ALGEBRA

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 Algebra e Almucabala, which is explained to denote the art of restitution and comparison, or opposition and comparison, or resolution 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 science, 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 six 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 Musa or son of Moses, 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. History.

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 Cosa, or "rule of the thing," where by Cosa, or the thing, was meant the first, or simple power of the unknown quantity.

7. Between the years 1470 and 1487 Lucas Paciolus 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 Proportioni et Proportionalia was printed. The part of this work, which relates to algebra, and which he calls l'Arte Maggiore; detta dal vulgo la Regola de la Cosa 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 numeral 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 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 resolution 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, shews 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 shewed 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 severally 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' Geometry, the science of algebra may be considered as having at-

tained some degree of perfection. It has, however, received many improvements from later writers, who, pursuing the paths struck out by Harriot and Descartes, have produced many new and beautiful theories, both in algebra and geometry. The writers upon algebra from this time became 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 Arithmetice, 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 these foreign mathematicians, Schooten, Hudde, Van-Heuraet, De Witte, Slufus, Huygens, &c. As to the algebraical writers of our own country, those whose labours have been most conspicuous were Wallis, and more especially Sir I. 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, are used to denote known quantities, while these to be found are represented by 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, $+a$ and $+b$ have like signs, but $+a$ and $-c$ 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 $+a$ and $-c$ 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 $\backslash times$ interpolated between every two of them. Thus $ab$, or $a\; \backslash times\; b$, denotes the product of $a$ and $b$; also $abc$, or $a\; \backslash times\; b\; \backslash times\; c$ 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 $\backslash times$ is interpolated between as before. Thus $a\; \backslash times\; \backslash overline\{c+d\}\; \backslash times\; \backslash overline\{e-f\}$ 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, $a\; \backslash times\; (c+d)\; \backslash times\; (e-f)$.

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 $\backslash frac\{12\}\{3\}$ denotes the quotient arising from the division of 12 by 3 or 4; $\backslash frac\{b\}\{a\}$ denotes the quotient arising from the division

of $b$ by $a$. This expression of a quotient is also called Addition 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.

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.

Sum, $5aa\; \backslash quad\; 0\; +\; 6xx$ Sum, $0$

Sum, $2a\; +\; 3b\; -\; 4c$

Sum, $ax\; +\; 2ay\; +\; bl\; -\; 3bz$

Remainder $3a\; -\; 7b$

Remainder $4x\; -\; 17y\; +\; 5$

$2xy\; +\; 6\; +\; 16x\; +\; 2y$

$aa\; -\; ax\; -\; xy\; -\; bb\; +\; by\; -\; xx$

Product $+ac$

$-20ab$

$-210abx$

$+6abcx$

Product $12aa\; -\; 6ab\; +\; 3ac$

$2xx\; +\; xy$

$2xx\; -\; 3xy\; -\; 2y^2$

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, sur-solid, 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 $a\; \backslash times\; a^3\; =\; a^4$, also $x^3\; \backslash times\; x^4\; =\; x^7$; and in general $a^m\; \backslash times\; a^n\; =\; a^\{m+n\}$.

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 expunge 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 $\backslash frac\{12abc\}\{3ac\}$; but the same quotient, by the rule just given, is more simply expressed thus $4b$.

Ex. 2. Divide $16a^3xy\; -\; 28a^2x^2y\; +\; 4a^2x^3$ by $4a^2x$.
The quotient is $4xy\; -\; 7x^2\; +\; x^2$.

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 $a^2$.

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 $2abc$ is the fraction $\backslash frac\{3ab^2\}\{2abc\}$.

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 the dividend are to be arranged according to the powers of some one of its last

terms, 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 divisor, 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 $8a^3\; +\; 2ab^2\; -\; 15b^3$ by $2a\; +\; 3b$; the operation will stand thus,

Here the terms of the divisor and dividend are arranged according to the powers of the quantity $a$. We now divide $8a^3$, 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 $8a^3\; +\; 12ab$ from the dividend; we thus get $-10ab^2\; -\; 15b^3$ 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.

Ex. 2.

41. Sometimes, as in this last example, the quotient will never terminate: in such a case it may either be considered as an infinite series, the law according to which the terms are formed being in general sufficiently obvious; or the quotient may be completed as in arithmetical division, by annexing to it a fraction, the numerator of which is the remainder, and denominator the divisor. Thus the quotient in last example may stand thus $1+x+x^2+\backslash frac\{x^3\}\{1-x\}$.

42. The reason of the rule for division is sufficiently manifest. For in the course of the operation, all the terms of the quotient obtained by it are multiplied by all the terms of the divisor, and the products successively subtracted from the dividend, till nothing remains; that therefore must evidently be the true quotient.

SECT. II. Of Fractions.

43. In the operation of division, the divisor may be sometimes less than the dividend, or may not be contained in it an exact number of times; in either case the quotient is expressed by means of a fraction. There can be no difficulty, however, in estimating the magnitude of such a quotient; if, for example it were the fraction $\backslash frac\{4\}\{7\}$, we may consider it as denoting either that some unit is divided into 7 equal parts, and that $\backslash frac\{4\}\{7\}$ of these are taken, or that $\backslash frac\{4\}\{7\}$ times the same unit is divided into 7 equal parts, and one of them taken.

44. In any fraction the upper number, or the dividend is called the numerator, and the lower number or the divisor is called the denominator. Thus in the fraction $\backslash frac\{a\}\{b\}$, $a$ is the numerator, and $b$ the denominator.

45. If the numerator be less than the denominator, such a fraction is called a proper fraction; but if the numerator be either equal to, or greater than the denominator, it is called an improper fraction; and if a quantity be made up of an integer and a fraction, it is called a mixed quantity. Thus $\backslash frac\{a\}\{a+x\}$ is a proper fraction; also $\backslash frac\{a+x\}\{a\}$ are both improper fractions; and $b+\backslash frac\{x\}\{a\}$ is a mixed quantity.

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 $\backslash frac\{b\}\{a\}$ is the reciprocal of the fraction $\backslash frac\{a\}\{b\}$.

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 $\backslash frac\{b\}\{a\}\; =\; c$; 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 $nb\; =\; nac$: let these equals be both divided by the same quantity $na$, and the quotients will be equal, that is $\backslash frac\{nb\}\{na\}\; =\; \backslash frac\{nac\}\{na\}$; 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 powers 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 remains, divide it also by its greatest simple divisor, and the last compound divisor by the resulting quantity: proceed in this way till nothing remains, 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 $a^3x-x^3$ and $a^3-2a^2x+ax^2$. The simple divisor $x$ being taken out of the former of these quantities, and $a$ out of the latter, they are reduced to $a^3-x^3$, and $a^3-2ax+x^2$, 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:

$a^3$

Hence it appears that $a\; -\; x$ is the greatest common measure required.

Ex. 2. Required the greatest common measure of $8a^3b^3\; -\; 10ab^3\; +\; 2b^4$, and $9a^2b\; -\; 9a^2b^2\; +\; 3a^2b^3\; -\; 3ab^4$.

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 $4a^3\; -\; 5ab\; +\; b^3$ and $3a^2\; -\; 3a^2b\; +\; ab^3\; -\; b^3$; 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:

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:

This remainder is to be divided by $-19b^2$, which being done, and the last divisor taken as a dividend as before, the rest of the operation will be as follows:

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 $np(a\; -\; x)$, $nq(a\; -\; x)$ 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

VOL. I. PART II.

by $ax$ and $bx$; let $q$ be the quotient, and the remainder will evidently be $bx\; -\; qax$, 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 $4a^3\; -\; 5ab\; +\; b^3$, and $12a^3\; -\; 12a^2b\; +\; 4ab^3\; -\; 4b^3$, the same must measure $3a^2b\; +\; ab^3\; -\; 4b^3$, the first remainder, as also $-19ab\; +\; 19ab^2$ 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 $3a^2b\; +\; ab^3\; -\; 4b^3$ the first remainder, therefore, by the third observation, $a\; -\; b$ must also be a divisor of $12a^3\; -\; 12ab\; +\; 3b^2$, or of $4a^3\; -\; 5ab\; +\; b^3$, the first divisor, and therefore also it must be a divisor of $12a^3\; -\; 12a^2b\; +\; 4ab^3\; -\; 4b^3$ 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. 1.

Ex. 1. Reduce $\backslash frac\{56a^3bc\}\{24a^2c^2\}$ 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 $\backslash frac\{56a^3bc\}\{24a^2c^2\}\; =\; \backslash frac\{7ab\}\{3ac\}$.

Ex. 2. Reduce $\backslash frac\{a^3x\; -\; x^3\}\{a^3\; -\; 2a^2x\; +\; ax^2\}$ to its lowest terms.

We have already found in the first example of prob. 1. that the greatest common measure of the numerator and denominator is $a\; -\; x$; and dividing both by this quantity we have

In like manner we find $\backslash frac\{9a^2b\; -\; 9a^2b^2\; +\; 3a^2b^3\; -\; 3ab^4\}\{8a^3b^3\; -\; 10ab^3\; +\; 2b^4\}\; =$

$\backslash frac\{9a^2\; +\; 3ab^2\}\{8ab\; -\; 2b^3\}$; the common measure being $b(a\; -\; b)$ as was shown in example 2. problem 1.

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.

4 H

Ex.

Fractions. Ex. 1. Let $x\; +\; \backslash frac\{x^2\}\{a\}$, and $x\; -\; \backslash frac\{a^2-x^2\}\{x\}$ be reduced to improper fractions.

Ex. 2. Reduce $a\; -\; x\; +\; \backslash frac\{x^2\}\{a+x\}$ to an improper fraction.

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 $\backslash frac\{ax+a^2\}\{x\}$ to a whole or mixed quantity.

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

And $\backslash frac\{x^2-y^2\}\{x-y\}\; =\; x+y$ 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 $\backslash frac\{a\}\{b\}$, $\backslash frac\{c\}\{d\}$ and $\backslash frac\{e\}\{f\}$ to fractions of equal value which have a common denominator.

Hence we find $\backslash frac\{a\}\{b\}\; =\; \backslash frac\{adf\}\{bdf\}$, $\backslash frac\{c\}\{d\}\; =\; \backslash frac\{cdf\}\{bdf\}$ and $\backslash frac\{e\}\{f\}\; =\; \backslash frac\{ebd\}\{bdf\}$, where the new fractions have a common denominator, as was required.

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

$\backslash frac\{ax(a+x)\}\{(a^2-x^2)(a-x)\}\; =\; \backslash frac\{a^2x+ax^2\}\{a^3-a^2x-ax^2+x^2\}$ new numerators. Fractions.

$a-x)(a+x)\; =\; a^2-x^2$ the common denominator.

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 $\backslash frac\{a\}\{b\}$, $\backslash frac\{c\}\{d\}$ and $\backslash frac\{e\}\{f\}$.

Hence $\backslash frac\{a\}\{b\}\; +\; \backslash frac\{c\}\{d\}\; +\; \backslash frac\{e\}\{f\}\; =\; \backslash frac\{adf+cdf+ebd\}\{bdf\}$ the sum required.

Ex. 2. From $\backslash frac\{a+x\}\{a\}$ subtract $\backslash frac\{a\}\{a+x\}$.

Ex. 3. Add together $\backslash frac\{x+2\}\{3\}$, $\backslash frac\{x\}\{4\}$ and $\backslash frac\{x-5\}\{2\}$.

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 $\backslash frac\{b\}\{a\}$ by $\backslash frac\{d\}\{c\}$.

Ex. 2. Multiply $\backslash frac\{a+b\}\{c\}$ by $\backslash frac\{a-b\}\{d\}$.

If it be required to multiply an integer by a fraction, the integer may be considered as having unity for a denominator. Thus $(a+x)\; \backslash times\; \backslash frac\{3d\}\{c\}\; =\; \backslash frac\{a+x\}\{1\}\; \backslash times\; \backslash frac\{3d\}\{c\}$

Mixed

Fractions. Mixed quantities may be multiplied after being reduced to the form of fractions by prob. 3. Thus

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

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

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 $\backslash frac\{a\}\{b\}$ by $\backslash frac\{c\}\{d\}$.

$\backslash frac\{c\}\{d\}\; \backslash div\; \backslash frac\{a\}\{b\}\; =\; \backslash frac\{ad\}\{bc\}$ the quotient required, or $\backslash frac\{a\}\{b\}\; \backslash times\; \backslash frac\{d\}\{c\}\; =\; \backslash frac\{ad\}\{bc\}$ as before.

Ex. 2. Divide $\backslash frac\{a^2\; +\; ab\}\{2x\}$ by $\backslash frac\{3a^2\}\{a-b\}$.

$\backslash frac\{3a^2\}\{a-b\}\; \backslash div\; \backslash frac\{a^2\; +\; ab\}\{2x\}\; =\; \backslash frac\{a^2\; -\; ab^2\}\{6a^2x\}\; =\; \backslash frac\{a^2\; -\; b^2\}\{6ax\}$ 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 $\backslash frac\{c\}\{d\}$ by $\backslash frac\{a\}\{b\}$. If

$\backslash frac\{c\}\{d\}$ is to be divided by $a$, the quotient is $\backslash frac\{c\}\{ad\}$, but if it is to be divided by $\backslash frac\{a\}\{b\}$, then the last quotient must be

multiplied by $b$; thus we have $\backslash frac\{cb\}\{ad\}$ for the quotient required. Or let $\backslash frac\{a\}\{b\}\; =\; m$, and $\backslash frac\{c\}\{d\}\; =\; n$, then $a\; =\; bm$ and $c\; =\; dn$; also $ad\; =\; bdm$ and $bc\; =\; bdn$; therefore $\backslash frac\{bdn\}\{bdm\}\; =\; \backslash frac\{n\}\{m\}\; =\; \backslash frac\{bc\}\{ad\}$.

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 $\backslash frac\{1\}\{a\}$, $\backslash frac\{1\}\{a^2\}$, and $\backslash frac\{1\}\{a^3\}$ denote the same powers of the root $\backslash frac\{1\}\{a\}$.

62. But before considering more particularly what relates to powers and roots, it will be proper to observe, that the quantities $\backslash frac\{1\}\{a\}$, $\backslash frac\{1\}\{a^2\}$, $\backslash frac\{1\}\{a^3\}$, &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 $\backslash frac\{1\}\{a\}$, $\backslash frac\{1\}\{a^2\}$, $\backslash frac\{1\}\{a^3\}$, &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 $\backslash frac\{1\}\{a\}$, $\backslash frac\{1\}\{a^2\}$, $\backslash frac\{1\}\{a^3\}$, 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 $\backslash frac\{1\}\{a\}$ as the quotient arising from the division of any power of $a$ by the next higher power, for example from the division of the 2d by the 3d, and so we have $\backslash frac\{1\}\{a\}\; =\; \backslash frac\{a^2\}\{a^3\}$; 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 $\backslash frac\{a^2\}\{a^3\}\; =\; a^\{2-3\}\; =\; a^\{-1\}$; therefore the fraction $\backslash frac\{1\}\{a\}$ may also be expressed thus, $a^\{-1\}$. By considering $\backslash frac\{1\}\{a^2\}$ as equal to $\backslash frac\{a^3\}\{a^5\}$, it will appear in the same manner that $\backslash frac\{1\}\{a^2\}\; =\; \backslash frac\{a^4\}\{a^6\}\; =\; a^\{-2\}$; and, proceeding in this way, we get $\backslash frac\{1\}\{a^3\}\; =\; \backslash frac\{a^4\}\{a^7\}\; =\; a^\{-3\}$, $\backslash frac\{1\}\{a^4\}\; =\; \backslash frac\{a^5\}\{a^9\}\; =\; a^\{-4\}$, &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 $1\; =\; \backslash frac\{a^1\}\{a^1\}\; =\; a^\{1-1\}\; =\; a^0$.

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 $\backslash frac\{1\}\{a^2\}\; \backslash times\; a^3$ or $a^\{-1\}\; \backslash times\; a^3\; =\; a^\{-1+3\}\; =\; a^2$, also $\backslash frac\{1\}\{x^3\}\; \backslash times$

Involution. $\backslash frac\{1\}\{x^3\}$ or $x^\{-3\}\; \backslash times\; x^\{-3\}\; =\; x^\{-6\}\; =\; \backslash frac\{1\}\{x^6\}$ and $\backslash frac\{1\}\{x^3\}\; \backslash times\; x^3$ or $x^\{-3\}\; \backslash times\; x^3\; =\; x^\{-3+3\}\; =\; x^0\; =\; 1$.

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 $\backslash frac\{a^4\}\{bc^3\}$ denotes the same thing as $\backslash frac\{a^4\; b^\{-1\}\}\{c^3\}$ or as $a^4\; b^\{-1\}\; c^\{-3\}$, also $\backslash frac\{a^4\}\{(x-1)^3\}$ may be otherwise expressed thus, $a^4(x-1)^\{-3\}$.

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 $2a^3x$. $(2a^3x)^3\; =\; 2\; \backslash times\; 2\; \backslash times\; 2\; \backslash times\; a^3\; \backslash times\; x^3\; \backslash times\; x^3\; =\; 8a^9x^6$, 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 $-\backslash frac\{2ax^2\}\{3b^2y\}$. $\backslash left(\backslash frac\{-2ax^2\}\{3b^2y\}\backslash right)^4\; =\; \backslash frac\{16a^4x^8\}\{81b^8y^4\}$ 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 $a+x$.

$a+x$ the root, or first power
$a+x$

$a^2+ax$
$+ax+x^2$

$a^3+2a^2x+ax^2$ the square, or second power
$a+x$

$a^3+2a^2x+ax^2$
$+a^2x+2ax^2+x^3$

$a^3+3a^2x+3ax^2+x^3$ the cube, or third power
$a+x$

$a^4+4a^3x+6a^2x^2+4ax^3$
$+a^2x+3a^2x^2+3ax^3+x^4$

$a^4+4a^3x+6a^2x^2+4ax^3+x^4$ the fourth power.

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

$a-x$ is the root, or first power;
$a^2-2ax+x^2$ the square, or 2d power;
$a^3-3a^2x+3ax^2-x^3$ the cube, or 3d power;
$a^4-4a^3x+6a^2x^2-4ax^3+x^4$ the 4th power.

