{c + \frac{1}{d}}}.$$

308. This manner of expressing a fraction enables us to find a series of other fractions, that approach in value to any given one, and each of them expressed in the smallest numbers possible. Thus, in the example $\frac{314159}{100000}$, which has been resolved into a continued fraction, § 306, and which is known to express nearly the proportion of the diameter of a circle to its circumference, if we take only the first two terms of the continued fraction, and put $\pi$ for $\frac{314159}{100000}$, we shall have

$$\pi = 3 + \frac{1}{7 + \frac{1}{15}} = 3 + \frac{1}{7 + \frac{1}{15}} = \frac{333}{106},$$

which is nearer the truth than the former.

And, by taking the first four terms, we have

$$\pi = 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1}}} = \frac{355}{113},$$

which is the proportion assigned by Metius, and is more exact than either of the preceding. The results are alternately greater and less than the truth.

309. Among continued fractions, those have been particularly distinguished in which the denominators, after a certain number of changes, are continually repeated in the same order. Such, for example, is the fraction

$$1 + \frac{1}{2 + \frac{1}{3 + \frac{1}{2 + \frac{1}{3 + \ldots}}}},$$

The amount of this fraction, though continued ad infinitum, may be easily found; for leaving out the first term, which is an integer, let us suppose

$$\pi = \frac{1}{2 + \frac{1}{3 + \frac{1}{2 + \frac{1}{3 + \ldots}}}},$$

Then, since after the second, all the terms return in the Continued the same order, it follows that their amount is also \( x \), thus we have

\[ x = \frac{1}{2} + \frac{1}{3 + x} \]

Hence \( x = \frac{3 + x}{6 + 2x + 1} \) and \( x^2 + 3x = \frac{3}{2} \) and \( x = \frac{-3 + \sqrt{15}}{2} \)

Therefore \( x + 1 \), or the sum of the series, \( = \frac{-1 + \sqrt{15}}{2} \)

In general if \( x = \frac{1}{a} + \frac{1}{b} + \frac{1}{a + b} + \ldots \), &c.

we find \( x = -\frac{b}{2} \pm \sqrt{\frac{b^2}{4} + \frac{b}{a}} \). Though the denominators did not return in the same order till after a greater interval, the value of the fraction would still be expressed by the root of a quadratic equation. And conversely, the roots of all quadratic equations may be expressed by periodical continued fractions, and may often by that means be very readily approximated in numbers, without the trouble of extracting the square root.

310. The reduction of a decimal into the form of a continued fraction sometimes renders the law of its continuation evident. Thus we know that \( \sqrt{2} = 1.41421356 \ldots \), but from the bare inspection of this decimal we discover no rule for its further continuation. If, however, it be reduced into a continued fraction, it becomes

\[ = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \ldots}}} \text{&c.} \]

and hence we see in what way it may be continued to any degree of accuracy.

311. When the root of any equation is found by the method explained in § 256, the value of the unknown quantity is evidently expressed by a continued fraction.

For if \( x \) be the root sought, we have \( x = a + \frac{1}{y}, y = b + \frac{1}{y'}, y' = b' + \frac{1}{y''}, y'' = b'' + \frac{1}{y'''} \), &c. where \( a, b, b', b'' \), \( b''' \), &c. denote the whole numbers, which are next less than the true values of \( x, y, y', y'' \), &c. If therefore in the value of \( x \) we substitute \( b + \frac{1}{y'} \) for \( y \), it becomes

\[ x = a + \frac{1}{b + \frac{1}{y'}} \]

Again, if in this second value of \( x \) we substitute \( b' + \frac{1}{y''} \) for \( y \) it becomes

\[ x = a + \frac{1}{b' + \frac{1}{y''}} \]

The next value of \( x \) is in like manner found to be

\[ x = a + \frac{1}{b' + \frac{1}{y'' + \frac{1}{y'''} \ldots}} \]

and so on continually.

Sect. XXII. Of Indeterminate Problems.

312. When the conditions of a question are such that the number of equations exceeds the number of unknown quantities, that question will admit of innumerable solutions, and is therefore said to be indeterminate. Thus, if it be required to find two numbers subject to no other limitation than that their sum be 10, we have two unknown quantities \( x \) and \( y \), and only one equation, viz. \( x + y = 10 \), which may evidently be satisfied by innumerable different values of \( x \) and \( y \), if fractional solutions be admitted. It is, however, usual in such questions as this, to restrict values of the numbers sought to positive integers, and therefore, in this case, we can have only these nine solutions;

\[ \begin{align*} x &= 1, 2, 3, 4, 5, 6, 7, 8, 9 \\ y &= 9, 8, 7, 6, 5, 4, 3, 2, 1 \end{align*} \]

which indeed may be reduced to five, for the first four become the same as the last four, by simply changing \( x \) into \( y \), and the contrary.

313. Indeterminate problems are of different orders according to the dimensions of the equation which is obtained after all the unknown quantities, but two, have been exterminated by means of the given equations. Those of the first order lead always to equations of this form,

\[ ax + by = c, \]

where \( a, b, c \) denote given whole numbers, and \( x, y \) two numbers to be found, so that both may be integers. That this condition may be fulfilled, it is necessary that the coefficients \( a, b \) have no common divisor which is not also a divisor of \( c \), for if \( a = md \) and \( b = me \), then \( ax + by = mdx + mey = c \), and \( dx + ey = \frac{c}{m} \); but \( d, e, x, y \) are supposed to be whole numbers, therefore \( \frac{c}{m} \) is a whole number, hence \( m \) must be a divisor of \( c \).

314. We proceed to illustrate the manner of resolving indeterminate equations of the first order by some numerical examples.

Ex. 1. Given \( 2x + 3y = 25 \), to determine \( x \) and \( y \) in whole positive numbers.

From the given equation we have \( x = \frac{25 - 3y}{2} \),

\[ -y + \frac{1}{2} ; \text{ now since } a \text{ must be a whole number, it follows that } \frac{1}{2} \text{ must be a whole number. Let us assume } \frac{1}{2} = z; \text{ then } 1 - y = 2z \text{ and } y = 1 - 2z, \text{ and since } x = 12 - y + \frac{1}{2} = 12 - y + z, \text{ therefore } x = 12 - 1 + 2z + z; \text{ hence we have } \]

\[ x = 11 + 3z, \quad y = 1 - 2z \] Indeterm. where $z$ might be any whole number whatever, if there were no limitation as to the signs of $x$ and $y$; but since these quantities are required to be positive, it is evident from the value of $y$, that $z$ must either be 0 or negative, and from the value of $x$, that, subtracting from the sign, it must be less than 4; hence $z$ may have these three values 0, -1, -2, -3.

If $z = 0$, $z = -1$, $z = -2$, $z = -3$.

Then $\begin{cases} x = 11, & y = 8 \\ x = 5, & y = 5 \end{cases}$

Ex. 2. It is required to divide 100 into such parts that the one may be divisible by 7 and the other by 11.

Let $7x$ be the first part, and $11y$ the second, then by the question $7x + 11y = 100$, and

$$x = \frac{100 - 11y}{7} = 14 - y + \frac{2 - 4y}{7};$$

hence it appears, that $\frac{2 - 4y}{7}$ must be a whole number. Let us assume $\frac{2 - 4y}{7} = z$, then $x = 14 - y + z$ and $4y = 2 - 7z$, or $y = \frac{2 - 7z}{4}$. Therefore $2 - 3z$ must be a whole number. Assume $\frac{2 - 3z}{4} = t$, then $y = t - z$, and $3z = 2 - 4t$, or $z = \frac{2 - 4t}{3}$. Therefore $\frac{2 - t}{3}$ must be a whole number.

Assume now $\frac{2 - t}{3} = v$, then $z = v - t$ and $t = 2 - 3v$, here it is evident $v$ may be any whole number taken at pleasure, so that to determine $x$ and $y$ we have the following series of equations:

$$\begin{align*} t &= 2 - 3v \\ z &= v - t = 4v - 2 \\ y &= t - z = 4 - 7v \\ x &= 14 - y + z = 119 + 8. \end{align*}$$

Now from the value of $y$ it appears, that $v$ must either be 0, or negative; but from the value of $x$ we find that $v$ cannot be a negative whole number, therefore $v$ can only be 0; hence the only values which $x$ and $y$ can have in whole numbers, are $x = 8$, $y = 4$.

Ex. 3. It is required to find all the possible ways in which 60l. can be paid in guineas and moidores only.

Let $x$ be the number of guineas and $y$ the number of moidores. Then the value of the guineas, expressed in shillings, is $21x$, and that of the moidores $27y$, therefore, from the nature of the question, $21x + 27y = 1200$, or, dividing the equation by 3, $7x + 9y = 400$, hence $x = \frac{400 - 9y}{7} = 57 - y + \frac{1 - 2y}{7}$, so that $\frac{1 - 2y}{7}$ must be a whole number.

Assume $\frac{1 - 2y}{7} = z$, then $x = 57 - y + z$, and $2y = 1$.

Therefore $\frac{1 - 2y}{7}$ must be a whole number.

Assume $\frac{1 - z}{2} = v$, then $y = v - 3z$, and $z = 1 - 2v$; therefore $v$ may be taken any whole number at pleasure, and $x$ and $y$ may be determined by the following equations:

$$\begin{align*} x &= 1 - 2v \\ y &= v - 3z = 7v - 3 \\ x &= 57 - y + z = 61 - 9v. \end{align*}$$

From the value of $x$, it appears that $v$ cannot exceed 6, and from the value of $y$, that it cannot be less than 1.

Hence if $v = 1, 2, 3, 4, 5, 6$, we have $x = 52, 43, 34, 25, 16, 7$, $y = 4, 11, 18, 25, 32, 39$.