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 $(a+x)^n\; =\; a^n\; +\; \backslash frac\{n\}\{1\}\; a^\{n-1\}x\; +\; \backslash frac\{n(n-1)\}\{1\; \backslash cdot\; 2\}\; a^\{n-2\}x^2\; +\; \backslash frac\{n(n-1)(n-2)\}\{1\; \backslash cdot\; 2\; \backslash cdot\; 3\}\; a^\{n-3\}x^3\; +\; \backslash frac\{n(n-1)(n-2)(n-3)\}\{1\; \backslash cdot\; 2\; \backslash cdot\; 3\; \backslash cdot\; 4\}\; a^\{n-4\}x^4\; +\; \backslash frac\{n(n-1)(n-2)(n-3)(n-4)\}\{1\; \backslash cdot\; 2\; \backslash cdot\; 3\; \backslash cdot\; 4\; \backslash cdot\; 5\}\; a^\{n-5\}x^5\; +\; \backslash dots$. This series will always terminate when $n$ is any whole positive number, by reason of some one of the factors $n-1$, $n-2$, &c. becoming $=0$; 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: $(a-x)^n\; =\; a^n\; -\; \backslash frac\{n\}\{1\}\; a^\{n-1\}x\; +\; \backslash frac\{n(n-1)\}\{1\; \backslash cdot\; 2\}\; a^\{n-2\}x^2\; -\; \backslash frac\{n(n-1)(n-2)\}\{1\; \backslash cdot\; 2\; \backslash cdot\; 3\}\; a^\{n-3\}x^3\; +\; \backslash frac\{n(n-1)(n-2)(n-3)\}\{1\; \backslash cdot\; 2\; \backslash cdot\; 3\; \backslash cdot\; 4\}\; a^\{n-4\}x^4\; -\; \backslash dots$, &c. 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 $\backslash frac\{n(n-1)\}\{1\; \backslash cdot\; 2\}\; a^\{n-2\}x^2\; =\; \backslash frac\{5\; \backslash times\; 4\}\{1\; \backslash times\; 2\}\; a^3x^2\; =\; 10a^3x^2$, the fourth $\backslash frac\{n(n-1)(n-2)\}\{1\; \backslash cdot\; 2\; \backslash cdot\; 3\}\; a^\{n-3\}x^3\; =\; \backslash frac\{5\; \backslash times\; 4\; \backslash times\; 3\}\{1\; \backslash times\; 2\; \backslash times\; 3\}\; a^2x^3\; =\; 10a^2x^3$, the fifth $\backslash frac\{n(n-1)(n-2)(n-3)\}\{1\; \backslash cdot\; 2\; \backslash cdot\; 3\; \backslash cdot\; 4\}\; a^\{n-4\}x^4\; =\; \backslash frac\{5\; \backslash times\; 4\; \backslash times\; 3\; \backslash times\; 2\}\{1\; \backslash times\; 2\; \backslash times\; 3\; \backslash times\; 4\}\; a^1x^4\; =\; 5ax^4$ and the sixth and last $\backslash frac\{n(n-1)(n-2)(n-3)(n-4)\}\{1\; \backslash cdot\; 2\; \backslash cdot\; 3\; \backslash cdot\; 4\; \backslash cdot\; 5\}\; a^\{n-5\}x^5\; =\; \backslash frac\{5\; \backslash times\; 4\; \backslash times\; 3\; \backslash times\; 2\; \backslash times\; 1\}\{1\; \backslash times\; 2\; \backslash times\; 3\; \backslash times\; 4\; \backslash times\; 5\}\; a^0x^5\; =\; x^5$; the remaining terms

Evolution. of the general theorem all vanish, by reason of the factor $n-1=0$ by which each of them is multiplied, so that we get $(a+x)^2\; =\; a^2\; +\; 2a^1x\; +\; 10a^1x^1\; +\; 10a^2x^1\; +\; 5a^3x^1\; +\; x^2$.

Ex. 2. It is required to raise $2d\; -\; \backslash frac\{2x\}\{3\}$ to the third power.

In this case $n=3$, so that if we put $a=2d$ and $x=\backslash frac\{2x\}\{3\}$ we have the first term of the general theorem, or $a^3\; =\; 8d^3$, the second $\backslash frac\{n\}\{1\}a^\{n-1\}x\; =\; 3\; \backslash times\; 4d^2\; \backslash times\; \backslash frac\{2x\}\{3\}\; =\; 6d^2x$, the third $\backslash frac\{n(n-1)\}\{1\; \backslash cdot\; 2\}\; a^\{n-2\}x^2\; =\; 3\; \backslash times\; 2d\; \backslash times\; \backslash frac\{4x^2\}\{9\}\; =\; \backslash frac\{2d^2x^2\}\{3\}$, and the fourth and last term $\backslash frac\{n(n-1)(n-2)\}\{1\; \backslash cdot\; 2\; \backslash cdot\; 3\}\; a^\{n-3\}x^3\; =\; \backslash frac\{3\; \backslash cdot\; 2\; \backslash cdot\; 1\}\{6\}\; \backslash cdot\; \backslash frac\{x^3\}\{27\}\; =\; \backslash frac\{x^3\}\{27\}$, and since the signs of the terms of any power of $a-x$ are $+$ and $-$ alternately we have $(2d\; -\; \backslash frac\{2x\}\{3\})^3\; =\; 8d^3\; -\; 6d^2x\; +\; \backslash frac\{2d^2x^2\}\{3\}\; -\; \backslash frac\{x^3\}\{27\}$.

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 $(p+q-r)^2\; =\; (a+b)^2\; =\; a^2\; +\; 2ab\; +\; b^2\; =\; a^2\; +\; 2p(q-r)\; +\; (q-r)^2$ but $2p(q-r)\; =\; 2pq\; -\; 2pr$, and by the general theorem $(q-r)^2\; =\; q^2\; -\; 2qr\; +\; r^2$, therefore, we get $(p+q-r)^2\; =\; a^2\; +\; 2pq\; -\; 2pr\; +\; q^2\; -\; 2qr\; +\; r^2$; 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 $\backslash sqrt\{\backslash phantom\{x\}\}$ (called the radical sign) is placed before it, and a small number placed over the sign to express the denomination of the root. Thus $\backslash sqrt\{a\}$ denotes the square root of $a$, $\backslash sqrt[3]\{a\}$ its cube root, $\backslash sqrt[4]\{a\}$ its fourth root, and in general, $\backslash sqrt[n]\{a\}$ 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 $\backslash sqrt\{a\}$ or $\backslash sqrt[2]\{a\}$ 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 Evolution.

• 2. The root of a negative quantity is also negative when the index of 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 $\backslash sqrt\{36a^2x^4\}\; =\; +6ax^2$ or $\backslash sqrt\{36a^2x^4\}\; =\; -6ax^2$, for either of these quantities, when multiplied by itself, produces $36a^2x^4$; so that the root required is $\backslash pm\; 6ax^2$, where the sign $\backslash pm$ 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 $\backslash sqrt[3]\{125a^6x^9\}\; =\; 5a^2x^3$ 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 $-a^3$, 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 $+a^6$, the root required is therefore said to be impossible, and may be expressed thus: $\backslash sqrt\{-a^3\}$.

The root of a fraction is found by extracting that root out of both numerator and denominator. Thus the

square root of $\backslash frac\{4a^2x^4\}\{9b^3y^2\}$ is $\backslash frac\{2ax^2\}\{3by\}$.

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

73. I. To extract the square root.

Range the terms of the quantity according 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 $a^2\; +\; 2ax\; +\; x^2$.

$a^2\; +\; 2ax\; +\; x^2$ $(a+x)$ the root required.

Ex. 2.

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 $c$.

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

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

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 $a^3\; +\; 3a^2x\; +\; 3ax^2\; +\; x^3$.

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.

Evolution. The operation as performed by the common rule (see ARITHMETIC) will stand thus:

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 $x^6\; +\; 6x^5\; -\; 40x^3\; +\; 96x\; -\; 64$

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 $x^3\; +\; 2x$ 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 $x^3\; +\; 2x$ to the second power, and multiplied the result by 3, we find 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 $x^3\; +\; 2x\; -\; 4$, when raised to the third power, gives a result the very same with the proposed power, we conclude $x^3\; +\; 2x\; -\; 4$ 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 assigned 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 $a^3\; +\; x^2$ will be found to be and the cube root of $a^3\; +\; x^3$ will stand thus 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 section 17, 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 $\backslash sqrt[3]\{a^2\}$, or thus $a^\{\backslash frac\{2\}\{3\}\}$.

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 often be more convenient to use the notation of fractional exponents; the following examples will shew how they may be expressed either way.

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 $a^\{\backslash frac\{m\}\{n\}\}\; =\; a^\{\backslash frac\{m\; \backslash cdot\; k\}\{n\; \backslash cdot\; k\}\}$. For let $a^\{\backslash frac\{m\}\{n\}\}\; =\; b$, then, raising both to the power $n$, $a^m\; =\; b^n$, and further raising both to the power $k$ we get $a^\{m\; \backslash cdot\; k\}\; =\; b^\{n\; \backslash cdot\; k\}$; let the root $kn$ be now taken and we find $a^\{\backslash frac\{m\; \backslash cdot\; k\}\{n\; \backslash cdot\; k\}\}\; =\; b^\{\backslash frac\{n\; \backslash cdot\; k\}\{n\; \backslash cdot\; k\}\}$.

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.

Here the exponent $\backslash frac\{1\}\{2\}$ must be reduced to the form of a fraction having 3 for a denominator, which will be the fraction $\backslash frac\{3\}\{6\}$; therefore $a^\{\backslash frac\{1\}\{2\}\}\; =\; a^\{\backslash frac\{3\}\{6\}\}\; =\; \backslash sqrt[6]\{a^3\}$.

Ex. 2. Reduce $5$ to the form of the cube root, and $3ab^2$ to the form of the square root.

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 $\backslash sqrt\{a\}$ and $\backslash sqrt[3]\{b^2\}$, or $a^\{\backslash frac\{1\}\{2\}\}$ and $b^\{\backslash frac\{2\}\{3\}\}$ to other equivalent surds of the same denomination.

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

Ex. 2. Reduce $3^\{\backslash frac\{1\}\{2\}\}$ and $2^\{\backslash frac\{1\}\{3\}\}$ to surds of the same denomination.

The new exponents are $\backslash frac\{1\}\{2\}$ and $\backslash frac\{1\}\{3\}$, therefore we have $3^\{\backslash frac\{1\}\{2\}\}\; =\; 3^\{\backslash frac\{3\}\{6\}\}\; =\; \backslash sqrt[6]\{3^3\}\; =\; \backslash sqrt[6]\{27\}$, and $2^\{\backslash frac\{1\}\{3\}\}\; =\; 2^\{\backslash frac\{2\}\{6\}\}\; =\; \backslash sqrt[6]\{2^2\}\; =\; \backslash sqrt[6]\{4\}$.

And in the same way the surds $A^\{\backslash frac\{1\}\{2\}\}$, $B^\{\backslash frac\{1\}\{3\}\}$ are reduced to these two $\backslash sqrt[6]\{A^3\}$ and $\backslash sqrt[6]\{B^2\}$.

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

Rule. Resolve 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 $\backslash sqrt\{48\}$ 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 $\backslash sqrt\{48\}\; =\; (4^2\; \backslash times\; 3)^\{\backslash frac\{1\}\{2\}\}\; =\; 4\; \backslash times\; 3^\{\backslash frac\{1\}\{2\}\}\; =\; 4\backslash sqrt\{3\}$.

Ex. 2. Reduce $\backslash sqrt\{98a^4x\}$, and $\backslash sqrt[3]\{24a^3x+40a^3x^2\}$, each to its most simple terms.

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

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

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, 3, 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 $\backslash sqrt\{27\}$ and $\backslash sqrt\{48\}$.

By prob. 3, we find $\backslash sqrt\{27\}\; =\; 3\backslash sqrt\{3\}$ and $\backslash sqrt\{48\}\; =\; 4\backslash sqrt\{3\}$, therefore $\backslash sqrt\{27\}\; +\; \backslash sqrt\{48\}\; =\; 3\backslash sqrt\{3\}\; +\; 4\backslash sqrt\{3\}\; =\; 7\backslash sqrt\{3\}$.

Ex. 2. Required the sum of $3\backslash sqrt[3]\{2\}$ and $5\backslash sqrt[3]\{2\}$.

$3\backslash sqrt[3]\{2\}\; =\; 3^\{\backslash frac\{1\}\{3\}\}\; \backslash sqrt[3]\{2\}\; =\; \backslash frac\{1\}\{3\}\; \backslash sqrt[3]\{2^3\}\; \backslash times\; \backslash sqrt[3]\{2\}\; =\; \backslash frac\{1\}\{3\}\; \backslash sqrt[3]\{2^4\}\; =\; \backslash frac\{1\}\{3\}\; \backslash sqrt[3]\{2^3\}\; \backslash sqrt[3]\{2\}$, therefore $3\backslash sqrt[3]\{2\}\; +\; 5\backslash sqrt[3]\{2\}\; =\; \backslash frac\{1\}\{3\}\; \backslash sqrt[3]\{2^3\}\; \backslash sqrt[3]\{2\}\; +\; \backslash frac\{1\}\{3\}\; \backslash sqrt[3]\{2^3\}\; \backslash sqrt[3]\{2\}\; =\; \backslash frac\{1\}\{3\}\; \backslash sqrt[3]\{2^3\}\; \backslash sqrt[3]\{2\}$.

Ex. 3. Required the difference between $\backslash sqrt\{80a^4x\}$ and $\backslash sqrt\{20a^3x^2\}$.

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

87. 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 $\backslash sqrt[3]\{a^2\}$ and $\backslash sqrt[5]\{a^3\}$.

Ex. 2. Divide $\backslash sqrt\{a^2\; -\; b^2\}$ by $\backslash sqrt[3]\{a+b\}$.

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

Ex. 3. Required the product of $5\backslash sqrt\{8\}$ and $3\backslash sqrt\{5\}$.

Ex. 4. Divide $8\backslash sqrt[3]\{56\}$ by $4\backslash sqrt[3]\{2\}$.

Ex. 5. Required the product of $x^\{\backslash frac\{1\}\{n\}\}$ and $x^\{\backslash frac\{1\}\{m\}\}$; also the quotient arising from the division of $a^\{\backslash frac\{1\}\{n\}\}$ by $b^\{\backslash frac\{1\}\{m\}\}$.

Surs 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 $3\backslash sqrt[3]\{3\}$ is $3\; \backslash times\; 3\; \backslash times\; (3)^\{\backslash frac\{1\}\{3\}\}\; =\; 9\backslash sqrt[3]\{3\}$. The $n$th power of $x^\{\backslash frac\{1\}\{n\}\}$ is $x^\{\backslash frac\{n\}\{n\}\}$. The cube root of $\backslash frac\{1\}\{8\}\backslash sqrt[4]\{2\}$ is $\backslash frac\{1\}\{2\}(2)^\{\backslash frac\{1\}\{4\}\}\; =\; \backslash sqrt[4]\{\backslash frac\{1\}\{2\}\}$ and the $n$th root of $x^\{\backslash frac\{1\}\{n\}\}$ is $x^\{\backslash frac\{1\}\{n\}\}$.

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

$9\; +\; 6\backslash sqrt\{5\}\; +\; 5\; =\; 14\; +\; 6\backslash sqrt\{5\}$ the square required.

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

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

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.

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 proportionals. Such, for example, are the numbers 2, 5, 9, 12; and, in general, the quantities $a,\; a+d,\; b,\; b+d$. If the two middle terms are equal, the quantities constitute what are called three arithmetical proportionals.

93. The most material property of four arithmetical proportionals 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 $a+d+b+d$; 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 proportionals 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 proportionals 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:

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 7, the last term is $a+6d$; and so on. Hence if $n$ denote the last term, $n$ the number of terms, and $a$ and $d$ express the first term, and common difference, we have $n=a+(n-1)d$.

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 $s=\backslash frac\{n\}\{2\}(a+x)$.

Geometrical Proportion. 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 $a=1$, $d=2$, $z=1+\backslash frac\{n\}\{2\}$, $(n-1)d=2n-1$, therefore $z=\backslash frac\{n\}\{2\}\; \backslash times\; 2n=n^2$.

II. Of Geometrical Proportion.

97. When of 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 12, 4, 15, 5, are four numbers in geometrical proportion; and in general, $na$, $a$, $nb$, $b$ may express any four proportionals, for $\backslash frac\{na\}\{a\}=n$, and also $\backslash frac\{nb\}\{b\}=n$.

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 $na$, $a$, $\backslash frac\{a\}\{n\}$. 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 $\backslash frac\{a\}\{b\}=\backslash frac\{c\}\{d\}$; let these equal quotients be multiplied by $bd$, and we have $\backslash frac\{abd\}\{b\}=\backslash frac\{cbd\}\{d\}$, 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 $a$, $x=\backslash frac\{bc\}\{a\}$. 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 be divided by $bd$, we get $\backslash frac\{ad\}\{bd\}=\backslash frac\{bc\}\{bd\}$ or $\backslash frac\{a\}\{b\}=\backslash frac\{c\}\{d\}$ hence it follows, from the definition

given of proportionals, (§ 97.) that $a:b::c:d$. From Geometrical Proportion. 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 $b:a::d:c$
• 2d, By alternation $a:c::b:d$*
• 3d, By composition $a+b:a::c+d:c$
  or $a+b:b::c+d:d$
• 4th, By division $a-b:b::c-d:c$
  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 $\backslash frac\{a\}\{n\}:\backslash frac\{b\}\{p\}::\backslash frac\{c\}\{n\}:\backslash frac\{d\}\{p\}$

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 $eh=fg$ (§ 100.), therefore, multiplying together these equal quantities $adeb=bcfg$, or $ae\; \backslash times\; db=bf\; \backslash times\; cg$, therefore by the second property (§ 100.), $ae:bf::cg:dh$.

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 $\backslash frac\{1\}\{2\},\; \backslash frac\{1\}\{4\},\; \backslash frac\{1\}\{8\},\; \backslash frac\{1\}\{16\}$, &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, $z=ar^\{n-1\}$.

* 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.

Reduction of Equations. 2. The product of the first and last term is equal to the product of any two terms equally distant from them; thus, supposing $ar^2$ the last term, it is evident that $a\; \backslash times\; ar^2\; =\; ar\; \backslash times\; ar^4\; =\; ar^2\; \backslash times\; ar^1$, &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 six, $s\; =\; a\; +\; ar\; +\; ar^2\; +\; ar^3\; +\; ar^4\; +\; ar^5$, and multiplying these equals by $r$, $rs\; =\; ar\; +\; ar^2\; +\; ar^3\; +\; ar^4\; +\; ar^5\; +\; ar^6$. If from the lower line, or $rs\; =\; ar\; +\; ar^2\; +\; \backslash dots\; +\; ar^5$, we subtract the upper line, or $s\; =\; a\; +\; ar\; +\; \backslash dots\; +\; ar^5$, the remainders will evidently be equal; but on the one side of the $r$, $n\; =$ we have $rs\; =\; a$, and on the other $ar^6\; =\; a$: therefore, $rs\; =\; s\; =\; ar^6\; =\; a$,

and dividing by $r-1$, $s\; =\; \backslash frac\{ar^6\; -\; a\}\{r-1\}$. 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, $s\; =\; \backslash frac\{ar^n\; -\; a\}\{r-1\}$, or $s\; =\; \backslash frac\{a(r^n\; -\; 1)\}\{r-1\}$.

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, $\backslash frac\{x\}\{2\}\; +\; \backslash frac\{x\}\{3\}\; =\; \backslash frac\{24\}\{x\}$, $2x\; +\; 3y\; =\; xy$; 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 $x^3\; -\; 2x^2\; +\; 4x\; =\; 10$, 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 resolution of simple equations.

115. Rule 1. Any quantity may be transposed 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 of Equations. the contrary without affecting the truth of the equation.

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.

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.

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.

Let all the terms be multiplied by 12, which is a multiple of 2, 3, and 4, and we have

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

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.

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.

121. Rule 5. 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.

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

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 $6x+6+4x+8=192-3x-9$, and hence, as before, $x=13$.

Ex. 4. Let $6x^3-20x^2=16x^2+2x^3$

Then dividing by $2x^3$, $3x-10=8+x$
And transposing, $3x-x=8+10$
Or $2x=18$
And therefore $x=9$.

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

Then $ax-b^2=cx$
And $ax-cx=b^2$
Whence $x=\backslash frac\{b^2\}\{a-c\}$.

Ex. 6. Let $x-6=\backslash frac\{x^3\}\{x+24\}$

Then $(x-6)(x+24)=x^3$
That is $x^3+18x-144=x^3$

Therefore $18x=144$
And $x=8$.

Ex. 7. Let $ax+b^2=\backslash frac\{ax^3+ac^2\}\{a+x\}$

Then $(a+x)(ax+b^2)=ax^3+ac^2$
Or $a^2x+ab^2+ax^3+b^2x=ax^3+ac^2$

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

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

Then $1-x=a+ax$
And $-x-ax=a-1$

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

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

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

Then by rule 4. $12+x=4+4\backslash sqrt\{x\}+x$
And by transposition $8=4\backslash sqrt\{x\}$
And by division $2=\backslash sqrt\{x\}$
And again by rule 4. $4=x$.

Ex. 10. Let $x+\backslash sqrt\{a^3+x^3\}=\backslash frac\{2a^3\}\{\backslash sqrt\{a^3+x^3\}\}$

Then, by rule 3. $x\backslash sqrt\{a^3+x^3\}+a^3+x^3=2a^3$
And by transposition, &c. $x\backslash sqrt\{a^3+x^3\}=a^3-x^3$

Therefore, by rule 4. $a^3x^3+x^3=a^3-2a^3x^3+x^3$

Whence $3a^3x^3=a^3$

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

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

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

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

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

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

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

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

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

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

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

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

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 $x=\backslash frac\{b^2-4a^2\}\{4a\}$.

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

Reduction of Equations. that quantity is wholly excluded. Let the same operation be now repeated with the new 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.

From the first equation $2x\; =\; 23\; -\; 3y$

From the second equation $5x\; =\; 10\; +\; 2y$

Let these values of $x$ be now put equal to each other.

And since $x\; =\; \backslash frac\{23-3y\}\{2\}$, or $x\; =\; \backslash frac\{10+2y\}\{5\}$, 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

From the first equation $x\; =\; \backslash frac\{23-3y\}\{2\}$

And this value of $x$ being substituted in the second equation, we have $5\; \backslash times\; \backslash frac\{23-3y\}\{2\}\; -\; 2y\; =\; 10$

127. Method 3. Let the given equations be multiplied or divided by such numbers or quantities, whether known or unknown, that the term which involves 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. Reduction of Equations.

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

and from these two equations we are to determine $x$ and $y$. To exterminate $x$, let the first equation be multiplied by 5, and the second by 2, thus we have

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

By adding these equations, we find

and therefore $x\; =\; 4$

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

By Method 1.

From the first equation we find $x\; =\; 32\; -\; \backslash frac\{2y\}\{3\}$

And from the second $x\; =\; 10\; +\; \backslash frac\{5y\}\{9\}$

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

By Method 2.

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

Reduction of Equations. 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

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

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 expeditious in practice.

It is required to determine $x$ and $y$.

From the 1st equation we have $4x-96=2y+64$.
And from the second, $12x+12y+20x-480=30y-15x+1620$.

These two equations when abridged become

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

To exterminate $y$, let the first equation be multiplied by $f$ and the second by $b$; and we have

Taking now the difference between these equations we find

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

and taking the difference as before, we get

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.

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

Let these values of $x$ be put equal to each other, thus we get the two following equations.

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

Here the given equations, when cleared from fractions, become

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:

Let

Reduction of Equations. Let the second equation be now subtracted from the first, and the third from the second, and we have

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

Subtracting now the latter equation from the former.

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

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

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 $x^2\; =\; b\; -\; y^2$.

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

Ex. 8. Given $\backslash begin\{cases\}\; axy\; +\; bx\; +\; cy\; =\; d\; \backslash \backslash \; fxy\; +\; gx\; +\; hy\; =\; k\; \backslash end\{cases\}$ To exterminate $y$.

From the first equation we find $y\; =\; \backslash frac\{d\; -\; bx\}\{ax\; +\; c\}$.

And from the second $y\; =\; \backslash frac\{k\; -\; gx\}\{fx\; +\; h\}$.

Therefore $\backslash frac\{d\; -\; bx\}\{ax\; +\; c\}\; =\; \backslash frac\{k\; -\; gx\}\{fx\; +\; h\}$, an equation in which the Simple Equations. unknown quantity $y$ is not found.

Ex. 9. Given $\backslash begin\{cases\}\; y^2\; -\; 3xy\; +\; ay\; =\; x^2\; \backslash \backslash \; y^2\; +\; 2ax\; -\; by\; =\; 4x^2\; -\; b^2\; \backslash end\{cases\}$ 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,

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

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

Therefore, taking the difference of these equations,

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

Therefore, subtracting as before, we get

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

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

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.

$(b^2\; +\; 2ax\; -\; 3x^2)^2\; =\; (a\; +\; b\; -\; 3x)(b^2\; -\; (a\; -\; 3x)(2ax\; -\; 4x^2\; +\; b^2))$; 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 relations 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 $x+y$ 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 $\backslash frac\{x-y\}\{y\}=3$, 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, $x+y=8$, $3x+6y=26$, 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,

VOL. I. PART II.

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 $=s$, 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 $x+d$, or by $x$, and $x-d$. If their product be given $=p$, the quantities are $x$, and $\backslash frac\{p\}\{x\}$; 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 $\backslash frac\{x\}\{2\}$, its third $\backslash frac\{x\}\{3\}$ and its fourth $\backslash frac\{x\}\{4\}$.

Therefore $x\; +\; \backslash frac\{x\}\{2\}\; +\; \backslash frac\{x\}\{3\}\; +\; \backslash frac\{x\}\{4\}\; =\; 50$.

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 $\backslash frac\{1\}\{2\}$ of its length in the mud, $\backslash frac\{1\}\{3\}$ 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 $\backslash frac\{x\}\{2\}$, and that in the water $\backslash frac\{x\}\{3\}$; therefore, from the nature of the question

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 $8x$ miles, and the other $7x$ 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 20d. but for every day he played, or was absent, he was to forfeit 8d.; now at the end

Simple Equations. of the time he had to received 11. 11s. 8d. It is required to find how many days he worked, and how many days he was idle.

Let $x$ be the number of days he worked.

Then will $40-x$ be the number of days he was idle.

Also $20\; \backslash times\; x\; =\; 20x$ = the sum he earned, in pence.

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

Now the difference of these two was 11. 11s. 8d. or 38d.

Therefore $20x\; -\; (320\; -\; 8x)\; =\; 38$

That is $28x\; =\; 700$

Hence $x\; =\; 25$ = the number of days he worked.

And $40\; -\; x\; =\; 15$ = 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 $\backslash frac\{x\}\{2\}$ be the price of the first sort,

And $\backslash frac\{x\}\{3\}$ = the price of the second sort.

Now the whole number being $2x$, we have

$5\; :\; 2x\; ::\; 2\; :\; \backslash frac\{4x\}\{5\}$ = price of both sorts at 5 for 2d.

Therefore $\backslash frac\{x\}\{2\}\; +\; \backslash frac\{x\}\{3\}\; -\; \backslash frac\{4x\}\{5\}\; =\; 4$, 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.

Therefore $12\; :\; 7\; ::\; x\; +\; 1920\; :\; x\; +\; 640$

And taking the product of the extremes, and means,

Hence $5x\; =\; 5760$ d. and $x\; =\; 1152$ d. = 4l. 16s.

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

Suppose he began play with $x$ shillings.

He lost $\backslash frac\{1\}\{2\}$ of his money, or $\backslash frac\{x\}\{2\}$, and had left $x\; -\; \backslash frac\{x\}\{2\}\; =\; \backslash frac\{x\}\{2\}$

He won 3s. and had then $\backslash frac\{3x\}\{4\}\; +\; 3\; =\; \backslash frac\{3x+12\}\{4\}$.

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

He won 2s. and had then $\backslash frac\{2x+8\}\{4\}\; +\; 2\; =\; \backslash frac\{2x+16\}\{4\}$.

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

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

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 $15\; :\; x\; ::\; 1\; :\; \backslash frac\{x\}\{15\}$ = the part of the work done by A.

And $10\; :\; x\; ::\; 1\; :\; \backslash frac\{x\}\{10\}$ = the part done by B.

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

Therefore, $\backslash frac\{x\}\{15\}\; +\; \backslash frac\{x\}\{10\}\; =\; 1$,

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 $x\; =\; \backslash frac\{s+d\}\{2\}$ and $y\; =\; \backslash frac\{s-d\}\{2\}$.

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.

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.

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 $x\; =\; \backslash frac\{y\}\{40\}$

And from the second $x\; +\; 5\; =\; \backslash frac\{y\; +\; 104\}\{36\}$

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

Hence $40y\; +\; 4160\; -\; 7200\; =\; 36y$, so that $4y\; =\; 3040$,

Therefore $y\; =\; 760$, and $x\; =\; \backslash frac\{y\}\{40\}\; =\; 19$.

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

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

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 $\backslash times\; 2$,

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

Hence $9x\; =\; 162$, and $x\; =\; 18$.

Therefore the numbers required are 18, 22, 10, and 45.

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 give 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.

H H

Then $x\; :\; a\; ::\; 1\; :\; \backslash frac\{a\}\{x\}\; =$ the part done by A } in $a$ hours.
$y\; :\; a\; ::\; 1\; :\; \backslash frac\{a\}\{y\}\; =$ the part done by B }

Also $x\; :\; b\; ::\; 1\; :\; \backslash frac\{b\}\{x\}\; =$ the part done by A } in $b$ hours.
$z\; :\; b\; ::\; 1\; :\; \backslash frac\{b\}\{z\}\; =$ the part done by C }

And $y\; :\; c\; ::\; 1\; :\; \backslash frac\{c\}\{y\}\; =$ the part done by B } in $c$ hours.
$z\; :\; c\; ::\; 1\; :\; \backslash frac\{c\}\{z\}\; =$ the part done by C }

Now by the question we have the three following equations.

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

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

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

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 $ax^2+b=c$ 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 $\backslash pm\; a$.

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 conditions 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. 1. $x^2+\backslash rho\; x=q$
2. 2. $x^2-\backslash rho\; x=q$
3. 3. $x^2-\backslash rho\; x=-q$

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. Resuming therefore the first equation, or $x^2+\backslash rho\; x=q$; 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 $x^2+\backslash rho\; x$ with $x^2+2ax+a^2=(x+a)^2$; and it will presently appear, that if we

suppose $\backslash rho=2a$, or $\backslash frac\{\backslash rho\}\{2\}=a$, the quantities $x^2+\backslash rho\; x$ 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 $x^2+\backslash rho\; x$ be completed into a square, by adding to it $\backslash frac\{\backslash rho^2\}\{4\}$, which is

equal to $a^2$; therefore, let $\backslash frac\{\backslash rho^2\}\{4\}$ be added to both sides of the equation $x^2+\backslash rho\; x=q$, and we have

and extracting the square root of each side, $x+\backslash frac\{\backslash rho\}\{2\}=\backslash pm\backslash sqrt\{\backslash frac\{\backslash rho^2\}\{4\}+q\}$; hence $x=\backslash frac\{\backslash rho\}\{2\}\backslash pm\backslash sqrt\{\backslash frac\{\backslash rho^2\}\{4\}+q\}$.

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 all 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 sign 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 $x+1=\backslash sqrt\{36\}=\backslash pm\; 6$. Hence $x=\backslash pm\; 6-1$, that is $x=+5$, or $x=-7$, and either of these numbers will be found to satisfy the equation for $5\; \backslash times\; 5+2\; \backslash times\; 5=35$, also $-7\; \backslash times\; -7+2\; \backslash times\; -7=35$.

Ex. 2. Given $\backslash frac\{x^2\}\{6\}-12=x$ 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, $x-3=\backslash pm\; 9$. And $x=\backslash pm\; 9+3$, therefore $x=+12$, or $x=-6$, and upon trial we find that each of these values satisfies

Ex

Quadratic Equations. $\backslash frac\{12\; \backslash times\; 12\}\{6\}\; =\; 12\; =\; 12$, also $\backslash frac\{6\; \backslash times\; -6\}\{6\}\; =\; -12\; =\; -6$.

Ex. 3. Given $x^2\; +\; 28\; =\; 11x$, to find $x$.

Then $x^2\; -\; 11x\; =\; -28$.

And, completing the square, $x^2\; -\; 11x\; +\; \backslash frac\{121\}\{4\}\; =\; \backslash frac\{121\}\{4\}\; -\; 28\; =\; \backslash frac\{9\}\{4\}$.

Therefore, by extracting the root, $x\; =\; \backslash frac\{11\}\{2\}\; \backslash pm\; \backslash frac\{3\}\{2\}$.

Hence $x\; =\; \backslash frac\{11\}\{2\}\; \backslash pm\; \backslash frac\{3\}\{2\}$, that is $x\; =\; +7$, or $x\; =\; +4$.

In the first two examples, we found one positive value for $x$ in each, and also one negative value, but in this example both the values of $x$ are positive, and upon trial each of them is found to satisfy the equation; for $7\; \backslash times\; 7\; +\; 28\; =\; 11\; \backslash times\; 7$, also $4\; \backslash times\; 4\; +\; 28\; =\; 11\; \backslash times\; 4$.

149. As at first sight it appears remarkable, that in every quadratic equation the unknown quantity admits always of two distinct values, or roots, it will be proper to consider a little farther the circumstances upon which this peculiarity depends. This is the more necessary, as the property of the unknown quantity admitting of several values is not peculiar to quadratics, but takes place also in equations of the higher degrees, where the cause of the ambiguity requires an explanation somewhat different from that which we have already given in the present case.

150. Let us again consider the equation $x^2\; +\; 2x\; =\; 35$, which forms the first of the three preceding examples; by transposing all the terms to one side, the same equation may be also expressed thus, $x^2\; +\; 2x\; -\; 35\; =\; 0$; so that we shall have determined $x$, when we have found such a number, as when substituted for it in the quantity $x^2\; +\; 2x\; -\; 35$ will render the result equal to 0. But $x^2\; +\; 2x\; -\; 35$ is the product of these two factors $x\; -\; 5$, and $x\; +\; 7$, as may be proved by actual multiplication; therefore to find $x$ we have $(x\; -\; 5)(x\; +\; 7)\; =\; 0$; and as a product can only become 0, when one of its factors is reduced to 0, it follows, that either of the two factors $x\; -\; 5$ and $x\; +\; 7$ may be assumed = 0; if $x\; -\; 5\; =\; 0$, then $x\; =\; 5$; but if $x\; +\; 7\; =\; 0$, then $x\; =\; -7$, so that the two values of $x$, or two roots of the equation $x^2\; +\; 2x\; =\; 35$ are $+5$ and $-7$, as we have already found in a different manner.

151. What has been just now shown in a particular case is true of any quadratic equation whatever, that is, if $x^2\; +\; px\; =\; q$, or by bringing all the terms to one side, $x^2\; +\; px\; -\; q\; =\; 0$, it is always possible to find two factors $x\; +\; a$, and $x\; -\; b$, such, that $x^2\; +\; px\; -\; q\; =\; (x\; +\; a)(x\; -\; b)$, where $a$ and $b$ are known quantities which depend only upon $p$ and $q$ the given numbers in the equation, and since that to have $(x\; +\; a)(x\; -\; b)\; =\; 0$, we may either assume $x\; +\; a\; =\; 0$, or $x\; -\; b\; =\; 0$, it evidently follows that the conditions of the equation $x^2\; +\; px\; -\; q\; =\; 0$, or $x^2\; +\; px\; =\; q$ are alike satisfied by taking $x\; =\; +a$ or $x\; =\; -b$.