315. In the foregoing examples the unknown quantities $x$ and $y$ have each a determinate number of positive values, and this will evidently be the case as often as the proposed equation is of this form, $ax + by = c$. If, however, $b$ be negative, that is, if the equation be of this form $ax - by = c$, or $ax = by + c$, we shall have questions of a different kind, admitting each of an infinite number of solutions; these, however, are to be resolved in the same manner as the preceding, as will appear from the following example.

Ex. 4. A person buys some horses and oxen, he pays 31 crowns for each horse, and 20 crowns for each ox, and he finds that the oxen cost him seven crowns more than the horses. How many did he buy of each?

Let $x$ be the number of horses, and $y$ that of the oxen; then, by the question,

$$20x = 31y + 7,$$

and $x = \frac{31y + 7}{20} = y + \frac{11y + 7}{20}.$

Therefore $\frac{11y + 7}{20}$ must be a whole number.

Let $\frac{11y + 7}{20} = v$, then $x = y + v$, and $y = \frac{20v - 7}{11}$.

Let $\frac{20v - 7}{11} = t$, then $y = v + t$ and $v = \frac{11t + 7}{9} = t + \frac{2t + 7}{9}$; therefore $\frac{2t + 7}{9}$ is a whole number.

Let $\frac{2t + 7}{9} = s$, then $v = t + s$ and $t = \frac{9s - 7}{2} = 4s + \frac{s - 7}{2}$; therefore $\frac{s - 7}{2}$ is a whole number.

Put $\frac{s - 7}{2} = r$, then $t = 4s + r$ and $s = 2r + 7$.

Having now no longer any fractions, we return to the values of $x$ and $y$ by the following series of equations, The least positive values of \( x \) and \( y \) will evidently be obtained by making \( r = -3 \), and innumerable other values will be had by putting \( r = -2, -1, 0, 1, \ldots \). Thus we have

\[ \begin{align*} x &= 5, 36, 67, 98, 129, 160, 191, 222, \ldots \\ y &= 3, 23, 43, 63, 83, 103, 123, 143, \ldots \end{align*} \]

each series forming an arithmetical progression, the common difference in the first being 31, and in the second 20.

316. If we consider the manner in which the numbers \( x, y \), in this example, are determined, from the succeeding quantities \( v, t, \ldots \), we shall immediately perceive that the coefficients of those quantities are the same as the succeeding quotients which arise in the arithmetical operation for finding the greatest common measure of 20 and 31, the coefficients of the given equation \( 20x = 31y + 7 \). The operation performed at length will stand thus:

\[ \begin{array}{c} 20)31(1 \\ -20 \\ \hline 11)20(1 \\ -11 \\ \hline 9)11(1 \\ -9 \\ \hline 2)9(4 \\ -8 \\ \hline 1)2(2 \\ -2 \\ \hline 0 \end{array} \]

Hence we may form a series of numeral equations which, when compared with the series of literal equations expressing the relations between \( x, y, v, \ldots \), as put down in the following table, will render the method of determining the latter from the former sufficiently obvious.

\[ \begin{align*} 31 &= 1 \times 20 + 11 \\ 20 &= 1 \times 11 + 9 \\ 11 &= 1 \times 9 + 2 \\ 9 &= 4 \times 2 + 1 \\ 2 &= 2 \times 1 + 0 \\ \end{align*} \]

And as every question of this kind may be analyzed in the same manner, we may hence form the following general rule for resolving indeterminate problems of the first order.

317. Let \( bx = ay + n \) be the proposed equation, in which \( a, b, n \) are given integers, and \( x, y \) numbers to be found. Let \( A \) denote the greatest multiple of \( b \) which is contained in \( a \), and \( c \) the remainder; also let \( B \) denote the greatest multiple of \( c \) contained in \( b \), and \( d \) the remainder; and \( C \) the greatest multiple of \( d \) contained in \( c \), and \( e \) the remainder; and so on, till one of the remainders be found equal to 0. The numbers \( A, B, C \) afford a series of equations from which another series may be derived, as in the following table:

\[ \begin{align*} a &= Ab + c \\ b &= Be + d \\ c &= Cd + e \\ d &= De + f \\ e &= Ef + g \\ f &= Fg + o \\ \end{align*} \]

and in the last equation of the second series any number whatever may be put for \( q \); it is also to be observed, that the given number \( n \) is to have the sign \( + \) prefixed to it, if the number of equations be odd, but \( - \) if that number be even. Having formed the second series of equations, the values of \( x \) and \( y \) may be thence found as in the foregoing examples. We proceed to shew the application of the rule.

Ex. 5. Required a number which being divided by 11 leaves the remainder 3, but being divided by 19 leaves the remainder 5.

Let \( N \) be the number, and \( x, y \) the quotients which arise from the respective divisions, then we have \( N = 11x + 3 \), also \( N = 19y + 5 \); hence \( 11x + 3 = 19y + 5 \), and \( 11x = 19y + 2 \), an equation which furnishes the following table:

\[ \begin{align*} 19 &= 1 \times 11 + 8 \\ 11 &= 1 \times 8 + 3 \\ 8 &= 2 \times 3 + 2 \\ 3 &= 1 \times 2 + 1 \\ 2 &= 2 \times 1 + 0 \\ \end{align*} \]

Here \( r \) may be assumed of any value whatever.

Hence we have

\[ \begin{align*} s &= 2r + 2 \\ t &= s + r = 3r + 2 \\ v &= 2t + s = 8r + 6 \\ y &= v + t = 11r + 8 \\ x &= y + v = 19r + 14 \\ \end{align*} \]

and the number required \( N = 200r + 157 \), where it is evident that the least number which can express \( N \) is 157.

Ex. 6. \[ \begin{cases} 3x + 5y + 7z = 560 \\ 9x + 25y + 49z = 2920 \end{cases} \] To determine \( x, y, z \) in whole numbers.

Given \[ \begin{cases} 3x + 5y + 7z = 560 \\ 9x + 25y + 49z = 2920 \end{cases} \]

From 7 times the first equation subtract the second; thus we have \( 12x + 10y = 1000 \), or \( 6x + 5y = 500 \); and from this last equation, by proceeding as in the foregoing example, we find

\[ \begin{align*} x &= 500 - 5v \\ y &= 6v - 500 \end{align*} \]

Let these values of \( x \) and \( y \) be substituted in either of the original equations; in the first, for example, as being the most simple, and we find \( 7x + 15y = 1560 \).

This last equation being resolved in the same manner, we find and hence it appears that the only values which \( t \) can have so as to give whole positive numbers for \( x, y, z \) are 209 and 210; thus we have

\[ x = 15, \quad y = 82, \quad z = 15 \\ \text{or} \quad x = 50, \quad y = 40, \quad z = 30. \]

318. If an equation was proposed involving three unknown quantities, as \( ax + by + cz = d \), by transposition we have \( ax + by = d - cz \), and, putting \( d - cz = d' \), \( ax + by = c' \). From this last equation we may find values of \( x \) and \( y \) of this form

\[ x = mr + nc', \quad y = m'r + n'(d - cz) \]

or \( x = mr + n(d - cz), \quad y = m'r + n'(d - cz) \)

where \( r \) and \( r' \) may be taken at pleasure, except in so far as the values of \( x, y, z \) may be required to be all positive, for from such restriction the values of \( z \) and \( r \) may be confined within certain limits to be determined from the given equation.

319. We proceed to indeterminate problems of the second degree. These produce equations of the three following forms,

I. \( y = \frac{a}{b + cx}, \quad \text{II. } y = \frac{a + bx}{c + dx}, \quad \text{III. } y = \sqrt{\frac{a + bx + cx^2}{b + cx}}. \)

In all these equations \( a, b, c \) denote given numbers; in the two first \( x \) is to be determined so that \( y \) may be an integer, and in the third \( x \) is to be determined so that \( y \) may be a rational quantity.

320. In the equation \( y = \frac{a}{b + cx} \) it is evident \( b + cx \) must be a divisor of \( a \); let \( d \) be one of its divisors, then \( b + cx = d \), and \( x = \frac{d - b}{c} \): hence, to find \( x \) we must search among the divisors of \( a \) for one such that if \( b \) be subtracted from it the remainder may be divisible by \( c \), and the quotient will be such a value of \( x \) as is required.

321. When \( y = \frac{a + bx}{c + dx} \), if \( d \) be a divisor of \( b \), \( x \) will be taken out of the numerator, if we divide it by \( c + dx \), and this form is then reduced to the preceding. But if \( d \) is not a divisor of \( b \), multiply both sides by \( d \), then \( dy = da + dbx \) or \( dy = b + ad - bc \), and so \( x \) is found by making \( c + dx \) equal to a divisor of \( ad - bc \).

Example. Given \( x + y + 2xy = 195 \), to determine \( x \) and \( y \) in whole numbers.

From the given equation \( y = \frac{195 - x}{1 + 2x} \); therefore

\[ 2y = \frac{390 - 2x}{1 + 2x} = -1 + \frac{391}{1 + 2x}. \]

Now \( 391 = 17 \times 23 \); hence we must assume \( 1 + 2x = 17 \), or \( 1 + 2x = 23 \): the first supposition gives us \( x = 8, y = 11 \); and the second \( x = 11, y = 8 \), the same result in effect as the former.

322. It remains to consider the formula \( y = \sqrt{\frac{a + bx + cx^2}{b + cx}} \), where \( x \) is to be found so that \( y \) may be a rational quantity; but as the condition of having \( x \) and \( y \) also integers would add greatly to the difficulty of the problem, and produce researches of a very intricate nature, we must be satisfied for the most part with fractional values. The possibility of rendering the proposed formula a square depends altogether upon the coefficients \( a, b, c \); and there are four cases of the problem, the solution of each of which is connected with some peculiarity in their nature.