From these considerations, it follows, that $x$ can have only two values in a quadratic equation; for if it could be supposed to have three, or more values,

then it would be possible to resolve $x^2\; +\; px\; -\; q$ into as many factors; $x\; -\; c$, $x\; -\; d$, &c. but the product of 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 resolution 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 $x^2\; +\; (b\; -\; a)x\; -\; ab$; and since this quantity must be equal to $x^2\; +\; px\; -\; q$, it follows, that $b\; -\; a\; =\; p$ 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 these 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

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

therefore, by extracting the square root, $v\; -\; x\; =\; \backslash pm\; \backslash sqrt\{p^2\; +\; 4q\}$; from this equation, and from the equation

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

Quadratic Equations. values of the roots $\backslash rho$ and $q$ are $\backslash frac\{-\backslash rho\; +\; \backslash sqrt\{\backslash rho^2\; +\; 4q\}\}\{2\}$ and

$\backslash frac\{-\backslash rho\; -\; \backslash sqrt\{\backslash rho^2\; +\; 4q\}\}\{2\}$, which agrees with the conclusion we have already found, (§ 148.)

155. It appears from what has been already shown, that the roots of a quadratic equation $x^2\; +\; \backslash rho\; x\; +\; q\; =\; 0$ always involve the quantity $\backslash sqrt\{\backslash rho^2\; +\; 4q\}$; hence it follows, that $\backslash rho^2\; +\; 4q$ must be a positive quantity; for if it were negative, as the square root of such a quantity could not be found, the value of $x$ could not possibly be obtained. If for example the value of $x$ were required from this equation $x^2\; +\; 13\; =\; 4x$, or $x^2\; -\; 4x\; =\; -13$, we should find $x\; =\; \backslash frac\{2\; \backslash pm\; \backslash sqrt\{-9\}\}\{2\}$; and as this expression for the roots requires us to extract the square root of $-9$, the equation from which it is derived must necessarily have involved some contradiction. It is not difficult to see wherein the absurdity consists, for since in this case $\backslash rho\; =\; -4$, and $q\; =\; -13$, the roots of the equation ought to be both positive (§ 154), and such that their sum $=\; 4$, while their product $=\; 13$, (§ 153), which is impossible.

156. Although imaginary quantities serve no other purpose in the resolution of quadratic equations, than to show that a particular problem cannot be resolved, by reason of some want of consistency in its data; yet they are not upon that account to be altogether rejected. By introducing them into mathematical investigations, many curious theories may be explained, and problems resolved in a more concise way, than can be done without the use of such quantities. This is particularly the case with respect to the higher parts of the mathematics.

157. The method which has been applied to the resolution of quadratic equations, properly so called, namely, such as are of this form $x^2\; +\; \backslash rho\; x\; +\; q\; =\; 0$, will also apply to all equations of this form,

Where the unknown quantity $x$ is found only in two terms, and such, that its exponent in the one term is double that in the other; for let us assume $x^n\; =\; y$, then $x^\{2n\}\; =\; y^2$, and therefore the equation

a quadratic equation, from which $y$ may be found, and thence $x$, by considering that $x^n\; =\; y$.

158. Before proceeding to give examples of questions producing quadratic equations, it is proper to observe, that although every such equation admits of two roots; yet it will frequently happen, that only one of them can be of use, the other being excluded by the conditions of the question. This will often be the case with respect to the negative root; as for example, when the unknown quantity denotes a number of men, a number of days, &c. And hence, in reckoning the cases of quadratic equations, it is common to neglect this one $x^2\; +\; \backslash rho\; x\; +\; q\; =\; 0$, where the roots are both negative; for an equation of this form can only be derived from a question which has some fault in its enunciation, and which, by a proper change in its form, will produce another equation having both its roots positive.

159. The remainder of this section shall be employed in solving some questions which produce quadratic equations.

Ex. 1. It is required to divide the number 10 into two such parts, that the sum of their squares may be 58. Quadratic Equations.

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, $x\; -\; 5\; =\; \backslash pm\; \backslash sqrt\{4\}\; =\; \backslash pm\; 2$.
And $x\; =\; 5\; \backslash pm\; 2\; =\; 7$
That is $x\; =\; 7$ 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 thrice 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

From the first equation we have $y\; =\; \backslash frac\{28\}\{x\}$.

And from the second $y\; =\; \backslash frac\{26\; -\; 2x\}\{3\}$.

Hence, $\backslash frac\{26\; -\; 2x\}\{3\}\; =\; \backslash frac\{28\}\{x\}$.

And, reducing, $26x\; -\; 2x^2\; =\; 84$

Or $2x^2\; -\; 26x\; =\; -84$
Hence $x^2\; -\; 13x\; =\; -42$

And comp. the sq. $x^2\; -\; 13x\; +\; \backslash frac\{169\}\{4\}\; =\; \backslash frac\{169\}\{4\}\; -\; 42\; =\; \backslash frac\{1\}\{4\}$

Hence, by extracting the root $x\; -\; \backslash frac\{13\}\{2\}\; =\; \backslash pm\; \backslash sqrt\{\backslash frac\{1\}\{4\}\}\; =\; \backslash pm\; \backslash frac\{1\}\{2\}$

Therefore $x\; =\; \backslash frac\{13\}\{2\}\; \backslash pm\; \backslash frac\{1\}\{2\}$

That is $x\; =\; 7$, or $x\; =\; 6$.

And since $y\; =\; \backslash frac\{28\}\{x\}$, we have $y\; =\; 4$, or $y\; =\; \backslash frac\{14\}\{3\}$.

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

$x\; =\; 7,\; y\; =\; 4$; Or $x\; =\; 6,\; y\; =\; \backslash frac\{14\}\{3\}$.

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 $\backslash frac\{175\}\{x\}$.

But,

Equations to explain some general properties, which belong to in general. equations of every degree; and also certain operations, which must frequently be performed upon equations, before they be 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:

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 $=0$, 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

$(x-a)(x^\{n-1\}\; +\; ax^\{n-2\}\; +\; a^2x^\{n-3\}\; +\; \backslash dots\; +\; a^\{n-2\}x\; +\; a^\{n-1\})$, as may be easily proved by multiplication; therefore, putting $n=4,\; 3$ and $2$ successively, we have

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 $(x-a)[(x^3\; +\; (p+a)x^2\; +\; (q+pa)x\; +\; a^2\; +\; pa\; +\; qa\; +\; r)]\; =\; 0$, so that putting $p\text{'}\; =\; p+a,\; q\text{'}\; =\; q+pa+q,\; r\text{'}\; =\; a^2\; +\; pa\; +\; qa\; +\; r$, we have

Hence, if the proposed equation $x^4\; +\; px^3\; +\; qx^2\; +\; rx\; +\; s$ be divided by $x-a$, the quotient will be $x^3\; +\; p\text{'}x^2\; +\; q\text{'}x\; +\; r\text{'}$, 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 $(x-a)(x^3\; +\; p\text{'}x^2\; +\; q\text{'}x\; +\; r\text{'})\; =\; 0$, 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 in general. also when $x^3\; +\; p\text{'}x^2\; +\; q\text{'}x\; +\; r\text{'}\; =\; 0$.

Let us now suppose that $b$ is a root of this equation, then by reasoning exactly as in last article, and putting $p\text{'}\; =\; b\; +\; p,\; q\text{'}\; =\; b^2\; +\; p\text{'}b\; +\; q\text{'}$, we shall have

and therefore

165. By proceeding in the same manner with the quadratic equation $x^2\; +\; p\text{'}x\; +\; q\text{'}\; =\; 0$, we shall find that if $c$ denote one of its roots, then

So that if we put $d\; =\; -(c+p\text{'})$, we at last find $x^4\; +\; px^3\; +\; qx^2\; +\; rx\; +\; s\; =\; (x-a)(x-b)(x-c)(x-d)$; 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 $x^4\; +\; px^3\; +\; qx^2\; +\; rx\; +\; s\; =\; 0$, 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 $x-a,\; x-b,\; x-c,\; \backslash dots$, &c. 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

we shall also have $(x-a)(x-b)(x-c)(x-d)\; =\; 0$; and therefore, by actual multiplication

168. If we compare together the coefficients of the same powers of $x$, we find the following series of equations:

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 in general. contrary sign, is equal to the sum of the roots multiplied together three and three, and so on for 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 shewn, 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 $\backslash frac\{n(n-1)\}\{1.2\}$ such combinations,

any equation will admit of $\backslash frac\{n(n-1)\}\{1.2\}$ divisors of the second degree.

171. For example, the equation $x^4\; +\; px^2\; +\; qx\; +\; r\; =\; 0$ which we have considered as equal to $(x-a)(x-b)(x-c)(x-d)\; =\; 0$ may be formed by the product of two factors of the second degree, in these six different ways.

Thus an equation of the fourth degree may have $4\; \backslash times\; 3\; =\; 6$ 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 $\backslash frac\{n(n-1)(n-2)\}\{1.2.3\}$; and so on.

173. When the roots of an equation are all positive, its simple factors will have this form and if for the sake of brevity we take only these three, the cubic equation which results from their product will have this form

where $p\; =\; a+b+c,\; q\; =\; ab+ac+bc,\; r\; =\; abc$, and here it appears that the signs of the terms are $+,\; +,\; -$ 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:

And hence we conclude, that when the roots are all negative, there is no change whatever in the signs.

VOL. I. PART II.

175. In general, if the roots of an equation be all real, that equation will have as many positive roots as 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 shewn.

When a cubic equation has one positive root, and the other two negative, its factors will be $x-a,\; x+b,\; x+c$, and the equation itself.

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.

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 shewn 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 $a\; +\; \backslash sqrt\{-b^2\},\; a\; -\; \backslash sqrt\{-b^2\}$, which quantities may also be expressed thus,

$a+b\backslash sqrt\{-1\},\; a-b\backslash sqrt\{-1\}$, so that we have these two factors
$x-a-b\backslash sqrt\{-1\},\; x-a+b\backslash sqrt\{-1\}$, 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 $x-a-b\backslash sqrt\{-1\},\; x-a+b\backslash sqrt\{-1\}$; 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 $x+a+b\backslash sqrt\{-1\},\; x+a-b\backslash sqrt\{-1\},\; x+a+c\backslash sqrt\{-1\},\; x+a-d\backslash sqrt\{-1\}$; 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 first. The truth of this remark will be evident, if we take two equations,

(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:

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

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 useful 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:

Now, that this equation may want its second term, it is evident, that we have only to suppose $3n+pn=0$, or $n=-\backslash frac\{p\}\{3\}$, for this assumption being made, and the value of $n$ substituted in the remaining terms, the equation becomes

or, putting $-\backslash frac\{p^2\}\{3\}+q=q\text{'}$, and $+\backslash frac\{2pn\}\{27\}-\backslash frac\{pn^3\}\{3\}+r=r\text{'}$ the same equation may also stand thus,

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; this 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 $y+\backslash frac\{5\}\{2\}$ for $x$, we find

Hence $y=\backslash pm\backslash frac\{5\}\{2\}$, and since $x=y+\backslash frac\{5\}\{2\}$, therefore $x=\backslash pm\backslash frac\{5\}\{2\}+\backslash frac\{5\}\{2\}=+3$ or $+2$.

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 abstracting

Cubic Equations. Extracting 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 that 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 $x^3\; +\; \backslash frac\{p\}\{a\}x^2\; +\; \backslash frac\{q\}\{b\}x\; +\; \backslash frac\{r\}\{c\}\; =\; 0$ denote an equation to be so transformed; and let us assume $y\; =\; abcx$; and therefore $x\; =\; \backslash frac\{y\}\{abc\}$, then by substitution, our equation becomes

and multiplying the whole equation by $a^3b^3c^3$, we have $y^3\; +\; bcp^2y^2\; +\; a^2bc^2qy\; +\; a^3b^3c^3r\; =\; 0$.

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,

it will be sufficient to assume $y\; =\; ax$, and therefore $x\; =\; \backslash frac\{y\}\{a\}$; for then we have

And $y^3\; +\; py^2\; +\; aqy\; +\; a^2r\; =\; 0$, 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 $x\; =\; \backslash sqrt[3]\{r\}$.

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

$x^2\; +\; cx\; =\; -c^2$, we find these values to be $\backslash frac\{-c\; +\; \backslash sqrt\{-3c^2\}\}\{2\}$

and $\backslash frac\{-c\; -\; \backslash sqrt\{-3c^2\}\}\{2\}$, or $\backslash frac\{-1\; +\; \backslash sqrt\{-3\}\}\{2\}c$ and $\backslash frac\{-1\; -\; \backslash sqrt\{-3\}\}\{2\}c$.

Thus it appears that any cubic equation of this form $x^3\; =\; c^3$, or $x^3\; -\; c^3\; =\; 0$ has these three roots

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 $x\; -\; c$, $x\; -\; \backslash frac\{-1\; +\; \backslash sqrt\{-3\}\}\{2\}c$

$c$, and $x\; -\; \backslash frac\{-1\; -\; \backslash sqrt\{-3\}\}\{2\}c$.

194. Let us now consider such cubic equations as have all their terms, and which are therefore of this form

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 $x\; =\; y\; -\; \backslash frac\{A\}\{3\}$, as directed in that article, then, by proper substitution, the above equation will be changed into another of this form

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 $x\; =\; y\; -\; \backslash frac\{A\}\{3\}$.

195. Refusing, therefore, the equation $y^3\; +\; qy\; +\; r\; =\; 0$, let us suppose $y\; =\; v\; +\; z$, and it becomes

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. and therefore $\backslash sqrt[3]\{\backslash frac\{2\}\{10+\backslash sqrt\{108\}\}\}\; =\; \backslash frac\{2\}\{2.732\}\; =\; .732$; 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 $\backslash sqrt[3]\{10+\backslash sqrt\{108\}\}$ 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 $a+\backslash sqrt\{b\}$ 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 $10+\backslash sqrt\{108\}$ is $1+\backslash sqrt\{3\}$, as may be proved by actually raising $1+\backslash sqrt\{3\}$ to the third power. Hence

The other two values of $y$ will be had by substituting $1+\backslash sqrt\{3\}$ and $1-\backslash sqrt\{3\}$ for $\backslash sqrt[3]\{A\}$ and $\backslash frac\{1\}\{\backslash sqrt[3]\{A\}\}$ in the second and third formulae of last article, also restoring the values of $\backslash alpha$ and $\backslash beta$. We thus have

So that the three values of $y$ are

and since $x\; =\; y+1$, the corresponding values of $x$ are

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 $x-1$ (§ 163). Accordingly the division being actually performed, and the quotient put $=0$, we have this quadratic equation

which, when resolved by the rule for quadratics, gives $x\; =\; -2\; \backslash pm\; \backslash sqrt\{-9\}$, the same imaginary values as before.

199. In the application of the preceding formulae (§ 196 and 197) to the resolution of the equation $y^3+9y+r=0$, it is necessary to find the square root of $\backslash frac\{1\}\{4\}q^3+\backslash frac\{1\}\{2\}r^2$, 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 $q$ and $r$ are independent of each other, it is evident that $q$ may be negative, and such that $\backslash frac\{1\}\{4\}q^3+\backslash frac\{1\}\{2\}r^2$ is greater than $\backslash frac\{1\}\{2\}r^2$, in this case the expression $\backslash sqrt[3]\{\backslash frac\{1\}\{4\}q^3+\backslash frac\{1\}\{2\}r^2\}$ 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 $q=-6$, $r=+4$, $\backslash frac\{1\}\{4\}q^3+\backslash frac\{1\}\{2\}r^2=-8$, $\backslash frac\{1\}\{2\}r^2=+4$, $\backslash sqrt[3]\{\backslash frac\{1\}\{4\}q^3+\backslash frac\{1\}\{2\}r^2\}=\backslash sqrt\{-4\}=2\backslash sqrt\{-1\}$, hence, by recurring to the formulae (§ 196), we have $A=2+2\backslash sqrt\{-1\}$, $B=2-\backslash sqrt\{-1\}$, and therefore the three roots of the equation expressed thus

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 shew 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 $2+2\backslash sqrt\{-1\}$ is the cube of this other imaginary expression $-1+\backslash sqrt\{-1\}$, and, in like manner, that $2-2\backslash sqrt\{-1\}$ is the cube of $-1-\backslash sqrt\{-1\}$, so that we have

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 $\backslash frac\{1\}\{4\}q^3$ greater than $\backslash frac\{1\}\{2\}r^2$; and, on the contrary, that if $\backslash frac\{1\}\{4\}q^3$ be greater than $\backslash frac\{1\}\{2\}r^2$ the roots are all real.

Let us suppose $a$ to be a real root of the proposed equation,

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

This quadratic equation is formed from the two remaining roots of the proposed equation, and by resolving it we find

And as, by hypothesis, all the roots are real, it is evident that $q$ must necessarily be negative, and greater than $\backslash frac\{1\}\{4\}a^2$; for otherwise the expression $\backslash sqrt\{-\backslash frac\{1\}\{4\}a^2-q\}$ would be imaginary. Let us change the sign of $q$, and put

Cubic Equations. put $q\; =\; \backslash frac\{1\}\{2\}a^3\; +\; d^3$; thus the roots of the equation $x^3\; +\; qx\; +\; r\; =\; 0$ will be

and here $d$ is a positive quantity.

To find an expression for $r$ in terms of $a$, and $d$, let $\backslash frac\{1\}\{2\}a^3\; +\; d$ be substituted for $q$ in the equation $a^3\; +\; qa\; +\; r\; =\; 0$; we thence find $r\; =\; -\backslash frac\{1\}\{2\}a^3\; +\; ad$; so that to compare together the quantities $q$ and $r$ we have these equations,

In order to make this comparison, let the cube of $\backslash frac\{1\}\{2\}q$ be taken, also the square of $\backslash frac\{1\}\{2\}r$, the results are

and therefore, by subtraction,

Now the square of any real quantity being always positive, it follows that $3d(\backslash frac\{1\}\{2\}a^3\; -\; \backslash frac\{1\}\{2\}d)^2$ will be positive when $d$ is positive; hence it is evident that in this case $\backslash frac\{1\}\{4\}q^3$ must be greater than $\backslash frac\{1\}\{4\}r^2$; and that the contrary cannot be true unless $d$ be negative, that is, unless that $-\backslash frac\{1\}\{2\}a\; +\; \backslash sqrt\{d\},\; -\backslash frac\{1\}\{2\}a\; -\; \backslash sqrt\{d\}$, the two other roots of the equation, are imaginary. If we suppose $d\; =\; 0$, then $\backslash frac\{1\}\{4\}q^3\; =\; \backslash frac\{1\}\{8\}a^9$, and the roots of the equations, which in this case are also real, are $a,\; -\backslash frac\{1\}\{2\}a,\; -\backslash frac\{1\}\{2\}a$.

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 $\backslash frac\{1\}\{4\}q^3$ greater than $\backslash frac\{1\}\{4\}r^2$; and consequently, that in this case the formulae for the roots must express real quantities notwithstanding their imaginary form.

201. Let $y^3\; -\; gy\; +\; r\; =\; 0$ 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 $a\; =\; -\backslash frac\{1\}\{2\}r$, $b\; =\; \backslash frac\{1\}\{2\}g^3\; -\; \backslash frac\{1\}\{2\}r^3$, one of the roots, as expressed by the first formula, § 196, will be

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 surds which compose the root are perfect cubes of the form $(A\; +\; B\backslash sqrt\{-1\})^3$ and $(A\; -\; B\backslash sqrt\{-1\})^3$, and then the value of $y$ becomes

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

and in general,

where $n$ is any power of 2, admits of being reduced to another form in which no impossible quantity is found.

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 surds be expanded into an infinite series and their sum be taken, the imaginary quantity $\backslash sqrt\{-1\}$ 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 $\backslash frac\{a\}\{3\}$, one-third of that arch is expressed by the following cubic equation.

Let us assume $\backslash text\{cos.\}\; \backslash frac\{a\}\{3\}\; =\; \backslash frac\{y\}\{n\}$, then, by substitution, the equation is transformed into the following

and in this cubic equation one of the roots is evidently $y\; =\; n\; \backslash times\; \backslash text\{cos.\}\; \backslash frac\{a\}\{3\}$: Now from the arithmetic of fines it appears that $\backslash text\{cos.\}\; a$, $\backslash text\{cos.\}\; (360^\backslash circ\; -\; a)$ and $\backslash text\{cos.\}\; (360^\backslash circ\; +\; a)$ are all expressed by the same quantity; therefore the equation must have for a root, not only $n\; \backslash times\; \backslash text\{cos.\}\; \backslash frac\{a\}\{3\}$, but

Therefore the three roots of the equation are

Let us next suppose that $y^3\; -\; gy\; +\; r\; =\; 0$ is a cubic equation

Cubic Equations. tion whose roots are required, and let us compare it with the former equation $y^3\; -\; \backslash frac\{3n^2\}\{4\}y\; =\; n^3\; \backslash times\; \backslash cos.\; a$; then it is evident that if we assume the quantities $n$ and $\backslash cos.\; a$, such, that

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

and since the cosine of an arch cannot exceed unity, therefore, $\backslash frac\{27r^3\}\{4q^3\}$ must be a proper fraction, that is $4q^3$ must exceed $27r^3$, or $\backslash frac\{1\}\{3\}q^3$ must exceed $\backslash frac\{1\}\{2\}r^3$; 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

The roots of the equation are

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 sine $=\; \backslash frac\{3r\backslash sqrt\{3\}\}\{2q\backslash sqrt\{q\}\}$

Then the roots of the equation are

The last formulae are derived from the equation

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 $x^3\; -\; 3x\; =\; 1$.

Ex. 2. It is required to find the roots of the equation $x^3\; -\; 3x\; =\; 1$.

SECT. XII. Of Biquadratic Equations.

207. WHEN a biquadratic equation contains all its terms, it has this form,

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 $x^4\; =\; b^4$, therefore $x^4\; -\; b^4\; =\; 0$; but $x^4\; -\; b^4$ may be resolved into two factors $x^2\; -\; b^2$ and $x^2\; +\; b^2$, each of which admit of a similar resolution; for $x^2\; -\; b^2\; =\; (x\; -\; b)(x\; +\; b)$ and $x^2\; +\; b^2\; =\; (x\; -\; b\backslash sqrt\{-1\})(x\; +\; b\backslash sqrt\{-1\})$. Hence it appears that the equation $x^4\; -\; b^4\; =\; 0$ may also be expressed thus:

so that $x$ may have these four values, $+b,\; -b,\; +b\backslash sqrt\{-1\},\; -b\backslash sqrt\{-1\}$, two of which are real and the others imaginary.

209. Next to pure biquadratic equations, in respect of ease of resolution, are such as want the second and fourth terms, and therefore have this form,

These may be resolved in the manner of quadratic equations; for if we put $y^2\; =\; x^2$ we have

from which we find $y\; =\; \backslash frac\{-q\; \backslash pm\; \backslash sqrt\{q^2\; -\; 4r\}\}\{2\}$, and there-

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 Petersburg Commentaries, and

Biquadratic and afterwards explained more fully in his Elements of 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,

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 $y\; =\; \backslash sqrt\{a\}\; +\; \backslash sqrt\{b\}\; +\; \backslash sqrt\{c\}$, and let us also suppose that the letters $a,\; b,\; c$, denote the roots of the cubic equation

then from the theory of equations we have

Let us now square the assumed formula

or substituting $-P$ for $a\; +\; b\; +\; c$, and transposing,

Let this equation be also squared and we have

and $\backslash sqrt\{ab\}\; +\; \backslash sqrt\{ac\}\; +\; \backslash sqrt\{bc\}\; =\; \backslash sqrt\{abc\}(\backslash sqrt\{a\}\; +\; \backslash sqrt\{b\}\; +\; \backslash sqrt\{c\})\; =\; \backslash sqrt\{Ry\}$; the same equation may be expressed thus:

Thus we have obtained the biquadratic equation

one of the roots of which $y\; =\; \backslash sqrt\{a\}\; +\; \backslash sqrt\{b\}\; +\; \backslash sqrt\{c\}$, and in which $a,\; b,\; c$ are the roots of the cubic equation $z^3\; +\; Pz^2\; +\; Qz\; -\; R\; =\; 0$.

212. That we may apply this resolution to the proposed equation $y^4\; +\; py^3\; +\; qy\; +\; r\; =\; 0$, 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;

and it immediately appears that $2P\; =\; p,\; -8\backslash sqrt\{R\}\; =\; q,\; P^2\; -\; 4Q\; =\; r$; and from these three equations we find

Hence it follows, that the roots of the proposed equation are generally expressed by the formula $y\; =\; \backslash sqrt\{a\}\; +\; \backslash sqrt\{b\}\; +\; \backslash sqrt\{c\}$; where $a,\; b,\; c$ denote the roots of this cubic equation

213. But to find each particular root, we must consider, that as the square root of a number may be either positive or negative, so each of the quantities $\backslash sqrt\{a\},\; \backslash sqrt\{b\},\; \backslash sqrt\{c\}$ 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

$\backslash sqrt\{a\},\; \backslash sqrt\{b\},\; \backslash sqrt\{c\}$ must be equal to $\backslash sqrt\{R\}$ or to $-\backslash sqrt\{R\}$, 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

$z\; =\; \backslash frac\{v\}\{4\}$. Thus the equation $z^3\; +\; \backslash frac\{p\}\{2\}z^2\; +\; \backslash frac\{p^2\; -\; 4r\}\{16\}z\; -\; \backslash frac\{q^2\}\{64\}\; =\; 0$ 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 $\backslash frac\{a\}\{4\},\; \backslash frac\{b\}\{4\},\; \backslash frac\{c\}\{4\}$, so that our rule may now be expressed thus:

Let $y^4\; +\; py^3\; +\; qy\; +\; r\; =\; 0$ be any biquadratic equation wanting its second term. Form this cubic equation

and find its roots, which let us denote by $a,\; b,\; c$. Then the roots of the proposed biquadratic equation are

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

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,

Biquadratic, and the remaining values of $y$ are in the first set of roots $y\; =\; -\backslash frac\{1\}\{2\}\backslash sqrt\{a\}$, $y\; =\; -\backslash frac\{1\}\{2\}\backslash sqrt\{a\}$, and in the second $y\; =\; +\backslash frac\{1\}\{2\}\backslash sqrt\{a\}$, $y\; =\; \backslash frac\{1\}\{2\}\backslash sqrt\{a\}$.

3. When the cubic equation has only one real, and two imaginary roots, its real root must necessarily be positive. For the imaginary roots can only come from a quadratic equation, having its last term positive, Sect. IX. and therefore of this form $v^2\; +\; Av\; +\; B\; =\; 0$, hence, the simple factor which contains the remaining root must have this form $v\; -\; y$, otherwise the last term of the cubic equation could not be negative.

By resolving the equation $v^2\; +\; Av\; +\; B\; =\; 0$, we find

here, the roots being supposed imaginary, $\backslash frac\{A^2\}\{4\}\; -\; B$ must be a negative quantity. That we may simplify the form of the roots, let us put $-\backslash frac\{A\}\{2\}\; =\; \backslash alpha$ and $\backslash frac\{A^2\}\{4\}\; -\; B\; =\; -\backslash beta^2$, then

so that in two of the four values of $y$, we have a quantity of this form

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

which is a real quantity. The two other roots involve this other expression

which, being treated in the same manner as the former, becomes

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 $v\; =\; u\; -\; \backslash frac\{2\backslash rho\}\{3\}$; thus it becomes

and in this equation the three roots will be real when

216. As an example of the method of resolving a biquadratic equation, let it be required to determine the roots of the following,

VOL. I. Part II.

By comparing this equation with the general formula, Reciprocal Equations, we have $\backslash rho\; =\; -25$, $q\; =\; +60$, $r\; =\; -36$, hence

and the cubic equation to be resolved is

the roots of which are found by the rules for cubics, to be 9, 16, and 25, so that we have $\backslash sqrt\{a\}\; =\; 3$, $\backslash sqrt\{b\}\; =\; 4$, $\backslash sqrt\{c\}\; =\; 5$. Now in this case $q$ is positive, therefore

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

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

where the given quantity $a$ and the unknown quantity $x$ are

Reciprocal $x$ are precisely alike concerned; for by substituting $\backslash frac\{1\}\{x\}$ for $x$, it becomes

and dividing by $a^4$,

an equation of the same kind as the former.

220. That we may effect the proposed transformation upon the equation

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

Let us assume $x\; +\; \backslash frac\{1\}\{x\}\; =\; z$

Then $x^2\; +\; 2\; +\; \backslash frac\{1\}\{x^2\}\; =\; z^2$ and $x^4\; +\; \backslash frac\{1\}\{x^4\}\; =\; z^4\; -\; 2$

Thus the equation $x^4\; +\; \backslash frac\{1\}\{x^4\}\; +\; p(x^2\; +\; \backslash frac\{1\}\{x^2\})\; +\; q\; =\; 0$

becomes $z^4\; +\; pz^2\; +\; q\; =\; 2$.

And since $x\; +\; \backslash frac\{1\}\{x\}\; =\; z$, therefore $x^2\; -\; zx\; +\; 1\; =\; 0$.

221. Hence upon the whole, to determine the roots of the biquadratic equation

we have the following rule.

Form this quadratic equation

and find its roots, which let us suppose denoted by $z\text{'}$ and $z\text{'}\text{'}$. Then the four roots of the proposed equation will be found by resolving two quadratic equations

222. It may be observed respecting these two quadratic equations, that since the last term of each is unity, if we put $a,\; a\text{'}$ to denote the roots of the one, and $b,\; b\text{'}$ those of the other, we have from the theory of equations $a\; a\text{'}\; =\; 1$, and therefore $a\text{'}\; =\; \backslash frac\{1\}\{a\}$, also $b\; b\text{'}\; =\; 1$, and $b\text{'}\; =\; \backslash frac\{1\}\{b\}$; now $a,\; a\text{'},\; b,\; b\text{'}$ are also the roots of the equation

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

which is of the sixth order, admits of a resolution in all respects similar to the former; for by putting it under this form

and putting also $x\; +\; \backslash frac\{1\}\{x\}\; =\; z$, so that $x^2\; -\; zx\; +\; 1\; =\; 0$, we have $x^3\; +\; \backslash frac\{1\}\{x^3\}\; =\; z^3\; -\; 3z$

Hence, by substitution, the proposed equation is transformed into the following cubic equation

Therefore, putting $z\text{'},\; z\text{'}\text{'},\; z\text{'}\text{'}\text{'}$ to denote its roots, the six roots of the proposed equation will be had by resolving these three quadratics

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 $\backslash frac\{1\}\{x\}$.

225. If the greatest exponent of the unknown quantity in a reciprocal equation is an odd number, as in this example

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

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

the roots of which may be generally denoted by $a,\; b,\; c$, and $d$. Thus we have, from the theory of equations,

Let us put

Then,

Equations with equal Roots. Then, by actual multiplication, we have

and taking the sum of these four equations

But since $a,\; b,\; c,\; d$ are the roots of the equation

Therefore, by substitution

228. Let us now suppose that the proposed biquadratic equation has two equal roots, or $a=b$, then $x-a=x-b$, and since one or other of these equal factors enters each of the four products $A,\; A\text{'},\; A\text{'}\text{'},\; A\text{'}\text{'}\text{'}$ it is evident that $A\; +\; A\text{'}\; +\; A\text{'}\text{'}\; +\; A\text{'}\text{'}\text{'}$ or $4x^3\; +\; 3px^2\; +\; 2qx\; +\; r$ must be divisible by $x-a$, or $x-b$. Thus it appears that if the proposed equation

has two equal roots, each of them must also be a root of this equation

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 two at least of the three equal factors $x-a,\; x-b,\; x-c$, must enter each of the four products $A,\; A\text{'},\; A\text{'}\text{'},\; A\text{'}\text{'}\text{'}$; so that in this case $A\; +\; A\text{'}\; +\; A\text{'}\text{'}\; +\; A\text{'}\text{'}\text{'}$, 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

230. Proceeding in the same manner, it may be shewn that whatever number of equal roots are in the proposed equation

they will all remain except one, in this equation

which is evidently derived from the former, by multiplying each of its terms by the exponent of $x$ in that 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 $x^n\; +\; Px^\{n-1\}\; +\; Qx^\{n-2\}\; +\; \backslash dots\; +\; Sx^2\; +\; Tx\; +\; U\; =\; 0$ an equation of the $n$th degree to have a divisor of this form

The equation

which is of the next lower degree, will have for a divisor

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

a new equation, which will have for a divisor

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

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 shewn § 228,

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

which is derived from it, as directed in § 228, have any common divisor; and by proceeding as before, we

Equations with rational Roots. find that they admit of this divisor $x-3$, which is also a factor of the last divisor $x^2-8x^2+21x-18$, 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 $(x-2)^3$ for one factor, and $(x-3)^3$ for another factor; so that the equation itself may be expressed thus, $(x-2)^3(x-3)^3=0$, 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 shewn 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.

Here the divisors which produce results equal to 0 are +1, -2 and +5, 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 so many substitutions, let us transform the equation, by putting $x+1$ for $y$.

is the transformed equation, and the divisors of the last term are +1, -1, +3, -3, +7, -7. These being put successively for $x$, we get +1 and +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 $x^4-6x^2-16x+21=0$ 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,

the roots of which are the imaginary quantities $-2+\backslash sqrt\{-3\}$ and $-2-\backslash sqrt\{-3\}$; so that since $y=x+1$, the roots of the equation $y^4-4y^3-8y+32=0$ are these, $y=+2$, $y=+4$, $y=-1+\backslash sqrt\{-3\}$, $y=-1-\backslash sqrt\{-3\}$.

If this literal equation were proposed

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 $x-1$, $x$, $x+1$ 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 $x^3-x^2-10x+6=0$.

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 $2\; +\; \backslash sqrt\{2\}$ and $2\; -\; \backslash sqrt\{2\}$.

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 $x^3\; -\; \backslash frac\{1\}\{2\}x^2\; +\; \backslash frac\{1\}\{4\}x\; -\; 6\; =\; 0$. Let us assume $x\; =\; \backslash frac\{y\}\{4\}$, thus the equation is transformed to

one root of which is $y=3$; hence $x\; =\; \backslash frac\{y\}\{4\}\; =\; \backslash frac\{3\}\{4\}$.

The proposed equation being now divided by $x\; -\; \backslash frac\{3\}\{4\}$ 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 ratio-

nal 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 $B\text{'}$ for the remainder, we have $\backslash frac\{B\}\{P\}\; =\; q\; +\; \backslash frac\{B\text{'}\}\{P\}$, and therefore

Hence it appears, that if $P$ be a divisor of $AB$, it is also a divisor of $AB\text{'}$. Now $B\text{'}$ is less than $P$, for it is the remainder which is found in dividing $B$ by $P$; therefore, seeing we cannot divide $B\text{'}$ by $P$, let $P$ be divided by $B\text{'}$, and $q\text{'}$ put for the quotient, also $B\text{'}\text{'}$ for the remainder; again let $P$ be divided by $B\text{'}\text{'}$, and $q\text{'}\text{'}$ put for the quotient, and $B\text{'}\text{'}\text{'}$ 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 a 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

Hence we have $AB\text{'}\; =\; \backslash frac\{AP\; -\; AB\text{'}\text{'}\}\{q\text{'}\}$, and

Now, if $AB$ be divisible by $P$, we have shown that $AB\text{'}$, and consequently $q\text{'}AB\text{'}$ is divisible by $P$; therefore, from the last equation, it appears that $AB\text{'}\text{'}$ must also be divisible by $P$.

Again, from the preceding series of equations, we have $AB\text{'}\text{'}\; =\; \backslash frac\{AP\; -\; AB\text{'}\text{'}\text{'}\}\{q\text{'}\text{'}\}$, and therefore

hence we conclude that $AB\text{'}\text{'}\text{'}$ is also divisible by $P$.