323. Case 1. Let \( a \) be a square number, then, putting \( s^2 \) for \( a \), we have \( y = \sqrt{\frac{s^2 + bx + cx^2}{b + cx}} \). Suppose \( s^2 + bx + cx^2 = g + mx \); then \( s^2 + bx + cx^2 = g + 2gmx + m^2x^2 \), or \( bx + cx^2 = 2gmx + m^2x^2 \), that is, \( b + cx = 2gm + m^2x \), hence

\[ x = \frac{2gm - b}{m^2}, \quad y = \sqrt{\frac{s^2 + bx + cx^2}{b + cx}} = \frac{cg - bm + gm^2}{m^2}. \]

Here \( m \) may be any rational quantity either whole or fractional.

324. Case 2. Let \( c \) be a square number \( = s^2 \), then putting \( \sqrt{a + bx + s^2x^2} = m + gx \), we find \( a + bx + s^2x^2 = m^2 + 2mgx + s^2x^2 \); or \( a + bx = m^2 + 2mgx \), hence we find

\[ x = \frac{m^2 - a}{b - 2mg}, \quad y = \sqrt{\frac{a + bx + s^2x^2}{b + cx}} = \frac{bm - gm^2 - ag}{b - 2mg}. \]

Here \( m \), as before, may be taken at pleasure.

325. Case 3. When neither \( a \) nor \( c \) are square numbers, yet, if the expression \( a + bx + cx^2 \) can be resolved into two simple factors, as \( f + gx \) and \( h + kx \), the irrationality may be taken away as follows.

Assume \( \sqrt{a + bx + cx^2} = \sqrt{(f + gx)(h + kx)} = m \); then \( (f + gx)(h + kx) = m^2(f + gx)^2 \), or \( h + kx = m^2(f + gx) \), hence we find

\[ x = \frac{fm^2 - h}{k - gm^2}, \quad y = \sqrt{(f + gx)(h + kx)} = \frac{(fk - gh)m}{k - gm^2} \]

and in these formulae \( m \) may be taken at pleasure.

326. Case 4. The expression \( a + bx + cx^2 \) may be transformed into a square as often as it can be resolved into two parts, one of which is a complete square, and the other a product of two simple factors; for then it has this form, \( p^2 + qr \), where \( p, q, r \) are quantities which contain no power of \( x \) higher than the first. Let us assume \( \sqrt{p^2 + qr} = p + mq \); thus we have \( p^2 + qr = p^2 + 2mpq + m^2q^2 \), and as this equation involves only the first power of \( x \), we may by proper reduction obtain from it rational values of \( x \) and \( y \), as in the three foregoing cases.

327. If we can by trials discover any one value of \( x \) which renders the expression \( \sqrt{a + bx + cx^2} \) rational, we may immediately reduce the quantity under the radical sign to the above-mentioned form, and thence find a general expression from which as many more values of \( x \) may be determined as we please. Thus let us suppose that \( p \) is a value of \( x \) which satisfies the condition determinate Problems.

\[ y^2 = a + bx + cx^2 \] \[ q^2 = a + bq + cp^2. \]

Therefore, by subtraction, \[ y^2 - q^2 = b(x - p) + c(x^2 - p^2) = (b + cp + cx)(x - p) \] and \( y = \sqrt{q^2 + (b + cp + cx)(x - p)}. \)

The quantity under the radical sign being now reduced to the prescribed form, it may be rendered rational by the substitution pointed out in the last article.

328. The application of the preceding general methods of resolution to any particular case is very easy; we shall therefore conclude with a very few examples.

Ex. 1. It is required to find two square numbers whose sum is a given square number.

Let \( a^2 \) be the given square number, and \( x^2, y^2 \), the numbers required. Then by the equation \( x^2 + y^2 = a^2 \), and \( y = \sqrt{a^2 - x^2} \). This equation is evidently of such a form as to be resolvable by the method employed in case 1. Accordingly, by comparing \( \sqrt{a^2 - x^2} \) with the general expression \( \sqrt{g^2 + bx + cx^2} \), we have \( g = a, b = 0, c = -1 \), and substituting these values in the formula of § 323, also \( n \) for \( m \), we find

\[ x = \frac{2an}{n^2 + 1}, \quad y = \frac{a(n^2 - 1)}{n^2 + 1}, \]

hence the numbers required are

\[ x^2 = \frac{4a^2n^2}{(n^2 + 1)^2}, \quad y^2 = \frac{a^2(n^2 - 1)^2}{(n^2 + 1)^2}. \]

If \( a = n + 1 \), where \( n \) is any number whatever, the square numbers \( x^2 \) and \( y^2 \) will both be integers, viz. \( x^2 = 4n^2 \) and \( y^2 = (n^2 - 1)^2 \). Let us suppose \( n = 2 \), then \( a = n^2 + 1 = 5 \), and \( a^2 = 25 \), hence \( x^2 = 4n^2 = 16 \), \( y^2 = (n^2 - 1)^2 = 9 \). Thus it appears that the square number 25 may be resolved into two other square numbers 9 and 16.

Ex. 2. It is required to find two square numbers whose difference shall be equal to a given square number \( b^2 \).

This question may be resolved in the same manner as the last. Or, without referring to any former investigation, let \( (x + n)^2 \) and \( x^2 \) be the numbers sought, then \( (x + n)^2 - x^2 = b^2 \), that is, \( 2nx + n^2 = b^2 \), hence

\[ x = \frac{b^2 - n^2}{2n}, \quad x + n = \frac{b^2 + n^2}{2n}. \]

So that the numbers sought are

\[ \left( \frac{b^2 + n^2}{4n^2} \right)^2, \quad \left( \frac{b^2 - n^2}{4n^2} \right)^2, \]

where \( n \) may be any number whatever. If, for example, \( b^2 = 25 \) and \( n = 1 \), then \( x = 12 \) and \( x + n = 13 \); so that the numbers required are 144 and 169.

Ex. 3. It is required to determine \( x \), so that \( \frac{x^2 + x}{2} \) may be a rational square.

Let \( y \) be the side of the square required, then \( \frac{x^2 + x}{2} = y^2 \) and \( 4x^2 + 4x + 1 = 8y^2 \). Let the first part of this equation be completed into a square by adding 1 to each side, then \( 4x^2 + 4x + 1 = 1 + 8y^2 \), and taking the root \( 2x + 1 = \sqrt{1 + 8y^2} \), so that we have to make \( 1 + 8y^2 \) a square. Assume

\[ 1 + 8y^2 = \left( 1 + \frac{p}{q} \right)^2 = 1 + \frac{2pq}{q^2} + \frac{p^2}{q^2} y^2, \]

then \( 8y = \frac{2pq}{q^2} + \frac{p^2}{q^2} y \). Hence by proper reduction \( y = \frac{2pq}{8q^2 - p^2} \), and

since \( 2x + 1 = \sqrt{1 + 8y^2} = \frac{8q^2 + p^2}{8q^2 - p^2} \),

therefore \( x = \frac{p^2}{8q^2 - p^2} \),

and \( \frac{x^2 + x}{2} = \frac{4p^2 q^2}{(8q^2 - p^2)^2} \), a rational square as was required.

Sect. XXIII. Of the Resolution of Geometrical Problems.

329. When a geometrical problem is to be resolved by algebra, the figure which is to be the subject of investigation must be drawn, so as to exhibit as well the known quantities, connected with the problem, as the unknown quantities, which are to be found. The conditions of the problem are next to be attentively considered, and such lines drawn, or produced, as may be judged necessary to its resolution. This done, the known quantities are to be denoted by symbols in the usual manner, and also such unknown quantities as can most easily be determined; which may be either those directly required, or others from which they can be readily found. We must next proceed to deduce from the known geometrical properties of the figure a series of equations, expressing the relations between the known and unknown quantities; these equations must be independent of each other, and as many in number as there are unknown quantities. Having obtained a suitable number of equations, the unknown quantities are to be determined in the same manner as in the resolution of numerical problems.

330. No general rule can be given for drawing the lines, and selecting the quantities most proper to be represented by symbols, so as to bring out the simplest conclusion; because different problems require different methods of solution. The best way to gain experience in this matter is to try the solution of the same problem in different ways, and then apply that which succeeds best to other cases of the same kind, when they afterwards occur. The following particular directions however may be of some use:

1. In preparing the figure by drawing lines, let them be either parallel or perpendicular to other lines in the figure, so as to form similar triangles. And if an angle be given, it will be proper to let the perpendicular be opposite to that angle, and to fall from one end of a given line, if possible.

2. In selecting the quantities for which symbols are to be substituted, those are to be chosen, whether required or not, which lie nearest the known or given parts of the figure, and by means of which the next adjacent parts may be expressed by addition and subtraction only, without the intervention of surds.

3. When two lines, or quantities, are alike related to other parts of the figure, or problem, the best way is to substitute for neither of them separately, but to substitute for their sum, or difference, or rectangle, or the sum of their alternate quotients, or some line or lines in the figure, to which they have both the same relation. 4. When the area or the perimeter of a figure is given, or such like parts of it as have only a remote relation to the parts required, it is sometimes of use to assume another figure similar to the proposed one, having one side equal to unity, or some other known quantity. For, from hence, the other parts of the figure may be found by the known proportions of like sides or parts, and so an equation will be obtained.

331. We shall now give the algebraical solutions of some geometrical problems.

Prob. 1. In a right-angled triangle, having given the base, and the sum of the hypotenuse and perpendicular, to find both these two sides.

Let ABC (Plate XIV. fig. 1.) represent the proposed triangle, right-angled at B. Let AB, the given base, be denoted by b, and AC + BC, the sum of the hypotenuse and perpendicular, by s; then if x be put for BC the perpendicular, the hypotenuse AC will be \(s - x\). But from the nature of a right-angled triangle \(AC^2 = AB^2 + BC^2\), that is,

\[b^2 + x^2 = (s - x)^2 = s^2 - 2sx + x^2.\]

Hence \(b^2 = s^2 - 2sx\), and \(x = \frac{s^2 - b^2}{2s} = BC\). Also

\[s - x = \frac{s^2 - b^2}{2s} = AC.\] Thus the perpendicular and hypotenuse are expressed by means of the known quantities \(b\) and \(s\), as required.