Proceeding in this manner, and observing that the series of quantities continually decrease till one of them $=1$, it is evident that we shall at last come to a product of this form $A\; \backslash times\; 1$, which must

Equations with rational Roots. 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 $\backslash frac\{b\}\{a\}$ 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 $a\; \backslash times\; a$ or $a^2$, and in like manner, that if a prime number is not a divisor of $b$, it cannot be a divisor of $b\; \backslash times\; b$, or $b^2$; therefore, it is evident that $a^2$ and $b^2$ have no common divisor, and thus the fraction $\backslash frac\{b^2\}\{a^2\}$ 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 $a\; \backslash times\; a^2$ or $a^3$, &c. 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 $a^n$; also, that if it is not a divisor of $b$, it cannot be a divisor of $b^n$, therefore if $\backslash frac\{b\}\{a\}$ is a fraction in its

lowest terms $\backslash frac\{b^n\}\{a^n\}$ 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

is any equation whatever, in which $P,\; Q,\; \backslash dots,\; T,\; U$ denote integer numbers; then if its roots are not integers they cannot possibly be rational fractions. For if possible,

let us suppose $x\; =\; \backslash frac\{b\}\{a\}$, a fraction reduced to its lowest terms, then, by substitution

and, reducing all the terms to a common denominator,

which equation may also be expressed thus

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 $b$, &c. 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

which, by collecting the positive terms into one sum, and the negative into another, may also be expressed thus

then, to determine a root of the equation, we must find such a number as when substituted for $x$ will render

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.

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

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

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

Approximation. 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 $N\text{'}$ 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 $-N\text{'}-1$.

Suppose the equation to be

which may be also expressed thus

Then, whatever be the values of the coefficients $p,\; q,\; r,\; s$, &c. it is evident that $x$ may be taken so great as to render each of the quantities $\backslash frac\{p\}\{x\},\; \backslash frac\{q\}\{x^2\},\; \backslash frac\{r\}\{x^3\},\; \backslash frac\{s\}\{x^4\}$ as small as

we please, and therefore their sum, or $\backslash frac\{p\}\{x\}\; +\; \backslash frac\{q\}\{x^2\}\; +\; \backslash frac\{r\}\{x^3\}\; +\; \backslash frac\{s\}\{x^4\}$

less than $1$; but in that case the quantity

will be positive, and such, that the first term $x^4$ is greater than the sum of all the remaining terms, therefore also $x^4\; +\; qx^2$ 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 $Nx^3\; +\; Nx^2\; +\; Nx\; +\; N$, or $N(x^3\; +\; x^2\; +\; x\; +\; 1)$; but $x^3\; +\; x^2\; +\; x\; +\; 1\; =\; \backslash frac\{x^4\; -\; 1\}\{x\; -\; 1\}$, 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 further appears that the same condition will also be fulfilled as often as $x^4\; >\; Nx^2$ or $x\; >\; N$, 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. Approximation. 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 numeral equation, and those being found, we may approximate to the roots continually, as in the following example:

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

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.

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

and $y\; =\; \backslash frac\{5\}\{19\}\; =\; .26$, therefore, for a first approximation, we have $x\; =\; 2.26$.

Let us next suppose $x\; =\; 2.26\; +\; y\text{'}$, then, rejecting as before the second and higher powers of $y\text{'}$ on account of their smallness, we have

Hence $y\text{'}\; =\; \backslash frac\{.135\}\{18.119\}\; =\; .0075$ and $x\; =\; 2.26\; +\; y\text{'}\; =\; 2.2675$.

This value of $x$ is true to the last figure, but a more accurate value may be obtained by supposing $x\; =\; 2.2675\; +\; y\text{'}\text{'}$, and finding the value of $y\text{'}\text{'}$ in the same manner as we have already found those of $y\text{'}$ and $y$; and thus the

* The sign $\backslash Rightarrow$ denotes that the quantities between which it is placed are unequal. Thus $a\; \backslash Rightarrow\; b$, signifies that $a$ is greater than $b$, and $a\; \backslash Leftarrow\; b$, that $a$ is less than $b$.

Approximation. 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 shewn 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 $x^2\; =\; 1\; +\; 2y$, and therefore $2\; =\; 1\; +\; 2y$, and $y\; =\; \backslash frac\{1\}\{2\}\; =\; .5$ nearly, hence $x\; =\; 1.5$ nearly.

Let us next assume $x\; =\; 1.3\; +\; y\text{'}$, then, proceeding as before, we find $2\; =\; 1.69\; +\; 5.07y\text{'}$, hence $y\text{'}\; =\; \backslash frac\{.197\}\{5.07\}\; =\; .039$, and $x\; =\; 1.3\; +\; .039\; =\; 1.339$ nearly.

To find a still nearer approximation let us suppose $x\; =\; 1.339\; +\; y\text{'}\text{'}$, then from this assumption we find $y\text{'}\text{'}\; =\; .000079$, and therefore $x\; =\; 1.339079$, 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

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

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

By substituting $\backslash frac\{1\}\{6\}$ 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 ad-

vantage 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 $\backslash frac\{1\}\{y\}$ a fraction, which when added to $a$,

completes the root, then $x\; =\; a\; +\; \backslash frac\{1\}\{y\}$. 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 $x\; =\; a\; +\; \backslash frac\{1\}\{y\}$. But $b$ being only an approximate value of $y$, in the same manner as $a$ is an approximate value of $x$, we may suppose $y\; =\; b\; +\; \backslash frac\{1\}\{y\text{'}\}$, then, by substituting $b\; +\; \backslash frac\{1\}\{y\text{'}\}$ for $y$, we shall have a new equation, involving only $y\text{'}$, which must be greater than unity; putting therefore $b\text{'}$ to denote the next whole number less than the root of the equation involving $y\text{'}$, we have $y\; =\; b\; +\; \backslash frac\{1\}\{y\text{'}\}\; =\; \backslash frac\{b\text{'}\; +\; 1\}\{b\text{'}\}$, and substituting this value in that of $x$ the result is

for a second approximate value of $x$.

To find a third value we may take $y\text{'}\; =\; b\text{'}\; +\; \backslash frac\{1\}\{y\text{'}\text{'}\}$, then if $b\text{'}\text{'}$ denote the next whole number less than $y\text{'}\text{'}$, we have $y\text{'}\; =\; b\text{'}\; +\; \backslash frac\{1\}\{y\text{'}\text{'}\}\; =\; \backslash frac\{b\text{'}\text{'}\; +\; 1\}\{b\text{'}\text{'}\}$ whence

and so on to obtain more accurate approximations.

257. We shall apply this method to the following example

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.

0, 1, 2, 3, 4, 5, 6, 7, 8.

Results.

+7, +1, +1, +13, +43, +97, +181, +301, +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 successive 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, ends

Approximation. 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 $x=1+\backslash frac\{1\}\{y\}$, thus, by substitution, we have

The limit of the positive roots of this last equation is 5, and by substituting 0, 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

To approach nearer to the first value of $y$, let us take

This last equation will be found to have only one real root between 2 and 3, from which it appears that $y=1+\backslash frac\{1\}\{2\}=\backslash frac\{3\}\{2\}$, and $x=1+\backslash frac\{1\}\{\backslash frac\{3\}\{2\}\}=\backslash frac\{5\}\{3\}$.

Let us next suppose $y=2+\backslash frac\{1\}\{y\}$, hence we find

and from this equation $y$ is found to be between 4 and 5. Taking the least limit we have

It is easy to continue this process by assuming $y=4+\backslash frac\{1\}\{y\}$, and so on, as far as may be judged necessary.

We return to the second value of $x$, which was found $=\backslash frac\{3\}\{2\}$ by the first approximation, and which corresponds to $y=2$. Putting $y=2+\backslash frac\{1\}\{y\}$, and substituting this value in the equation $y^3\; -\; 4y^2\; +\; 3y\; +\; 1\; =\; 0$, which was formerly found, we get

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

Put $y=1+\backslash frac\{1\}\{y\}$, and we thence find by substitution

an equation which gives $y$ between 4 and 5, hence as before,

That we may proceed in the approximation we have only to suppose $y=4+\backslash frac\{1\}\{y\}$, and so on. The equation $x^3\; -\; 7x\; +\; 7$ has also a negative root between $-3$ and

$-4$, and to find a nearer value we may put $x=-3-\backslash frac\{1\}\{y\}$, hence we have $y^3\; -\; 20y^2\; -\; 9y\; -\; 1\; =\; 0$, and $y\; >\; 20$,

and therefore, for the first approximation, $x\; =\; -3\; -\; \backslash frac\{1\}\{20\}\; =\; -3.05$. By putting $y\; =\; 20\; +\; \backslash frac\{1\}\{y\}$, &c. we may

obtain successive values of $x$, each of which will be more exact than that which preceded it.

258. The successive equations which involve $y,\; y\text{'},\; y\text{'}\text{'},$ &c. have never more than one root greater than unity, unless that two or more roots of the proposed equation are contained between the limits $a$, and $a+1$, but when that circumstance has place, as in the preceding example, some one of the equations involving $y,\; y\text{'},$ &c. will have more than one root greater than unity, and from each root a series of equations may be derived, by which we may approximate to the particular roots of the proposed equation contained between the limits $a$ and $a+1$.

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 $\backslash frac\{ax\}\{a-x\}$ into an infinite series

Operation.

Thus it appears, that

Here the law of the series being evident, the terms may be continued at pleasure.

Ex. 2. It is required to convert $\backslash frac\{a^2\}\{(a+x)^2\}$ into an infinite series

Therefore , the law of continuation being evident.

261. A second method by which algebraic quantities, whether rational or irrational, may be converted

Now the quantities A, B, C, D, &c. 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, &c. which is contrary to what we have all along supposed. To determine the quantities A, B, C, D, &c. therefore, we have this series of equations

Here the law of relation which takes place among the quantities A, B, C, D, &c. is evident, viz. that if P, Q, R, denote any three coefficients which immediately follow each other

and from this equation, by means of the coefficients already determined, we find $F\; =\; 0$, $G\; =\; \backslash frac\{1\}\{a^6\}$, $H\; =\; -\backslash frac\{1\}\{a^7\}$, $K\; =\; 0$, &c.

Therefore, resuming the assumed equation, and substituting for A, B, C, &c. their respective values, we have

262. As a second example of the method of indeter-

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 $\backslash frac\{a^2\}\{a^2\; +\; ax\; +\; x^2\}$ is to be converted into a series proceeding by the powers of x; we are therefore to assume

where A denotes these terms of the series into which x does not at all enter; B the terms which contain only the first power of x; C 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

minate coefficients, let it be required to express the square root of $a^2\; -\; x^2$ by means of a series. For this purpose we might assume

but as we would find the coefficients of the odd powers of x, each = 0, let us rather assume

then, squaring both sides, and transposing, we have

Hence $A^2\; -\; a^2\; =\; 0$ and $A\; =\; a$

&c. &c.

and substituting for A, B, C, &c. their values

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 $\backslash frac\{1\}\{\backslash sqrt\{a^2\; +\; x^2\}\}$, or $\backslash frac\{\backslash sqrt\{a^2\; -\; x^2\}\}\{\backslash sqrt\{a^2\; +\; x^2\}\}$.

263. The binomial theorem affords a third method of resolving quantities into series, but before we explain this method it will be proper to show how the theorem itself may be investigated.

Let $a\; +\; x$ be any binomial quantity, which is to be raised

raised to a power denoted by $\backslash frac\{m\}\{n\}$, where $m$ and $n$ denote any numbers either positive or negative. Or because $a+x\; =\; a\backslash left(1+\backslash frac\{x\}\{a\}\backslash right)$, if we put $\backslash frac\{x\}\{a\}=y$, then $(a+x)^\{\backslash frac\{m\}\{n\}\}\; =\; a^\{\backslash frac\{m\}\{n\}\}\; \backslash times\; (1+y)^\{\backslash frac\{m\}\{n\}\}$; 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.

it appears that the powers of $1+x$ have this form

where the coefficients 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 $\backslash frac\{Q\}\{x^r\}$, 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

Then we have also

Let us put $(1+y)^\{\backslash frac\{m\}\{n\}\}\; =\; u$, $(1+x)^\{\backslash frac\{m\}\{n\}\}\; =\; v$, and therefore $(1+y)^\{\backslash frac\{m\}\{n\}\}\; =\; u^\{\backslash frac\{n\}\{m\}\}$, $(1+x)^\{\backslash frac\{m\}\{n\}\}\; =\; v^\{\backslash frac\{n\}\{m\}\}$, then, taking the difference between the two series, we have

Because $u^\{\backslash frac\{n\}\{m\}\}\; =\; 1+y$ and $v^\{\backslash frac\{n\}\{m\}\}\; =\; 1+x$, by subtracting the latter equation from the former, $u^\{\backslash frac\{n\}\{m\}\}\; -\; v^\{\backslash frac\{n\}\{m\}\}\; =\; y-x$, hence, and from the last series, we have

266. But every expression of the form $u^\{\backslash frac\{n\}\{m\}\}\; -\; v^\{\backslash frac\{n\}\{m\}\}$ is divisible by $u-v$, when $m$ is a whole number, thus we have

so that if we substitute for $\backslash frac\{u^\{\backslash frac\{n\}\{m\}\}\; -\; v^\{\backslash frac\{n\}\{m\}\}\}\{u^\{\backslash frac\{n\}\{m\}\}\; -\; v^\{\backslash frac\{n\}\{m\}\}\}$ its value, as found

from these equations, and divide each term of the series by the denominator $y-x$ we have

Now as this last equation must be true, whatever be the values of $y$ and $z$, we may suppose $y=x$, but in that case $1+y=1+x$ or $u=v$, and therefore $u=v$. Thus the equation is reduced to

or to the following:

so that, putting for $u^\{\backslash frac\{n\}\{m\}\}$ and $u^\{\backslash frac\{n\}\{m\}\}$ their values $(1+y)^\{\backslash frac\{m\}\{n\}\}$ and $1+y$ we have

But from the equation originally assumed we have

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 we have the following series of equations:

Infinite
Series.

&c.

&c.

Or, substituting for A, B, C, &c. their values as determined from the preceding equations:

&c.

267. Refusing now the assumed equation,

And observing that $\backslash frac\{x\}\{a\}\; =\; y$ and $(a+x)^\{\backslash frac\{m\}\{n\}\}\; =\; a^\{\backslash frac\{m\}\{n\}\}(1+y)^\{\backslash frac\{m\}\{n\}\}$ we have

where A, B, C, &c. denote the coefficients of the preceding terms, or

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 $\backslash frac\{u^m\; -\; v^m\}\{u\; -\; v\}$ when $u\; =\; v$ it has been taken for granted that

$\backslash frac\{m\}\{n\}$ is positive, but the same conclusion will be obtained when $\backslash frac\{m\}\{n\}$ is negative. For, changing $+m$ into $-m$, and observing that

we have

Now we have already found, that when $u\; =\; v$, the fraction $\backslash frac\{u^m\; -\; v^m\}\{u^n\; -\; v^n\}$ becomes $\backslash frac\{m\; u^\{m-n\}\}\{n\; u^\{n-1\}\}$, therefore in the same case

and from this last expression we derive the same value for $u^m$ or $(1+y)^\{\backslash frac\{m\}\{n\}\}$ 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 $\backslash frac\{m\}\{n\}$ 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. 1. It is required to express $\backslash frac\{r^3\}\{(r+x)^3\}$ by means of a series.

Let $\backslash left(1\; +\; \backslash frac\{x\}\{r\}\backslash right)^\{-3\}$ be compared with $(a+x)^\{\backslash frac\{m\}\{n\}\}$ and we have

Hence, by substituting these values of $a,\; x,\; m,\; n$ in the first general formula of (§267) we have

Ex. 2. It is required to express $\backslash sqrt[3]\{a+b\}$ by the form of a series.

By comparing $\backslash left(1\; +\; \backslash frac\{b\}\{a\}\backslash right)^\{\backslash frac\{1\}\{3\}\}$ with $(a+x)^\{\backslash frac\{m\}\{n\}\}$ we have $a=1$, $x=\backslash frac\{b\}\{a\}$, $m=1$, $n=3$,

and substituting as in last example

Ex. 3. It is required to resolve $\backslash frac\{r^2\}\{(r^2+z^2)^\{\backslash frac\{3\}\{2\}\}\}$ into a series.

Because $\backslash frac\{r^2\}\{(r^2+z^2)^\{\backslash frac\{3\}\{2\}\}\}\; =\; r^2\; \backslash times\; (r^2+z^2)^\{-\backslash frac\{3\}\{2\}\}$ if we raise $r^2+z^2$ to the $-\backslash frac\{3\}\{2\}$ power, and multiply the resulting series

by $r^2$, we shall have the series required. Or the given quantity may be reduced to a more simple form thus;

Ex. 4. It is required to find a series equal to $\backslash frac\{\backslash sqrt\{a^2+x^2\}\}\{\backslash sqrt\{a^2-x^2\}\}$.

First by the binomial theorem we have

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

SECT. XVIII. Of the Reversion of Series.

271. THE method of indeterminate coefficients, which we have already employed when treating of infinite se-

ries, 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 $y\; =\; n\; +\; ax\; +\; bx^2\; +\; cx^3\; +\; dx^4\; +\; \backslash dots$

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

Now when $x=0$, it is evident that $z=0$, therefore we may assume for $x$ a series of this form.

where the coefficients $A,\; B,\; C,\; D,\; \backslash dots$ denote quantities as yet unknown, but which are entirely independent of the quantity $x$. To determine those quantities let the first, second, third, &c. powers of the series

be

Of Logarithms, &c. be found by multiplication, and substituted for respectively, in the equation

thus we have

and, putting the coefficients of each = 0,

these equations give

&c.

273. As an example of the application of this formula, let it be required to determine $x$ from the equation

In this case we have

Therefore, substituting these values, we have

274. In the equation

in which both sides are expressed by series, and it is required to find $y$ in terms of $x$, we must assume, as before,

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 may be determined.

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 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, &c. 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:

In like manner may fractions be expressed by the powers of 2. Thus

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

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 $y\text{'}$ be two numbers, and $x$ and $x\text{'}$ their logarithms, so that $r^x\; =\; y$, and $r^\{x\text{'}\}\; =\; y\text{'}$, then $r^x\; \backslash times\; r^\{x\text{'}\}\; =\; yy\text{'}$, or $r^\{x+x\text{'}\}\; =\; yy\text{'}$, hence, from the definition, $x+x\text{'}$ is the logarithm of $yy\text{'}$, that is the sum of the logarithms of $y$ and $y\text{'}$ is the logarithm of $yy\text{'}$.

2. The difference of the logarithms of two numbers is equal to the logarithm of their quotient; for if

Of Logarithms, &c. if $r^x\; =\; y$ and $r^\{x\text{'}\}\; =\; y\text{'}$, then $\backslash frac\{r^x\}\{r^\{x\text{'}\}\}\; =\; \backslash frac\{y\}\{y\text{'}\}$ or $r^\{x-x\text{'}\}\; =\; \backslash frac\{y\}\{y\text{'}\}$,

therefore, by the definition, $x-x\text{'}$ is the logarithm of $\backslash frac\{y\}\{y\text{'}\}$; that is the difference of the logarithms of $y$ and $y\text{'}$ is the logarithm of $\backslash frac\{y\}\{y\text{'}\}$.

3. Let $n$ be any number whatever, then, $\backslash log.\; N^n\; =\; n\; \backslash times\; \backslash log.\; N$. For $N^n$ is $N$ multiplied into itself $n$ times, therefore the logarithm of $N^n$ is equal to the logarithm of $N$ added to itself $n$ times, or to $n\; \backslash times\; \backslash log.\; N$.

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 2.7182818 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 question 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 resume 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

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 $(1+a)^\{nx\}\; =\; (1+v)^n$, and resolving both sides of the equation into series by means of the binomial theorem,

Therefore, subtracting unity from both sides, and dividing by $n$, we have

and by supposing the factors which constitute the terms of each series to be actually multiplied, and the products arranged according to the powers of $n$, the last equation will have this form

Here the coefficients of the powers of $n$, viz. also 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,

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. $n$, that is

Let us now put $A$ to denote the constant multiplier

and substitute for $v$, its value $y-1$, thus we at last find

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:

But $\backslash log.\; \backslash frac\{n+x\}\{n\}\; =\; \backslash log.\; (n+x)\; -\; \backslash log.\; n$, therefore

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 $\backslash log.\; y$ 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. $\backslash frac\{1\}\{A\}$ 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 $\backslash frac\{1\}\{A\}\; =\; 1$ 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)

and that of $r\_1$ the radical number of any system is

1

Because Of Logarithms, &c.

By substituting $-v$ for $+v$ we have

Now, $\backslash log.\; (1+v)\; -\; \backslash log.\; (1-v)\; =\; \backslash log.\; \backslash frac\{1+v\}\{1-v\}$, therefore, subtracting the latter series from the former we have

Put $\backslash frac\{1+v\}\{1-v\}\; =\; y$, then $v\; =\; \backslash frac\{y-1\}\{y+1\}$ and the last series becomes

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 $\backslash frac\{n+x\}\{n\}$ for $y$ in the last formula we have

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 $l,\; l\text{'}$ the logarithms of the same number according to two other systems whose moduli are $m$ and $m\text{'}$; then

therefore $\backslash frac\{l\}\{m\}\; =\; \backslash frac\{l\text{'}\}\{m\text{'}\}$ and $m\; :\; m\text{'}\; ::\; l\; :\; l\text{'}$

That is, the logarithms of the same number, according to different systems, are directly proportional to the moduli of those 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, &c. 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

but the resulting series $\backslash frac\{2\; \backslash cdot\; 9\}\{11\}\; +\; \backslash frac\{2\; \backslash cdot\; 9^3\}\{3\; \backslash cdot\; 11^3\}\; +\; \backslash frac\{2\; \backslash cdot\; 9^5\}\{5\; \backslash cdot\; 11^5\}\; +\; \backslash dots$ 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

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 .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 $4\; =\; 2^2$ is also given § 279. Therefore, substituting $\backslash log.\; 4\; =\; 2\; \backslash log.\; 2$ for $\backslash log.\; n$, and 1 for $x$, in the series

The first three terms of this series are sufficient to give the result true to the seventh decimal, so that we have $\backslash text\{hyp.\; log.\; \}\; 5\; =\; 1.6094379$, and

Hence the modulus of the common system of logarithms, or $\backslash frac\{1\}\{\backslash text\{hyp.\; log.\; \}\; 10\}$ is found = .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

Having found the hyp. log. of 2 to be .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

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 $x\; \backslash times\; \backslash log.\; a\; =\; \backslash log.\; b$. Hence it follows,

VOL. I. PART II.

that $x\; =\; \backslash frac\{\backslash log.\; b\}\{\backslash log.\; a\}$. 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:

where $n$ is an indefinite quantity, which is to disappear in the course of the investigation.

By the binomial theorem we have

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

where it will readily appear that

as to the values of $B$, $C$, &c. it is of no importance to know them, for they will all disappear in the course of the investigation. Hence, by substituting for $(1\; +\; a)^x$ its value, as expressed by the last series, we have

and expanding the latter part of this equation by means of the binomial theorem it becomes

But $An\; +\; Bn^2\; +\; \backslash dots\; =\; n(A\; +\; Bn\; +\; \backslash dots)$ also $(An\; +\; Bn^2\; +\; \backslash dots)^2\; =\; n^2(A\; +\; Bn\; +\; \backslash dots)^2$, and $(An\; +\; Bn^2\; +\; \backslash dots)^3\; =\; n^3(A\; +\; Bn\; +\; \backslash dots)^3$, &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

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 $x$ ought therefore to destroy each other; and this being the case, the equation is reduced to

and since we have found

It is evident from § 288 that $\backslash Lambda$ 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

Here the radical number is expressed by means of its hyperbolic logarithm. Again, if we suppose $x\; =\; \backslash frac\{1\}\{\backslash Lambda\}$, then

Thus it appears that the quantity $r^\{\backslash frac\{1\}\{\backslash Lambda\}\}$ is equal to a constant number, which, by taking the sum of a sufficient number of terms of the series, will be found $=\; 2.718281828459045\backslash dots$. Let us denote this number by $e$, then $r^\{\backslash frac\{1\}\{\backslash Lambda\}\}\; =\; e$, and hence $r\; =\; e^\{\backslash Lambda\}$. Now if we remark that $\backslash Lambda$ 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 $r^\{\backslash frac\{1\}\{\backslash Lambda\}\}\; =\; e$, therefore $\backslash frac\{1\}\{\backslash Lambda\}\; \backslash times\; \backslash log.\; r\; =\; \backslash log.\; e$ and $\backslash Lambda\; =\; \backslash frac\{\backslash log.\; r\}\{\backslash log.\; e\}$, here $\backslash log.\; r$ and $\backslash log.\; e$ denote logarithms taken according to any system whatever.

295. If we now resume the equation,

and substitute for $\backslash Lambda$ its value $\backslash frac\{\backslash log.\; r\}\{\backslash log.\; e\}$ we shall have the following general expression for any exponential quantity whatever

which by supposing $r=e$ becomes

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 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 those 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

from which we find

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

$1\; :\; r\; ::\; R\; :\; rR$ = the interest of $R$ for a year, and $R\; +\; rR\; =\; (1+r)R\; =\; R^2$ = amount of $R$ 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 $R^2\; +\; rR^2\; =\; (1+r)R^2\; =\; R^3$; 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 $pR^t$, hence we have this formula

which

Annuities, which from the nature of logarithms may also be expressed thus:

Hence we find

or, by logarithms,

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 $p$.

therefore $p\; =\; 64.42.\; =\; 64.\; 8s.\; 5d.$ 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 $t$.

301. In treating of compound interest we have supposed the interest to be joined to the principal at the end of every 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 any thing 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 re-

quires 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;

let $r$ 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 years 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 years annuity, which is simply $A$. Therefore, the whole amount of the improved annuities is the geometrical series.

the sum of which, by § 106, is $A\; \backslash frac\{R^t\; -\; 1\}\{R\; -\; 1\}\; =\; A\; \backslash frac\{R^t\; -\; 1\}\{r\}$;

and since this sum must be equal to the amount of the purchase money, or $PR^t$, we have

and from this equation, we find

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

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 $\backslash frac\{1\}\{R^t\}\; =\; 0$, the formula

And if the annuity is to commence after $n$ years and

continue for ever, the formula $P\; =\; \backslash frac\{A\}\{rR^n\}\; \backslash left(\; 1\; -\; \backslash frac\{1\}\{R^t\}\; \backslash right)$ be-

comes $P\; =\; \backslash frac\{A\}\{rR^n\}$.

SECT. XXI. Of continued Fractions.

306. Every quantity which admits of being expressed by a common fraction may also be expressed in

Continued Fractions. 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 $\backslash frac\{314159\}\{100000\}$, or which is the same $3\; +\; \backslash frac\{14159\}\{100000\}$. Since $100000\; =\; 7\; \backslash times\; 14159\; +\; 887$, therefore $\backslash frac\{14159\}\{100000\}\; =\; \backslash frac\{14159\}\{7\; \backslash times\; 14159\; +\; 887\}\; =\; \backslash frac\{1\}\{7\; +\; \backslash frac\{887\}\{14159\}\}$ and $\backslash frac\{314159\}\{100000\}\; =\; 3\; +\; \backslash frac\{1\}\{7\; +\; \backslash frac\{887\}\{14159\}\}$.

Now $\backslash frac\{887\}\{14159\}\; =\; \backslash frac\{887\}\{15\; \backslash times\; 887\; +\; 854\}\; =\; \backslash frac\{1\}\{15\; +\; \backslash frac\{854\}\{887\}\}$, and substituting this for $\backslash frac\{887\}\{14159\}$, in the value of $\backslash frac\{314159\}\{100000\}$ already found we have $\backslash frac\{314159\}\{100000\}\; =\; 3\; +\; \backslash frac\{1\}\{7\; +\; \backslash frac\{1\}\{15\; +\; \backslash frac\{854\}\{887\}\}\}$.

Again, $\backslash frac\{854\}\{887\}\; =\; \backslash frac\{854\}\{854\; +\; 33\}\; =\; \backslash frac\{1\}\{1\; +\; \backslash frac\{33\}\{854\}\}$, which being substituted as before, gives $\backslash frac\{314159\}\{100000\}\; =\; 3\; +\; \backslash frac\{1\}\{7\; +\; \backslash frac\{1\}\{15\; +\; \backslash frac\{1\}\{1\; +\; \backslash frac\{33\}\{854\}\}\}\}$.

By operations similar to the preceding we find $\backslash frac\{33\}\{854\}\; =\; \backslash frac\{1\}\{25\; +\; \backslash frac\{29\}\{33\}\}$, $\backslash frac\{29\}\{33\}\; =\; \backslash frac\{1\}\{1\; +\; \backslash frac\{4\}\{29\}\}$, $\backslash frac\{4\}\{29\}\; =\; \backslash frac\{1\}\{7\; +\; \backslash frac\{1\}\{4\}\}$, therefore, by substitution

$\backslash frac\{314159\}\{100000\}\; =\; 3\; +\; \backslash frac\{1\}\{7\; +\; \backslash frac\{1\}\{15\; +\; \backslash frac\{1\}\{1\; +\; \backslash frac\{1\}\{25\; +\; \backslash frac\{1\}\{1\; +\; \backslash frac\{1\}\{7\; +\; \backslash frac\{1\}\{4\}\}\}\}\}\}$.

By an operation, in all respects the same as has been just now performed may any fraction whatever be reduced to the form $a\; +\; \backslash frac\{1\}\{b\; +\; \backslash frac\{1\}\{c\; +\; \backslash frac\{1\}\{d\; +\; \backslash dots\}\}\}$, &c.

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 $a\; +\; \backslash frac\{1\}\{b\; +\; \backslash frac\{1\}\{c\; +\; \backslash frac\{1\}\{d\; +\; \backslash dots\}\}\}$, &c.

Then since after the second, all the terms return in the

it will evidently be reduced to a common fraction by adding the reciprocal of $d$ to $b$, and the reciprocal of that sum to $c$, and again the reciprocal of this last sum to $a$; now the reciprocal of $d$, or $\backslash frac\{1\}\{d\}$, added to $c$ is $c\; +\; \backslash frac\{1\}\{d\}\; =\; \backslash frac\{cd+1\}\{d\}$, again the reciprocal of this sum, or $\backslash frac\{d\}\{cd+1\}$, added to $b$ is $b\; +\; \backslash frac\{d\}\{cd+1\}\; =\; \backslash frac\{bcd+b+d\}\{cd+1\}$, and the reciprocal of this last quantity, viz. $\backslash frac\{cd+1\}\{bcd+b+d\}$ when added to $a$ gives $\backslash frac\{abcd+ab+ad+cd+1\}\{bcd+b+d\}\; =\; a\; +\; \backslash frac\{1\}\{b\; +\; \backslash frac\{1\}\{c\; +\; \backslash frac\{1\}\{d\}\}\}$.

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 $\backslash frac\{314159\}\{100000\}$ 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 $\backslash pi$ for $\backslash frac\{314159\}\{100000\}$, we shall have $\backslash pi\; =\; 3\; +\; \backslash frac\{1\}\{7\}\; =\; \backslash frac\{22\}\{7\}$ nearly, and this is the proportion which was found by Archimedes. Again by taking the first three terms we have $\backslash pi\; =\; 3\; +\; \backslash frac\{1\}\{7\; +\; \backslash frac\{1\}\{15\}\}\; =\; 3\; +\; \backslash frac\{15\}\{106\}\; =\; \backslash frac\{333\}\{106\}$ which is nearer the truth than the former. And by taking the first four terms we have $\backslash pi\; =\; 3\; +\; \backslash frac\{1\}\{7\; +\; \backslash frac\{1\}\{15\; +\; \backslash frac\{1\}\{1\}\}\}\; =\; \backslash frac\{355\}\{113\}$ which is the proportion assigned by Metius, and is more exact than either of the preceding. These 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 $1\; +\; \backslash frac\{1\}\{2\; +\; \backslash frac\{1\}\{3\; +\; \backslash frac\{1\}\{2\; +\; \backslash frac\{1\}\{3\; +\; \backslash dots\}\}\}\}$, &c.

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 $\backslash pi\; =\; \backslash frac\{1\}\{2\; +\; \backslash frac\{1\}\{3\; +\; \backslash frac\{1\}\{2\; +\; \backslash frac\{1\}\{3\; +\; \backslash dots\}\}\}\}$, &c.

Then since after the second, all the terms return in the

Continued the same order, it follows that their amount is also $=x$,
Fractions, thus we have

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 $\backslash sqrt\{2\}\; =\; 1.41421356\; \backslash dots$ 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

and hence we see it 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 $x\; =\; a\; +\; \backslash frac\{1\}\{y\}$, $y\; =\; b\; +\; \backslash frac\{1\}\{y\text{'}\}$, $y\text{'}\; =\; b\text{'}\; +\; \backslash frac\{1\}\{y\text{'}\text{'}\}$, $y\text{'}\text{'}\; =\; b\text{'}\text{'}\; +\; \backslash frac\{1\}\{y\text{'}\text{'}\text{'}\}$, &c. where $a,\; b,\; b\text{'},\; b\text{'}\text{'}$, &c. denote the whole numbers, which are next less than the true values of $x,\; y,\; y\text{'},\; y\text{'}\text{'}$, &c. If therefore in the value of $x$ we substitute $b\; +\; \backslash frac\{1\}\{y\text{'}\}$ for $y$, it becomes

Again, if in this second value of $x$ we substitute $b\text{'}\; +\; \backslash frac\{1\}\{y\text{'}\text{'}\}$ for $y\text{'}$ it becomes

The next value of $x$ is in like manner found to be

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 the values of the numbers sought to positive integers, and therefore, in this case, we can have only these nine solutions;

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,

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 $dx\; +\; ey\; =\; \backslash frac\{c\}\{m\}$; but $d,\; e,\; x,\; y$ are supposed to be whole numbers, therefore $\backslash frac\{c\}\{m\}$ 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 $x\; =\; \backslash frac\{25-3y\}\{2\}\; =\; 12\; -\; \backslash frac\{3y\}\{2\}$;

now since $x$ must be a whole number it follows that $\backslash frac\{1-3y\}\{2\}$ must be a whole number. Let us assume $\backslash frac\{1-3y\}\{2\}\; =\; z$, then $1-3y\; =\; 2z$ and $y\; =\; 1-2z$, and since $x\; =\; 12\; -\; y\; +\; \backslash frac\{1-3y\}\{2\}\; =\; 12\; -\; y\; +\; z$, therefore $x\; =\; 12\; -\; 1\; +\; 2z\; +\; z$; hence we have

Indeterminate Problems. 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, abstracting 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

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

hence it appears that $\backslash frac\{2\; -\; 4y\}\{7\}$ must be a whole number. Let us assume $\backslash frac\{2\; -\; 4y\}\{7\}\; =\; z$, then $x\; =\; 14\; -\; y\; +\; z$ and $4y\; =\; 2\; -\; 7z$ or $y\; =\; \backslash frac\{2\; -\; 7z\}\{4\}\; =\; \backslash frac\{2\; -\; 3z\}\{4\}\; -\; z$, therefore $\backslash frac\{2\; -\; 3z\}\{4\}$ must be a whole number. Assume $\backslash frac\{2\; -\; 3z\}\{4\}\; =\; t$, then $y\; =\; t\; -\; z$, and $3z\; =\; 2\; -\; 4t$, or $z\; =\; \backslash frac\{2\; -\; 4t\}\{3\}\; =\; \backslash frac\{2\; -\; t\}\{3\}\; -\; t$, therefore $\backslash frac\{2\; -\; t\}\{3\}$ must be a whole number.

Assume now $\backslash frac\{2\; -\; t\}\{3\}\; =\; v$, 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:

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 601. 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 $x\; =\; \backslash frac\{400\; -\; 9y\}\{7\}\; =\; 57\; -\; y\; +\; \backslash frac\{1\; -\; 2y\}\{7\}$, so that $\backslash frac\{1\; -\; 2y\}\{7\}$ must be a whole number.

Assume $\backslash frac\{1\; -\; 2y\}\{7\}\; =\; z$, then $x\; =\; 57\; -\; y\; +\; z$ and $2y\; =\; 1$

$-7z$ or $y\; =\; \backslash frac\{1\; -\; 7z\}\{2\}\; =\; \backslash frac\{1\; -\; z\}\{2\}\; -\; 3z$ therefore $\backslash frac\{1\; -\; z\}\{2\}$ must be a whole number. Indeterminate Problems.

Assume $\backslash frac\{1\; -\; z\}\{2\}\; =\; v$, 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

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 $ax\; -\; ty\; =\; c$, or $ax\; =\; ty\; +\; c$, 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

Therefore $\backslash frac\{11y\; +\; 7\}\{20\}$ must be a whole number.