If a solution in numbers be required, we may suppose \(AB = b = 3\), and \(AC + CB = s = 9\), then

\[BC = \frac{s^2 - b^2}{2s} = 4,\] and \(AC = \frac{s^2 + b^2}{2s} = 5\).

Prob. 2. In a right-angled triangle, having given the hypotenuse, also the sum of the base and perpendicular, it is required to determine both these two sides.

Let ABC (fig. 1.) represent the proposed triangle, right-angled at B. Put \(a = AC\) the given hypotenuse, and \(s = AB + BC\) the given sum of the sides, then if \(x\) be put for \(AB\), the base, \(s - x\) will denote \(BC\) the perpendicular.

Now, from the nature of right-angled triangles, \(AC^2 = AB^2 + BC^2\); therefore \(x^2 + (s - x)^2 = a^2\), or \(x^2 + s^2 - 2sx + x^2 = a^2\); hence we have this quadratic equation \(x^2 - sx + \frac{a^2 - s^2}{2}\), which being resolved, by completing the square, we find \(x = \frac{s \pm \sqrt{2a^2 - s^2}}{2} = AB\), and

\[s - x = \frac{s \mp \sqrt{2a^2 - s^2}}{2} = BC.\] Thus it appears, that either of the two quantities \(\frac{s + \sqrt{2a^2 - s^2}}{2}, \frac{s - \sqrt{2a^2 - s^2}}{2}\) may be taken for \(AB\); but whichever of the two be taken, the remaining one is necessarily equal to \(BC\).

Prob. 3. It is required to inscribe a square in a given triangle.

Let ABC (fig. 2.) be the given triangle, and EFGH the inscribed square. Draw the perpendicular AD cutting EF the side of the square in K; then, because the triangle is given, the perpendicular AD may be considered as given. Let BC = b, AD = p, and, considering AK as the unknown quantity, (because from it the square may be readily determined), let AK = x; then KD = EF = p - x.

The triangles ABC, AEF, are similar; therefore \(AB : BC :: AK : EF\), that is, \(p : b :: x : p - x\). Hence, by taking the product of the extremes and means, \(p^2 - px = bx\), and \(x = \frac{p^2}{p + b} = AK\). If the side of the square be required, it may be immediately found by subtracting AK from AD the perpendicular. Thus we have \(p - \frac{p^2}{p + b} = \frac{pb}{p + b} = KD = EF\). Hence it appears that we may either take AK, a third proportional to AD + BC and AD, or take DK, a fourth proportional to AD + BC, AD and BC, and the point K being found, the manner of constructing the square is sufficiently obvious.

Prob. 4. Having given the area of a rectangle inscribed in a given triangle, it is required to determine the sides of the rectangle.

Let ABC (fig. 3.) be the given triangle, and EDGF the rectangle whose sides are required. Draw the perpendicular CI cutting DG in H. Put \(AB = b, CI = p, DG = EF = x, DE = HI = y\), then \(CH = p - y\). Let \(a^2\) denote the given area.

The triangles CDG, CAB are similar; hence

\[CH : DG :: CI : AB,\] or \(p - y : x :: p : b.\)

So that to determine \(x\) and \(y\), we have these two equations,

\[xy = a^2, \quad bp - by = px.\]

From the first equation we find \(y = \frac{a^2}{x}\), and from the second \(y = \frac{bp - px}{b}\), therefore \(\frac{bp - px}{b} = \frac{a^2}{x}\); hence \(x^2 - bx = \frac{a^2 b}{p}\), and, from this quadratic equation, by completing the square, &c. we find

\[x = \frac{b}{2} \pm \frac{\sqrt{b^2 - \frac{a^2 b}{p}}}{2}, \quad y = \frac{a^2}{x} = \frac{p}{2} \pm \frac{\sqrt{p^2 - \frac{pa^2}{b}}}{2}.\]

Hence it appears, that if \(\frac{a^2 b}{p}\) be less than \(\frac{b^2}{4}\), that is, if \(a^2\) be less than \(\frac{pb}{4}\), there are two different rectangles, having the same area, which may be inscribed in the given triangle. It also appears that, to render the problem possible, the given space \(a^2\) must not be greater than \(\frac{pb}{4}\), that is, than half the area of the given triangle.

Prob. 5. In a triangle, there are given the base, the vertical angle, and the sum of the sides about that angle, to determine each of these sides.

Let us suppose that ABC (fig. 4.) is the triangle, of which there is given the base AC, the vertical angle \(ABC\). Resolution \( \triangle ABC \), and the sum of the sides \( AB, BC \). Put \( AC = a \), \( AB + BC = b \), cofine of \( \angle ABC = c \), and let \( AB, BC \), the sides required, be denoted by \( x \) and \( y \).

Let \( CD \) be drawn from either of the angles at the base perpendicular to the opposite side \( AB \); then rad. : cof. \( B :: CB : BD \); therefore \( BD = \text{cof. } B \times CB = cy \).

Now, from the principles of geometry, \( AC^2 = AB^2 + BC^2 - 2AB \times BD \). Hence, and from the question, we have these two equations,

\[ x + y = b, \quad x^2 - 2xy + y^2 = a^2. \]

From the square of the first of these equations, viz. \( x^2 + 2xy + y^2 = b^2 \), let the second be subtracted, thus we have \( 2(1+c)xy = b^2 - a^2 \), and \( xy = \frac{b^2 - a^2}{1+c} \). Again, from the square of the first equation let the double of this last equation, viz. \( 4xy = 2(b^2 - a^2) \), be subtracted, and the result is \( x^2 - 2xy + y^2 = \frac{2a^2 - (1+c)b^2}{1+c} \), so that by taking the square root of this last equation we obtain

\[ x - y = \sqrt{\frac{2a^2 - (1+c)b^2}{1+c}}. \]

Thus we have found the difference between the sides, now their sum is given \( = b \); hence, by adding \( \frac{1}{2} \) the difference to \( \frac{1}{2} \) the sum, we find

\[ x = \frac{b}{2} + \frac{1}{2} \sqrt{\frac{2a^2 - (1+c)b^2}{1+c}}; \]

and subtracting \( \frac{1}{2} \) the difference from \( \frac{1}{2} \) the sum,

\[ y = \frac{b}{2} - \frac{1}{2} \sqrt{\frac{2a^2 - (1+c)b^2}{1+c}}. \]

If the angle at \( B \) be a right angle, this problem becomes the same as prob. 2.

332. By a method of investigation, in all respects similar to that which has been employed in these examples, any proposed geometrical problem may be reduced to an algebraic equation, the roots of which will exhibit arithmetical values of that geometrical magnitude which constitutes the unknown quantity in the equation. But the roots of algebraic equations may also be expressed by geometrical magnitudes, and hence a geometrical construction of a problem may be derived from its algebraic solution. For example, quadratic equations, which all belong to one or other of these three forms,

\[ x^2 + ax = bc, \quad x^2 - ax = bc, \quad x^2 - ax = -bc, \]

or, \( x(x+a) = bc, \quad x(x-a) = bc, \quad x(a-x) = bc \),

may be constructed as follow.

333. Construction of the first and second forms. Let a circle \( EABD \) (fig. 5.) be described with a radius \( = \frac{1}{2}a \), in which, from any point \( A \) in the circumference apply a chord \( AB = b - c \) (\( b \) being supposed greater than \( c \)), and produce \( AB \) so that \( BC = c \); then \( AC = b \).

Let \( H \) be the centre of the circle, then \( CH \) cutting the circumference in \( D \) and \( E \), then, in the first case, the positive value of \( x \) shall be represented by \( CD \), and in the second by \( CE \). For by construction \( DE = a \), therefore, if \( CD \) be called \( x \), then \( CE = x + a \), but if \( CE = x \), then \( CD = x - a \). Now, by the elements of geometry, \( EC \times CD = AC \times CB \), that is, \( x(x-a) = bc \), or \( x^2 - ax = bc \), which equation comprehends the first and second cases.

If the negative roots be required, that of the first case will be \( CE \) and that of the second \( CD \).

When \( b \) and \( c \) are equal the construction will be rather more simple, for then \( AB \) vanishing, \( AC \) will coincide with the tangent \( CF \). Therefore if a right-angled triangle \( HFC \) be constructed, whose legs \( HF \) and \( FC \) are equal respectively to \( \frac{1}{2}a \) and \( b \), then will \( CD \), the value of \( x \) in the first case, be equal to \( CH - HF \), and \( CE \), the value of \( x \) in the latter, \( = CH + HF \).

334. Construction of the third form.—Let a circle \( EADB \) (fig. 6.) be described with a radius \( = \frac{1}{2}a \) as before, in which apply a chord \( AB = b + c \), and take \( AC = b \). Through \( C \) draw the diameter \( DCE \), then either \( DC \) or \( EC \) will be positive roots of the equation. For since \( ED = a \), if either \( EC \) or \( CD = x \), the remaining part of the diameter shall be \( a - x \); now by the nature of the circle \( EC \times CD = AC \times CB \), that is, \( x(a-x) = bc \), or \( x^2 - ax = -bc \), hence it is evident that the roots are rightly determined.

If \( b \) and \( c \) are equal, the construction will be the same, only it will then not be necessary to describe the whole circle; for since \( AC \) will be perpendicular to the diameter, if a right-angled triangle \( HCA \) be constructed, having its hypotenuse \( HA = \frac{1}{2}a \) and base \( AC = b \), the roots of the equation will be expressed by \( AH + HC \) and \( AH - HC \).

335. If \( b \) and \( c \) be unequal, that \( b - c \) in the first two cases, or \( b + c \) in the third, is greater than \( a \), then instead of these quantities, \( \frac{1}{2}b \) and \( 2c \), or in general \( \frac{b}{n} \) and \( nc \) (where \( n \) is any number whatever) may be used. Or a mean proportional may be found between \( b \) and \( c \), and the construction performed as directed in each case when \( b \) and \( c \) are equal.

336. It appears from § 333 and 334, that every geometrical problem which produces a quadratic equation may be constructed by means of a straight line and a circle, or is a plane problem; hence, on the contrary, if a problem can be constructed by straight lines and circles, its algebraic resolution will not produce an equation higher than a quadratic. Cubic and biquadratic equations may be constructed geometrically by means of any two conic sections; hence it follows that every geometrical problem which requires for its construction two conic sections, will, when resolved by algebra, produce a cubic or biquadratic equation.

Sect. XXIV. Of the Loci of Equations.

337. When an equation contains two indeterminate quantities \( x \) and \( y \), then for each particular value of \( x \) there may be as many values of \( y \) as it has dimensions in that equation. So that if in an indefinite line \( AE \) (fig. 7.) there be taken a part \( AP \) to represent \( x \), and a perpendicular \( PM \) be drawn to represent \( y \), there will be as many points \( M, M', \ldots \) &c., the extremities of these perpendiculars, as there are dimensions of \( y \) in the proposed equation. And the values of \( PM, PM', \ldots \) will be the roots of the equation which are found by substituting for \( x \) its value in any particular case. Loci of Equations. Hence it appears that in any particular equation we may determine as many points M, as we please, and a line which passes through all these points is called the locus of the equation. The line AP which expresses any value of x is called an abscissa; and PM which expresses the corresponding value of y is called an ordinate. Any two corresponding values of x and y are also called co-ordinates.

338. When the equation that arises by substituting for x any particular value AP has all its roots positive, the points M, M', &c. will lie all on one side of AE, but if any of them be negative, they must be set off on the other side of AE towards m.

If x be supposed to become negative, then the line AP which represents it is to be taken in a direction opposite to that which represents the positive values of x; the points M, m, are to be taken as before, and the locus is only complete when it passes through all the points M, m, so as to exhibit a value of y corresponding to every possible value of x.

If in any case one of the values of y vanishes, then the point M coincides with P, and the locus meets AE in that point. If one of the values of y becomes infinite, then it flows that the curve has an infinite arc, and in that case the line PM becomes an asymptote to the curve, or touches it at an infinite distance, if AP itself is finite.

If when x is supposed infinitely great, a value of y vanishes, then the curve approaches to AE as an asymptote.

If any values of y become impossible, then so many points M vanish.

339. From these observations and the theory of equations, it appears that when an equation is proposed involving two indeterminate quantities x and y, there may be as many intersections of the curve that is the locus of the equation and of the line PM, as there are dimensions of y in the equation; and as many intersections of the curve and the line AE as there are dimensions of x in the equation.

340. A curve line is called geometrical or algebraic, when the equation which expresses the relation between x and y, any abscissa and its corresponding ordinate, consists of a finite number of terms, and contains besides these quantities only known quantities. Algebraic curves are divided into orders according to the dimensions of the equations which express the relations between their abscissas and ordinates, or according to the number of points in which they can intersect a straight line.

341. Straight lines themselves constitute the first order of lines, and when the equation expressing the relation between x and y is only of one dimension, the points M must be all found in a straight line which contains with AE a given angle. Suppose for example that the given equation is ay = bx - cd = 0, and that its locus is required.

Since \( y = \frac{bx + cd}{a} \), it follows that APM (fig. 8.) being a right angle, if AN be drawn making the angle NAP such that its cosine is to its sine as a to b, and drawing AD parallel to the ordinates PM, and equal to \( \frac{cd}{a} \), if DF be drawn parallel to AN, then will DF be the locus required; where it is to be observed that Loci of AD and PN are to be taken on the same side of AE Equations, if bx and cd have the same sign, but on opposite sides of AE if they have contrary signs.

342. These curves whose equations are of two dimensions constitute the second order of lines, and the first kind of curves. Their intersections with a straight line can never exceed two (\$ 339.)

The curves whose equations are of three dimensions form the third order of lines, and the second kind of curves; and their intersections with a straight line can never exceed three, and after the same manner curves of the higher orders are denominated.

Some curves, if they were completely described, would cut a straight line in an infinite number of points, but these belong to none of the orders we have mentioned, for the relation between their ordinates and abscissas cannot be expressed by a finite equation, involving only ordinates and abscissas with determinate quantities. Curves of this kind are called mechanical or transcendental.

343. As the roots of an equation become impossible always in pairs, so the intersections of a curve and its ordinate PM must vanish in pairs if any of them vanish. Let PM (fig. 9.) cut the curve in the points M and m, and by moving parallel to itself come to touch it in the point N, then the two points of intersection M and m go to form one point of contact N. If PM still move on parallel to itself, the points of intersection will beyond N become imaginary, as the two roots of an equation first become equal and then imaginary.

344. The curves of the 3d, 5th, 7th, orders, and all whose dimensions are odd numbers, have always one real root at least, and consequently for every value of x the equation by which y is determined must have at least one real root; so that as x, or AP, may be increased in infinitum on both sides, it follows that M must go off in infinitum on both sides without limit.

In curves whose dimensions are even numbers, as the roots of their equations may become all impossible, it follows that the figure of the curve may be like a circle or oval that is limited within certain bounds, beyond which it cannot extend.

345. When two roots of the equation by which y is determined become equal, either the ordinate PM touches the curve, two points of intersection in that case going into a point of contact, or the point M is a punctum duplex in the curve, two of its arcs intersecting each other there; or some oval that belongs to that kind of curve becoming infinitely little in M, it vanishes into what is called a punctum conjugatum.

If in the equation y be supposed = 0, then the roots of the equation by which x is determined, will give the distances of the points where the curve meets AE from A, and if two of those roots be found equal, then either the curve touches the line AE, or AE passes through a punctum duplex in the curve. When y is supposed = 0, if one of the values of x vanishes, the curve in that case passes through A. If two vanish, then either AE touches the curve in A, or A is a punctum duplex.

As a punctum duplex is determined from the equality of two roots, so is a punctum tripulum from the equality of three roots.

346. To To illustrate these observations we shall take a few examples.

Ex. 1. It is required to describe the line that is the locus of this equation \( y^2 = ax + ab \), or \( y^2 - ax - ab = 0 \), where \( a \) and \( b \) denote given quantities. Since \( y^2 = \pm \sqrt{ax + ab} \), if \( AP = x \) (fig. 10.) be assumed of a known value, and PM, PM' let off on each side equal to \( \sqrt{ax + ab} \), the points M, m will belong to the locus required; and for every positive value of AP there may thus be found a point of the locus on each side. The greater AP, or \( x \), is taken, the greater does \( \sqrt{ax + ab} \) become, and consequently PM and PM' the greater; and if AP be supposed infinitely great, PM and PM' will also become infinitely great, therefore the locus has two infinite arcs that go off to an infinite distance from AE and from AD. If \( x \) be supposed to vanish, then \( y = \pm \sqrt{ab} \), so that \( y \) does not vanish in that case, but passes through D and d, taking AD and Ad each \( = \sqrt{ab} \).

If P be supposed to move to the other side of A, then \( x \) becomes negative, and \( y = \pm \sqrt{ab - ax} \), so that \( y \) will have two values as before, while \( x \) is less than \( b \); but if \( AB = b \), and the point P be supposed to come to B, then \( ab = ax \), and \( y = \pm \sqrt{ab - ax} = 0 \); that is, PM and PM' vanish, and the curve there meets the line AE. If P be supposed to move from A beyond B, then \( x \) becomes greater than \( b \), and \( ax \) greater than \( ab \), so that \( ab - ax \) being negative, \( \sqrt{ab - ax} \) becomes imaginary; that is, beyond B there are no ordinates which meet the curve, and consequently on that side the curve is limited in B.

All this agrees very well with what is known by other methods, that the curve whose equation is \( y^2 = ax + ab \) is a parabola whose vertex is B, axis BE, and parameter equal to \( a \). For since \( b = x - BP \) and \( y = PM \), from the equation \( ab = ax + y^2 \), or \( a(b - x) = y^2 \), we have \( a \times BP = PM^2 \), which is the well-known property of the parabola.

Ex. 2. It is required to describe the line that is the locus of the equation \( xy + ay + cy = bc + bx \),

or \( y = \frac{bc + bx}{a + c + x} \).

Here it is evident (fig. 11.) that the ordinate PM can meet the curve in one point only, there being but one value of \( y \) corresponding to each value of \( x \). When \( x = 0 \), then \( y = \frac{bc}{a + c} \), so that the curve does not pass through A. If \( x \) be supposed to increase, then \( y \) will increase, but will never become equal to \( b \), since \( y = b \times \frac{c + x}{a + c + x} \), and \( a + c + x \) is always greater than \( c + x \).

If \( x \) be supposed infinite, then the terms \( a \) and \( c \) vanish compared with \( x \), and consequently \( y = b \times \frac{x}{x} = b \); from which it appears, that taking AD = \( b \), and drawing GD parallel to AE, it will be an asymptote, and touch the curve at an infinite distance. If \( x \) be now supposed negative, and AP be taken on the other side of A,

then \( y = b \times \frac{c - x}{a + c - x} \), and if \( x \) be taken on that side \( = c \), then \( y = b \times \frac{c - c}{a} = 0 \), so that the curve must pass through B if \( AB = c \). If \( x \) be supposed greater than \( c \), then \( c - x \) become negative, and the ordinate will become negative, and lie on the other side of AE, till \( x \) become equal to \( a + c \), and then \( y = b \times \frac{a}{c} \), that is, because the denominator is \( c \), \( x \) becomes infinite, so that if \( AK \) be taken \( = a + c \), the ordinate \( K_4 \) will be an asymptote to the curve.

If \( x \) be taken greater than \( a + c \) or AP greater than \( AK \), then both \( c - x \) and \( a + c - x \) become negative, and consequently \( y = b \times \frac{x - c}{x - a - c} \) becomes a positive quantity; and since \( x - c \) is always greater than \( x - a - c \), it follows that \( y \) will be always greater than \( b \) or \( KG \), and consequently the rest of the curve lies in the angle FGH. And as \( x \) increases, since the ratio of \( x - c \) to \( x - a - c \) approaches still nearer to a ratio of equality, it follows that PM approaches to an equality with PN, therefore the curve approaches to its asymptote GH on that side also.

The curve is the common hyperbola, for since \( b(c + x) = y(a + c + x) \), by adding \( ab \) to both sides, \( b(a + c + x) = y(a + c + x) + ab \), and \( (b - y)(a + c + x) = ab \), that is \( NM \times GN = GC \times BC \), which is the property of the common hyperbola.

Ex. 3. It is required to describe the locus of the equation \( ay^2 - xy = x^3 + bx^2 \).

Here \( y^2 = \frac{x^3 + bx^2}{a - x} \), and therefore \( y = \pm \sqrt{\frac{x^3 + bx^2}{a - x}} \),

hence PM and PM' (fig. 12.) are to be taken on each side, and equal to \( \sqrt{\frac{x^3 + bx^2}{a - x}} \); this expression, by supposing \( x = a \), becomes infinite because its denominator is then \( = 0 \), therefore if \( AB \) be taken \( = a \), and BK be drawn perpendicular to \( AB \), the line \( BK \) shall be an asymptote to the curve. If \( x \) be supposed greater than \( a \), or AP greater than \( AB \), then \( a - x \) being negative, the fraction \( \frac{x^3 + bx^2}{a - x} \) will become negative, and its square root impossible; so that no part of the locus can lie beyond B. If \( x \) be supposed negative, or \( P \) taken on the other side of A, then \( y = \pm \sqrt{\frac{-x^3 + bx^2}{c + x}} \), hence the values of \( y \) will be real and equal as long as \( x \) is less than \( b \), but if \( x = b \), then \( y = \pm \sqrt{\frac{-b^3 + b^3}{a - b}} = 0 \), and consequently if \( AD \) be taken \( = b \), the curve will pass through D, and there touch... Loci of the ordinate. If \( x \) be taken greater than \( b \), then Equations.

\[ \pm \sqrt{-x^3 - ax^2 + bx} = 0 \]

becomes imaginary, so that no part of the curve is found beyond D. The portion between A and D is called a nodus. If \( y \) be supposed \( = 0 \), then will \( x^3 + bx^2 = 0 \) be an equation whose roots are \( -b, 0, 0 \), from which it appears that the curve passes twice through A, and has in A a punctum duplex. This locus is a line of the 3d order.

If \( b \) is supposed to vanish in the proposed equation, so that \( ay^2 - xy^3 = s^3 \), then will A and D coincide (fig. 13.), and the nodus vanishes, and the curve will have in the point A a cuspis, the two arcs AM and AM, in this case, touching one another in that point. This is the same curve which the ancients called the Clifford of Diocles.

If instead of supposing \( b \) positive, or equal to 0, we suppose it negative, the equation will be \( ay^2 - xy^3 = s^3 - bx^2 \), the curve will in this case pass through D as before (fig. 14.), and taking AB \( = a \), BK will be its asymptote. It will have a punctum conjugatum in A, because when \( y \) vanishes two values of \( x \) vanish, and the third becomes \( -b \) or AD. The whole curve, besides this point, lies between DQ and BK. These remarks are demonstrated after the same manner as in the first case.

347. If an equation have this form,

\[ y = ax^n + bx^{n-1} + cx^{n-2} + \ldots \]

and \( n \) is an even number, then will the locus of the equation have two infinite arcs lying on the same side of AE, (fig. 13.) ; for if \( x \) becomes infinite, whether positive or negative, \( x^n \) will be positive and \( ax^n \) have the same sign in either case, and as \( ax^n \) becomes infinitely greater than the other terms \( bx^{n-1}, \ldots \), it follows that the infinite values of \( y \) will have the same sign in these cases, and consequently the two infinite arcs of the curve will lie on the same side of AE.

But if \( n \) be an odd number, then when \( x \) is negative \( x^n \) will be negative, and \( ax^n \) will have the contrary sign to what it had when \( x \) is positive, and therefore the two infinite arcs will in this case lie on different sides of AE, as in fig. 16, and tend towards parts directly opposite.

348. If an equation have this form \( yx = a^{n+1} \), and \( n \) be an odd number, then when \( x \) is positive \( y = \frac{a^{n+1}}{x^n} \), but when \( x \) is negative \( y = -\frac{a^{n+1}}{x^n} \), so that this curve must all lie in the vertically opposite angles KAE, FAE, (fig. 17.) as the common hyperbola, FK, Fe being asymptotes.

But if \( n \) be an even number, then \( y \) is always positive whether \( x \) be positive or negative, because \( x^n \) in this case is always positive, and therefore the curve must all lie in the two adjacent angles KAE and KAE, (fig. 18.) and have AK and AE for its asymptotes.

349. If an equation be such as can be reduced into two other equations of lower dimensions, without affecting \( y \) or \( x \) with any radical sign, then the locus shall consist of the two loci of those inferior equations. Thus the locus of the equation \( y^2 - 2xy + by + x^2 - bx = 0 \),

which may be resolved into these two, \( x - y = 0 \), \( y - x \) Arithmetic \( + b = 0 \), is found to be two straight lines cutting the absciss AE (fig. 19.) in angles of 45° in the points A, B, whose distance \( AB = b \). In like manner some cubic equations can be resolved into three simple equations, and then the locus is three straight lines, or may be resolved into a quadratic and simple equation, and then the locus is a straight line and a conic section. In general, curves of the superior orders include all the curves of the inferior orders, and what is demonstrated generally of any one order is also true of the inferior orders. Thus, for example, any general property of the conic sections holds true of two straight lines as well as a conic section, particularly that the rectangles of the segments of parallels bounded by them will always be to one another in a given ratio.

350. From the analogy which subsists between algebraic equations and geometrical curves, it is easy to see that the properties of the former must suggest corresponding properties of the latter. Hence the principles of algebra admit of the most extensive application to the theory of curve lines. It may be demonstrated, for example, that the locus of every equation of the second order is a conic section; and, on the contrary, the various properties of the diameters, ordinates, tangents, &c., of the conic sections may be readily deduced from the theory of equations.

Sect. XXV. Of the Arithmetic of Sines.

351. The relations which subsist between the fines and cofines of any arches of a circle, and those of their sums, or differences, &c., constitute what is called the arithmetic of fines. This branch of calculation has its origin in the application of algebra to geometry, and is of great importance in the more difficult parts of the mathematics, as well as in their application to physics.

352. In treating this subject, it is necessary to attend to the following observations.

1. If the fines of all arches between 0° and 180° be supposed positive, the fines of arches between 180° and 360° must be considered as negative; again, the fines of arches between 360° and 540° will be positive, and those of arches between 540° and 720° negative, and so on.

2. If the cofines of arches between 0° and 90° be supposed positive, the cofines of arches between 90° and 270° must be considered as negative, and the cofines of arches between 270° and 450° positive, and so on.

3. When an arch changes from + to —, or from — to +, its fine undergoes a like change, but its cofine is the same as before.

The truth of these observations must be evident from this consideration, that when a line, taken in a certain direction, decreases till it becomes = 0, and afterwards increases, but in a contrary direction; then, if in the former state it was considered as positive, it must be negative in the latter, and contrariwise.

353. The following proposition may be considered as the foundation of the arithmetic of fines.

Let \( a \) and \( b \) denote any two arches of a circle.

Then, if radius be supposed = 1.

\[ \sin(a + b) = \sin a \times \cos b + \cos a \times \sin b. \] Let C be the centre of the circle (fig. 20.), and AB, BD the arches denoted by \(a\) and \(b\); then AD = \(a + b\); draw the radii CA, CB, CD, and the lines BE, BF, DG; then BE, BF, DG, are the fines of \(a\), \(b\), and \(a + b\), respectively; and CE, CF, CG their cosines. Join EF, and draw FH parallel to DG. Because the angles CEB, CFB are right angles, the points C, E, B, F are in the circumference of a circle, hence the angle FCB is equal to FEB; that is, to the alternate angle EFH; now CFB, EHF are both right angles, therefore the triangles CFB, EHF are similar, hence CF : CB = (CD) : FH : FE; but CF : CD :: FH : DG; therefore FH : FE :: FH : DG, hence FF = DG = fin. \((a + b)\). Because BFBC is a quadrilateral inscribed in a circle, from the elements of geometry, we have BC × EF = BE × CF + BF × CE; but BE = fin. \(a\), CF = cof. \(b\), BF = fin. \(b\), CE = cof. \(a\), BC = 1, and EF = DG = fin. \((a + b)\), therefore fin. \((a + b)\) = fin. \(a\) × cof. \(b\) + cof. \(a\) × fin. \(b\), as was to be proved.

354. If in the preceding theorem we suppose the arch \(b\) to become negative, then fin. \(b\) will also become negative. Thus we obtain a second theorem, viz.

\[ \text{Sin. } (a - b) = \text{fin. } a \times \text{cof. } b - \text{cof. } a \times \text{fin. } b. \]

Because cof. \((a + b)\) = fin. \((90^\circ - a) - b)\), and by the second theorem fin. \((90^\circ - a) - b)\) = fin. \((90^\circ - a)\) × cof. \(b\) — cof. \((90^\circ - a)\) × fin. \(b\) = cof. \(a\) × cof. \(b\) — fin. \(a\) × fin. \(b\), therefore

\[ \text{Theor. IX. } 2 \text{Cof. } a \times \text{fin. } na = \text{fin. } (n + 1) a + \text{fin. } (n - 1) a \] \[ \text{Theor. X. } 2 \text{Sin. } a \times \text{cof. } na = \text{fin. } (n + 1) a - \text{fin. } (n - 1) a \] \[ \text{Theor. XI. } 2 \text{Cof. } a \times \text{cof. } na = \text{cof. } (n + 1) a + \text{cof. } (n - 1) a \] \[ \text{Theor. XII. } 2 \text{Sin. } a \times \text{fin. } na = -\text{cof. } (n + 1) a + \text{cof. } (n - 1) a. \]

356. By means of the four last theorems, the powers and products of the fines and cosines of arches may be expressed in terms of the sums and differences of certain multiples of those arches.

Thus, if in Theor. XII. we suppose \(n = 1\), it becomes

\[ 2 \text{Sin.}^2 a = -\text{cof. } 2 a + 1. \]

To find the third power of fin. \(a\), let both sides of this equation be multiplied by 2 fin. \(a\), then 4 fin. \(a\) = 2 fin. \(a\) (—cof. \(2 a + 1\)), but 2 fin. \(a\) × cof. \(2 a = \text{fin. } 3 a - \text{fin. } a\), Theor. X. Therefore

\[ 4 \text{Sin.}^3 a = -\text{fin. } 3 a + 3 \text{fin. } a. \]

Again, for the fourth power, let both sides of the last equation be multiplied by 2 fin. \(a\), then 8 fin. \(a\) = 2 fin. \(a\) (—fin. \(3 a + 3 \text{fin. } a\)); but 2 fin. \(a\) × fin. \(3 a = -\text{cof. } 4 a + \text{cof. } 2 a\), and 2 fin. \(a\) × fin. \(a = -\text{cof. } 2 a + 1\); Theor. XII.; therefore by substitution

\[ 8 \text{Sin.}^4 a = \text{cof. } 4 a - 4 \text{cof. } 2 a + 3. \]

Proceeding in this way, the successive powers of fin. \(a\) may be calculated as in the following table:

| Sin. \(a\) | fin. \(a\) | |------------|------------| | 2 Sin. \(a\) | -cof. \(2 a + 1\) | | 4 Sin. \(a\) | -fin. \(3 a + 3 \text{fin. } a\) | | 8 Sin. \(a\) | cof. \(4 a - 4 \text{cof. } 2 a + 3\) | | 16 Sin. \(a\) | fin. \(5 a - 5 \text{fin. } 3 a + 10 \text{fin. } a\) | | 32 Sin. \(a\) | -cof. \(6 a + 6 \text{cof. } 4 a - 15 \text{cof. } 2 a + 10\) | | 64 Sin. \(a\) | -fin. \(7 a + 7 \text{fin. } 5 a - 21 \text{fin. } 3 a + 35 \text{fin. } a\) |

&c. The successive powers of the cosines may be found in the same manner. Thus,

\[ \begin{align*} \text{Cof. } a &= \text{col. } a \\ 2 \text{ Cof. }^2 a &= \text{col. } 2a + 1 \\ 4 \text{ Cof. }^3 a &= \text{col. } 3a + 3 \text{ col. } a \\ 8 \text{ Cof. }^4 a &= \text{col. } 4a + 4 \text{ col. } 2a + 3 \\ 16 \text{ Cof. }^5 a &= \text{col. } 5a + 5 \text{ col. } 3a + 10 \text{ col. } a \\ 32 \text{ Cof. }^6 a &= \text{col. } 6a + 6 \text{ col. } 4a + 15 \text{ col. } 2a + 10 \\ 64 \text{ Cof. }^7 a &= \text{col. } 7a + 7 \text{ col. } 5a + 21 \text{ col. } 3a + 35 \text{ col. } a \\ &\text{etc.} \end{align*} \]

357. As an example of the products of the fines and cosines of an arch, let it be proposed to express fin. \(3a\) \(x\) col. \(a\) by the fines or cosines of multiples of \(a\). We have already found \(4 \text{ fin. }^3 a = -3 \text{ fin. } 3a + 3 \text{ fin. } a\), therefore

\[ \begin{align*} 16 \text{ fin. }^3 a \times \text{ col. }^2 a &= 2 \text{ col. } a \times 2 \text{ col. } a(-3 \text{ fin. } 3a + 3 \text{ fin. } a) \\ &= 2 \text{ col. } a(-\text{ fin. } 4a + 2 \text{ fin. } 2a) \\ &= -\text{ fin. } 5a + \text{ fin. } 3a + 2 \text{ fin. } a. \end{align*} \]

Thus it appears that all positive integer powers of the fine and cosine of an arch, or any product of those powers, may be expressed in finite terms by the fines and cosines of multiples of that arch.

358. On the contrary, the fine and cosine of any arch may be expressed by the powers of the fine and cosine of an arch whereof it is a multiple. For it appears from the 9th and 11th theorems that

\[ \begin{align*} \text{Sin. } (n+1)a &= 2 \text{ col. } a \times \text{ fin. } na - \text{ fin. } (n-1)a, \\ \text{Cof. } (n+1)a &= 2 \text{ col. } a \times \text{ col. } na - \text{ col. } (n-1)a, \end{align*} \]

therefore, by taking \(n = 0, 1, 2, 3, \ldots\) successively, we have

\[ \begin{align*} \text{Sin. } a &= \text{ fin. } a \\ \text{Sin. } 2a &= 2 \text{ col. } a \times \text{ fin. } a \\ \text{Sin. } 3a &= 2 \text{ col. } a \times \text{ fin. } 2a - \text{ fin. } a \\ \text{Sin. } 4a &= 2 \text{ col. } a \times \text{ fin. } 3a - \text{ fin. } 2a \\ \text{Sin. } 5a &= 2 \text{ col. } a \times \text{ fin. } 4a - \text{ fin. } 3a \\ &\text{etc.} \end{align*} \]

\[ \begin{align*} \text{Cof. } a &= \text{ col. } a \\ \text{Cof. } 2a &= 2 \text{ col. } a \times \text{ col. } a - 1 \\ \text{Cof. } 3a &= 2 \text{ col. } a \times \text{ col. } 2a - \text{ col. } a \\ \text{Cof. } 4a &= 2 \text{ col. } a \times \text{ col. } 3a - \text{ col. } 2a \\ \text{Cof. } 5a &= 2 \text{ col. } a \times \text{ col. } 4a - \text{ col. } 3a \\ &\text{etc.} \end{align*} \]

So that, putting \(s\) for the fine, and \(c\) for the cosine of the arch \(a\), and remarking that \(c^2 = 1 - s^2\),

\[ \begin{align*} \text{Sin. } a &= s \\ \text{Sin. } 2a &= 2cs \\ \text{Sin. } 3a &= 4c^3s - 4s^3 + 3s \\ \text{Sin. } 4a &= 8c^3s - 4c^3s(-8s^3 + 4s) \\ \text{Sin. } 5a &= 16c^3s - 12c^3s + s = 16s^3 - 20s^3 + 5s \\ &\text{etc.} \end{align*} \]

\[ \begin{align*} \text{Cof. } a &= c \\ \text{Cof. } 2a &= 2c^2 - 1 \\ \text{Cof. } 3a &= 4c^3 - 3c \\ \text{Cof. } 4a &= 8c^3 - 8c^3 + 1 \\ \text{Cof. } 5a &= 16c^3 - 20c^3 + 5c \\ &\text{etc.} \end{align*} \]

356. If it be required to find the fine or cosine of an arch, from having given the fine or cosine of some multiple of that arch, it may be found by resolving an equation of an order denoted by the numerical coefficient of the multiple arch. Thus if the cosine of half the arch, let \(C\) denote the given cosine, and \(x\) that which is required, then the equation \(c = 2x^2 - 1\) becomes \(C = 2x^2 - 1\), which equation being resolved, gives

\[ x = \pm \sqrt{\frac{1+C}{2}}. \]

If the fine be required, from that of twice the arch being given, it may be found from the equation \(2a = 2c\), which, by putting \(S\) for the given fine, and \(y\) for the fine required, becomes \(S = 2y\sqrt{1-y^2}\), or, by squaring both sides, and reducing,

\[ y^2 - y^4 = S^2; \quad \text{whence} \quad y^2 = \frac{1 \pm \sqrt{1-S^2}}{2}, \]

and

\[ y = \pm \sqrt{\frac{1 \pm \sqrt{1-S^2}}{2}}. \]

The two values of \(x\) indicate that there are two arches, the one as much less than \(90^\circ\), as the other exceeds \(90^\circ\), such, that the cosine of the double of each is expressed by the same number. And the four values of \(y\) show that there are four arches, viz. two positive and two negative, such, that the fine of the double of each is expressed by the same number.

Suppose now that the cosine of an arch is given to find the cosine of one-third of that arch, then, putting \(C\) to denote the given cosine, and \(x\) that which is required, the equation to be resolved is

\[ 4x^3 - 3x = C, \quad \text{or} \quad x^3 - \frac{3}{4}x - \frac{C}{4} = 0. \]

By comparing this cubic equation with the general equation \(x^3 + qx + r = 0\), it appears that \(q\) is negative and such that \(4q^3 = 27r^2\), for \(C\) is always less than unity; hence it follows that the equation belongs to the irreducible case, or that which cannot be resolved by Cardan's rule. The equation \(4 \text{ fin. }^3 - 3 \text{ fin. } a = -\text{ fin. } 3a\) is also of the same form; in order, therefore, to find either the fine or cosine of one-third of a given arch recourse must be had to the methods of approximation explained in Sect. XVI.

360. The sum of any powers of the fines or cosines of arches which constitute the arithmetical progression \(a, a+d, a+2d, a+3d, \ldots\) to \(a+nd\), may be be found as follows. We have already found, therefore, by substituting \(a, a+d, a+2d, \ldots\) successively for \(p\), we obtain the following series of equations:

\[ \begin{align*} \sin a &= \sin a \\ \sin (a+d) &= 2 \cos d \times \sin a - \sin (a-d) \\ \sin (a+2d) &= 2 \cos d \times \sin (a+d) - \sin a \\ \sin (a+3d) &= 2 \cos d \times \sin (a+2d) - \sin (a+d), \\ &\vdots \end{align*} \]

Therefore, if we substitute

\[ S = \sin a + \sin (a+d) + \sin (a+2d) + \ldots + \sin (a+nd), \]

by taking the sum of all the equations, it is evident that

\[ S + \sin (a+(n+1)d) = \sin a + 2 \cos d \times S - \sin (a-d) - (S - \sin (a+nd)), \]

which equation, by proper reduction, becomes

\[ S = \frac{\sin a - \sin (a+(n+1)d) + \sin (a+nd) - \sin (a-d)}{2(1-\cos d)}. \]

By proceeding in the same manner with Theor. VII. viz.

\[ \cos (p+d) = 2 \cos d \times \cos p - \cos (p-d), \]

and substituting \(a, a+d, a+2d, \ldots\) successively for \(p\); also putting

\[ C = \cos a + \cos (a+d) + \cos (a+2d) + \ldots + \cos (a+nd), \]

we obtain this other theorem

\[ C = \frac{\cos a - \cos (a+(n+1)d) + \cos (a+nd) - \cos (a-d)}{2(1-\cos d)}. \]

361. It is worthy of remark, that if the arch \(d\) is contained \(n+1\) times, either in the whole circumference, or any number of circumferences, that is, if \((n+1)d=q \times 360^\circ\), where \(q\) is any whole number; then \(nd=q \times 360^\circ-d\). Thus we have \(\sin (a+(n+1)d)\)

\[ = \sin (a+q \times 360^\circ) = \sin a, \text{ also } \sin (a+nd) = \sin (a-d+q \times 360^\circ) = \sin (a-d); \]

for the sine of any arch is equal to the sine of the same arch increased by any number of circumferences, and the same is true also of the cosine of an arch. Hence it appears that in these circumstances the terms in the numerators of the fractions, which are equal to \(S\) and \(C\), destroy one another, and thus \(S\) and \(C\) are both \(=0\); that is, the positive sines and cosines are equal to the negative sines and cosines, respectively. Now if the circumference of a circle be divided into \(n+1\) equal parts at the points \(A, A', A'', A'''\), &c. (fig. 21.), and any diameter \(BC\) drawn, then, if the arch \(BA=a\), and the arch \(AA'=d\), the arches \(BA', BA'A'', \ldots\) will be equal to \(a+d, a+2d, \ldots\) respectively; and, supposing the extremity of the diameter to fall between \(A\) and \(A'\), the arch \(BA, \ldots A'''\) will be equal to \(a+nd\). Hence we derive the following remarkable property of the circle. Let the circumference of a circle be divided into any number of equal parts at the points \(A, A', A'', \ldots\) and from the points of division let the fines \(AD, A'D', A''D'', \ldots\) be drawn upon any diameter \(BCE\); then, the sum of \(AD, A'D', \ldots\) the fines on one side of the diameter, shall be equal to the sum of \(A''D'', A'''D'''\), &c. the fines on the other side of the diameter. Also the sum of \(CD, C'D', \ldots\) the cosines on the side of the centre, shall be equal to the sum of \(C'D', C''D'', \ldots\) the cosines on the other side of the centre.

362. Let us next investigate the sum of the squares of the fines of the arches \(a, a+d, a+2d, \ldots\) For this purpose we may form a series of equations from the theorem

\[ 2 \sin^2 a = 1 - \cos 2a. \]

Thus we have

\[ \begin{align*} 2 \sin^2 a &= 1 - \cos 2a \\ 2 \sin^2 (a+d) &= 1 - \cos 2(a+d) \\ 2 \sin^2 (a+2d) &= 1 - \cos 2(a+2d), \\ &\vdots \end{align*} \]

Let Let $S' = \sin^2 a + \sin^2 (a+d) + \sin^2 (a+2d) + \ldots + \sin^2 (a+nd)$.

Then by addition, and observing that $\cos^2 2a + \cos^2 2(a+d) + \ldots + \cos^2 2(a+nd)$ is, by § 365,

$$\frac{\cos^2 2a - \cos^2 2(a+(n+1)d) + \cos^2 2(a+nd) - \cos^2 2(a-d)}{2(1-\cos^2 2d)},$$

we have

$$2S' = n - \frac{\cos^2 2a - \cos^2 2(a+(n+1)d) + \cos^2 2(a+nd) - \cos^2 2(a-d)}{2(1-\cos^2 2d)}.$$

In the same manner, by forming a series of equations from this theorem, $2 \cos^2 a = 1 + \cos 2a$, and, putting

$$\cos^2 a + \cos^2 (n+d) + \cos^2 (a+2d) + \ldots + \cos^2 (a+nd),$$

we find

$$-2C' = n + \frac{\cos^2 2a - \cos^2 2(a+(n+1)d) + \cos^2 2(a+nd) - \cos^2 2(a-d)}{2(1-\cos^2 2d)}.$$

363. If we now suppose $d$ to be such an arch that $(d+1)d = \text{the whole circumference} = 360^\circ$, then

$$\cos^2 2(a+(n+1)d) = \cos^2 (2a+2\times360^\circ) = \cos^2 2a,$$

also

$$\cos^2 2(a+nd) = \cos^2 (2a-d+2\times360^\circ) = \cos^2 2(a-d).$$

Thus it appears, that in this particular case the numerators of the fractional parts of the values of $2S'$ and $2C'$ are each $= 0$; and hence $2S'$ and $2C'$ are each $= n$.

We must except, however, the case of $n = 1$, for then $d = 180^\circ$, and $\cos^2 2d = 1$, so that the denominator of each fraction vanishing, as well as the numerator, it would be wrong to conclude that the fractions themselves vanish.

Now if the circumference of a circle be divided into $n+1$ equal parts at the points $A, A', A''$, &c. (fig. 21.), and any diameter $BE$, as also the lines $AD, A'D', A''D''$, &c., be drawn, then, if the arch $BA = a$, and the arch $AA' = d$, we have, as in § 361, $AD = \sin a$, and $A'D' = \sin (a+d)$, $A''D'' = \sin (a+2d)$, &c., and supposing the point $B$ to fall between $A$ and $A'$, $A''D'' = \sin (a+nd)$. Hence we derive the following very elegant and general theorem relating to the circle.

Let the circumference of a circle be divided into $n$ equal parts (where $n$ is any greater number than 2), at the points $A, A', A''$, &c.; and from the points of division let the lines $AD, A'D', A''D''$, &c., be drawn perpendicular to any diameter whatever. Twice the sum of the squares of the lines $AD, A'D', A''D''$, &c., is equal to $n$ times the square of the radius of the circle. Also twice the sum of the squares of the cosines $CD, CD', CD''$, &c., is equal to $n$ times the square of the radius of the circle.

364. We might now proceed to find the sum of the cubes of the lines of the arches $a, a+d, a+2d$, &c., from the equation.

$$\text{ch. } (n+1)a = \text{ch. sup. } a \times \text{ch. } na - \text{ch. sup. } (n-1)a$$

$$\text{ch. sup. } (n+1)a = \text{ch. sup. } a \times \text{ch. sup. } (n-1)a$$

367. Let $x = \text{chord of } a$, and $y = \text{chord of its supplement}$, then putting $0, 1, 2, 3$, &c. successively for $n$, and observing that $\text{ch. } o = 0$, we obtain from the first of these theorems the following series of equations:

$$\begin{align*} \text{ch. } a &= x \\ \text{ch. } 2a &= y \\ \text{ch. } 3a &= x(y^2 - 1) \\ \text{ch. } 4a &= x(y^3 - 2y) \\ \text{ch. } 5a &= x(y^4 - y^2 + 1) \\ \text{ch. } 6a &= x(y^5 - y^3 + 3y) \\ \text{ch. } 7a &= x(y^6 - 5y^4 + 6y^2 - 1), \\ &\text{&c.} \end{align*}$$

If $4 = x^2$, and the powers of that quantity be substituted for $y^2$, and its powers, in the chords of $3a, 5a, 7a$, Arithmetic of Sines.

7a, &c. also in the chords of the supplements of 2a, 4a, 6a, &c. we shall obtain the following series of equations, expressing the relations between the chord of any arch, and the chords of the multiples of that arch, if those multiples be odd numbers, or the chords of their supplements, if they be even numbers.

ch. \(a = +x\) ch. sup. \(2a = -x^2 + 2\) ch. \(3a = -x^3 + 3x\) ch. sup. \(4a = +x^4 - 4x^2 + 2\) ch. \(5a = +x^5 - cx^3 + 5x\) ch. sup. \(6a = -x^6 + 6x^4 - 9x^2 + 2\) ch. \(7a = -x^7 + 7x^5 - 14x^3 + 7x\) &c.

These equations are the foundation of the theory of angular sections, or method of dividing a given angle, or arch of a circle, into any proposed number of equal parts; a problem which evidently requires, for its general algebraic solution, the determination of the roots of an equation of a degree equal to the number of parts into which the arch is to be divided. By means of the same series of equations, we may also find the side of any regular polygon inscribed in a circle, and in this case the multiple arch, being equal to the whole circumference, will have its chord \(= 0\).

368. The relation between the tangents of any two arches, and that of their sum, may be readily found by means of the 1st and 3d theorems of this section. For since \(\tan(a + b) = \tan a \times \tan b + \cot a \times \cot b\), and \(\cot(a + b) = \cot a \times \cot b - \tan a \times \tan b\); therefore, dividing the former equation by the latter,

\[ \tan(a + b) = \frac{\tan a \times \tan b + \cot a \times \cot b}{\cot a \times \cot b - \tan a \times \tan b} \]

and hence the tangent of an arch being given, the tangent of any part of that arch, as its half, third, &c. may be found by the resolution of an equation.