Let $\backslash frac\{11y\; +\; 7\}\{20\}\; =\; v$, then $x\; =\; y\; +\; v$ and $y\; =\; \backslash frac\{20v\; -\; 7\}\{11\}\; =\; v\; -\; \backslash frac\{7\}\{11\}$; hence $\backslash frac\{7\}\{11\}$ must be a whole number.

Let $\backslash frac\{7\}\{11\}\; =\; t$, then $y\; =\; v\; +\; t$ and $v\; =\; \backslash frac\{11t\; +\; 7\}\{9\}\; =\; t\; +\; \backslash frac\{2t\; +\; 7\}\{9\}$; therefore $\backslash frac\{2t\; +\; 7\}\{9\}$ is a whole number.

Let $\backslash frac\{2t\; +\; 7\}\{9\}\; =\; s$, then $v\; =\; t\; +\; s$ and $t\; =\; \backslash frac\{9s\; -\; 7\}\{2\}\; =\; 4s\; -\; \backslash frac\{7\}\{2\}$; therefore $\backslash frac\{7\}\{2\}$ is a whole number.

Put $\backslash frac\{7\}\{2\}\; =\; r$, then $t\; =\; 4s\; +\; r$ and $v\; =\; 2s\; +\; 7$.

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 $r\; =\; -2,\; r\; =\; -1,\; r\; =\; 0,\; r\; =\; +1$, &c. Thus we have

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 $v,\; t$, &c. we shall immediately perceive that the coefficients of those quantities are the same as the successive 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

Hence we may form a series of numeral equations which, when compared with the series of literal equations expressing the relations between $x,\; y,\; v$, &c. as put down in the following table, will render the method of determining the latter from the former sufficiently obvious

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 $a$ be the greatest of the two numbers $a,\; b$, and 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.

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 show 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.

Here $r$ may be assumed of any value whatever, Hence we have

and the number required $N\; =\; 209r\; +\; 157$ where it is evident that the least number which can express $N$ is 157.

Ex. 6. $\backslash left.\; \backslash begin\{array\}\{l\}\; 3x\; +\; 5y\; +\; 7z\; =\; 560\; \backslash \backslash \; 9x\; +\; 25y\; +\; 49z\; =\; 2920\; \backslash end\{array\}\; \backslash right\backslash \}$ To determine $x,\; y,\; z$ in whole numbers.

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 examples we find

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 $7z\; +\; 15v\; =\; 1560$. This last equation being resolved in the same manner we find

and hence it appears that the only values which $x$ can have so as to give whole positive numbers for $x,\; y,\; z$ are 209 and 210, thus we have

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 $d-cz=c\text{'}$, $ax+by=c\text{'}$. From this last equation we may find values of $x$ and $y$ of this form

where $m$ 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 $m$ 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,

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 $y\; =\; \backslash frac\{a\}\{b+cx\}$ it is evident $b+cx$ must be a divisor of $a$, let $d$ be one of its divisors, then $b+cx=d$, and $x\; =\; \backslash frac\{d-b\}\{c\}$, 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 $y\; =\; \backslash frac\{a+bx\}\{c+dx\}$, 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 $dy\; =\; \backslash frac\{da+dbx\}\{c+dx\}$ or $dy\; =\; b\; +\; \backslash frac\{ad-bc\}\{c+dx\}$, 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 $y\; =\; \backslash frac\{195-x\}\{1+2x\}$, therefore

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 $y\; =\; \backslash sqrt\{a+bx+cx^2\}$ 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 $g^2$ for $a$, we have $y\; =\; \backslash sqrt\{g^2+bx+cx^2\}$. Suppose $\backslash sqrt\{g^2+bx+cx^2\}\; =\; g+mx$ then $g^2+bx+cx^2\; =\; g^2\; +\; 2gmx\; +\; m^2x^2$, or $bx+cx^2\; =\; 2gmx\; +\; m^2x^2$, that is $b+cx\; =\; 2gm\; +\; m^2x$, hence

Here $m$ may be any rational quantity either whole or fractional.

324. Case 2. Let $c$ be a square number $=g^2$, then putting $\backslash sqrt\{a+bx+g^2x^2\}\; =\; m+gx$, we find $a+bx+g^2x^2\; =\; m^2\; +\; 2mgx\; +\; g^2x^2$, or $a+bx\; =\; m^2\; +\; 2mgx$, hence we find

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 $b+lx$ the irrationality may be taken away as follows.

Assume $\backslash sqrt\{a+bx+cx^2\}\; =\; \backslash sqrt\{(f+gx)(b+lx)\}\; =\; m(f+gx)$, then $(f+gx)(b+lx)\; =\; m^2(f+gx)^2$, or $b+lx\; =\; m^2(f+gx)$, hence we find

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 $p,\; q$, and $r$ are quantities which contain no power of $x$ higher than the first. Let us assume $\backslash sqrt\{p^2+qr\}\; =\; p+mq$; thus we have $p^2+qr\; =\; p^2\; +\; 2mpq\; +\; m^2q^2$ and $r\; =\; 2mp\; +\; m^2q$, 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 $\backslash sqrt\{a+bx+cx^2\}$ 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

Indeterminate required, and that $q$ is the corresponding value of $y$, then

Therefore, by subtraction,

$y^2\; -\; q^2\; =\; b(x\; -\; p)\; +\; c(x^2\; -\; p^2)\; =\; (b\; +\; cp\; +\; cx)(x\; -\; p)$ and $y\; =\; \backslash sqrt\{q^2\; +\; (b\; +\; cp\; +\; cx)(x\; -\; p)\}$. The quantity under the radical sign being now reduced to the preferred form, it may be rendered rational by the substitution pointed out in 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 question $x^2\; +\; y^2\; =\; a^2$, and $y\; =\; \backslash sqrt\{a^2\; -\; x^2\}$. This equation is evidently of such a form as to be resolvable by the method employed in case 1. Accordingly by comparing $\backslash sqrt\{a^2\; -\; x^2\}$ with the general expression $\backslash sqrt\{g^2\; +\; bx\; +\; cx^2\}$ we have $g\; =\; a,\; b\; =\; 0,\; c\; =\; -1$, and substituting these, values in the formulae of § 323. also for $+m$, we find

$x\; =\; \backslash frac\{2an\}\{n^2\; +\; 1\},\; y\; =\; \backslash frac\{a(n^2\; -\; 1)\}\{n^2\; +\; 1\}$, hence the numbers required are

If $a\; =\; n^2\; +\; 1$, 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

$x\; =\; \backslash frac\{b^2\; -\; n^2\}\{2n\}$ and $x\; +\; n\; =\; \backslash frac\{b^2\; +\; n^2\}\{2n\}$. So that the numbers

sought are $\backslash frac\{(b^2\; +\; n^2)^2\}\{4n^2\},\; \backslash frac\{(b^2\; -\; n^2)^2\}\{4n^2\}$

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 $\backslash frac\{x^2\; +\; x\}\{2\}$ may be a rational square.

Let $y$ be the side of the square required, then $\backslash frac\{x^2\; +\; x\}\{2\}\; =\; y^2$ and $4x^2\; +\; 4x\; =\; 8y^2$. 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 $2x\; +\; 1\; =\; \backslash sqrt\{1\; +\; 8y^2\}$, so that we have to make $1\; +\; 8y^2$ a square. Assume

VOL. I. Part II.

$1\; +\; 8y^2\; =\; \backslash left(1\; +\; \backslash frac\{p\}\{q\}y\backslash right)^2\; =\; 1\; +\; \backslash frac\{2p\}\{q\}y\; +\; \backslash frac\{p^2\}\{q^2\}y^2$, then $8y\; =\; \backslash frac\{2p\}\{q\}y\; +\; \backslash frac\{p^2\}\{q^2\}y^2$. Hence by proper reduction $y\; =\; \backslash frac\{2pq\}\{8q^2\; -\; p^2\}$ and

since $2x\; +\; 1\; =\; \backslash sqrt\{1\; +\; 8y^2\}\; =\; \backslash frac\{8q^2\; +\; p^2\}\{8q^2\; -\; p^2\}$ therefore $x\; =\; \backslash frac\{p^2\}\{8q^2\; -\; p^2\}$

and $\backslash frac\{x^2\; +\; x\}\{2\}\; =\; \backslash frac\{4p^2q^2\}\{(8q^2\; -\; p^2)^2\}$, 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, or 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.

Resolution of Geometrical Problems.

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

Hence $b^2=s^2-2sx$, and $x=\backslash frac\{s^2-b^2\}\{2s\}=BC$. Also

$s-x=\backslash frac\{s^2-b^2\}\{2s\}=\backslash frac\{s^2+b^2\}\{2s\}=AC$. 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

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 $x^2-sx=\backslash frac\{a^2-s^2\}\{2\}$, which being resolved, by completing the square, we find $x=\backslash frac\{s\backslash pm\backslash sqrt\{2a^2-s^2\}\}\{2\}=AB$, and

$s-x=\backslash frac\{s\backslash mp\backslash sqrt\{2a^2-s^2\}\}\{2\}=BC$. Thus it appears that either of the two quantities $\backslash frac\{s+\backslash sqrt\{2a^2-s^2\}\}\{2\}$, $\backslash frac\{s-\backslash sqrt\{2a^2-s^2\}\}\{2\}$

may be taken for AB, but which ever 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 $AD:BC::AK:EF$; that is $p:b::x:p-x$. Hence by taking the product of the extremes and means, $p^2=px+bx$, and $x=\backslash frac\{p^2\}\{p+b\}=AK$. If the side of the square be required, it may be immediately found by subtracting AK from AD the perpendicular. Thus we have $p-\backslash frac\{p^2\}\{p+b\}=\backslash frac\{pb\}\{p+b\}=KD=EF$. 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

So that to determine $x$ and $y$ we have these two equations

From the first equation we find $y=\backslash frac\{a^2\}\{x\}$, and from the second $y=\backslash frac\{bp-px\}\{b\}$, therefore $\backslash frac\{bp-px\}\{b\}=\backslash frac\{a^2\}\{x\}$ hence $x^2-bx=\backslash frac\{a^2b\}\{p\}$, and from this quadratic equation, by completing the square, &c. we find

Hence it appears that if $\backslash frac\{a^2b\}\{p\}$ be less than $\backslash frac\{b^2\}\{4\}$, that is if $a^2$ be less than $\backslash frac\{b^2p\}\{4\}$, 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 $\backslash frac\{b^2p\}\{4\}$, 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 of Geometrical Problems. ABC and the sum of the sides AB, BC. Put AC = a, AB + BC = b, cosine of $\backslash angle\; ABC\; =\; c$, 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, $\backslash text\{rad.\}\; :\; \backslash text\{cos.\; \}\; B\; ::\; CB\; :\; BD$; therefore $BD\; =\; \backslash text\{cos.\; \}\; B\; \backslash times\; CB\; =\; cy$.

Now, from the principles of geometry, $AC^2\; =\; AB^2\; +\; BC^2\; -\; 2AB\; \backslash times\; BC\; \backslash cos\; B$. Hence, and from the question, we have these two equations

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 $2xy\; =\; \backslash frac\{b^2\; -\; a^2\}\{1+c\}$. Again, from the square of the first equation let the double of this last equation, viz. $4xy\; =\; \backslash frac\{2(b^2\; -\; a^2)\}\{1+c\}$, be subtracted, and the result is $x^2\; -\; 2xy\; +\; y^2\; =\; \backslash frac\{2a^2\; -\; (1+c)b^2\}\{1+c\}$, so that by taking the square root of this last equation we obtain

Thus we have found the difference between the sides, now their sum is given = b, hence, by adding $\backslash frac\{1\}\{2\}$ the difference to $\backslash frac\{1\}\{2\}$ the sum we find

and subtracting $\backslash frac\{1\}\{2\}$ the difference from $\backslash frac\{1\}\{2\}$ the sum

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,

may be constructed as follows.

333. Construction of the first and second forms. Let a circle EABD (fig. 5.) be described with a radius $=\; \backslash frac\{1\}\{2\}a$, 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 = a.

Let H be the centre of the circle, join 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, there-

fore, if CD be called x, then CE = x + a, but if CE = x, then CD = x - a. Now by the elements of geometry $EC\; \backslash times\; CD\; =\; AC\; \backslash times\; CB$, 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 $\backslash frac\{1\}\{2\}a$ 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 $=\; \backslash frac\{1\}\{2\}a$ 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 $EC\; \backslash times\; CD\; =\; AC\; \backslash times\; CB$, 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 = $\backslash frac\{1\}\{2\}a$ and base AC = b, the roots of the equation will be expressed by AH + HC and AH - HC.

335. If b and c be so unequal, that b - c in the first two cases, or b + c in the third, is greater than a, then, instead of these quantities, $\backslash frac\{1\}\{2\}b$ and 2c, or in general $\backslash frac\{b\}\{n\}$ 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 bi-quadratic 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 bi-quadratic 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', &c. the extremities of these perpendiculars, as there are dimensions of y in the proposed equation. And the values of PM, PM', &c. will be the roots of the equation which are found by substituting for x its value in any particular

case. 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 $M,\; M\text{'},\; \backslash dots$ will lie all on one side of $AE$, but if any of them be negative, these 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 the 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 $y$ vanish, then the point $M$ coincides with $P$, and the locus meets $AE$ in that point. If one of the values of $y$ become infinite, then it shows 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 $y$ vanish, 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 abscissas 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 $y\; =\; \backslash frac\{bx\; +\; cd\}\{a\}$, 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 $\backslash frac\{cd\}\{a\}$, if $DE$ be drawn parallel to $AN$, then will $DE$

be the locus required; where it is to be observed that $AD$ and $PN$ are to be taken on the same side of $AE$ 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 $x$ vanish, 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 triplex from the equality of three roots.

346. 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 $y^2\; =\; \backslash pm\; \backslash sqrt\{ax\; +\; ab\}$, if $AP\; =\; x$ (fig. 10.) be assumed of a known value and $PM,\; PM\text{'}$ set off on each side equal to $\backslash sqrt\{ax\; +\; ab\}$ 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 $\backslash sqrt\{ax\; +\; ab\}$ become, and consequently $PM$ and $PM\text{'}$ the greater, and if $AP$ be supposed infinitely great, $PM$ and $PM\text{'}$ 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 $y\; =\; \backslash pm\; \backslash sqrt\{ab\}$, so that $y$ does not vanish in that case, but passes through $D$ and $d$, taking $AD$ and $Ad$ each $=\; \backslash sqrt\{ab\}$.

If $P$ be supposed to move to the other side of $A$, then $x$ becomes negative, and $y\; =\; \backslash pm\; \backslash sqrt\{ab\; -\; ax\}$, 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 $y\; =\; \backslash pm\; \backslash sqrt\{ab\; -\; ax\}\; =\; 0$; that is $PM$ and $PM\text{'}$ 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, $\backslash sqrt\{ab\; -\; ax\}$ 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 $b\; =\; \backslash pm\; x\; =\; BP$ and $y\; =\; PM$, from the equation $ab\; =\; ax\; +\; y^2$, or $a(b\; -\; x)\; =\; y^2$, we have $a\; \backslash times\; BP\; =\; PM^2$, 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 $xy\; +\; ay\; +\; cy\; =\; bc\; +\; bx$,

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 $y\; =\; \backslash frac\{bc\}\{a\; +\; c\}$ 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 $y\; =\; b\; \backslash times\; \backslash frac\{c\; +\; x\}\{a\; +\; c\; +\; x\}$, 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 $y\; =\; b\; \backslash times\; \backslash frac\{x\}\{x\}\; =\; b$, 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 $y\; =\; b\; \backslash times\; \backslash frac\{c\; -\; x\}\{a\; +\; c\; -\; x\}$, and if $x$ be taken on that side

$=\; c$, then $y\; =\; b\; \backslash times\; \backslash frac\{c\; -\; c\}\{a\}\; =\; 0$, so that the curve must pass through $B$ if $AB\; =\; c$. If $x$ be supposed greater than $c$, then $c\; -\; x$ becomes negative, and the ordinate will become negative, and lie on the other side of $AE$,

till $x$ become equal to $a\; +\; c$, and then $y\; =\; b\; \backslash times\; \backslash frac\{-a\}\{0\}$, that is, because the denominator is $0$, $x$ becomes infinite, so that if $A$ be taken $=\; a\; +\; c$, the ordinate $GD$ 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 $y\; =\; b\; \backslash times\; \backslash frac\{x\; -\; c\}\{x\; -\; a\; -\; c\}$ 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.

This 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 $NM\; \backslash times\; GN\; =\; GC\; \backslash times\; BC$ which is the property of the common hyperbola.

Ex. 3. It is required to describe the locus of the equation $xy^2\; -\; xy\; =\; x^3\; +\; bx^2$.

Here $y^2\; =\; \backslash frac\{x^3\; +\; bx^2\}\{a\; -\; x\}$, and therefore $y\; =\; \backslash pm\; \backslash sqrt\{\backslash frac\{x^3\; +\; bx^2\}\{a\; -\; x\}\}$, whence $PM$ and $PM\text{'}$ (fig. 12.) are to be taken on each side, and equal to $\backslash sqrt\{\backslash frac\{x^3\; +\; bx^2\}\{a\; -\; x\}\}$; 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 $\backslash frac\{x^3\; +\; bx^2\}\{a\; -\; x\}$ 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 $y\; =\; \backslash pm\; \backslash sqrt\{\backslash frac\{-x^3\; +\; bx^2\}\{c\; +\; x\}\}$, hence the values of $y$ will be real and equal as long as $x$ is less than $b$, but if $x\; =\; b$, then $y\; =\; \backslash sqrt\{\backslash frac\{-b^3\; +\; b^2\}\{a\; -\; b\}\}$, and consequently if $AD$ be taken $=\; b$, the curve will pass through $D$, and there touch the

the ordinate. If $x$ be taken greater than $b$, then $\backslash pm\; \backslash sqrt\{\backslash frac\{-x^2\; -\; ax^2\}\{c+x\}\}$ 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 $x^2\; +\; bx^2\; =\; 0$ 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 $ay^2\; -\; xy^2\; =\; x^3$, then will $A$ and $D$ coincide (fig. 13.) and the nodus vanish, 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 Cissoid of Diocles.

If instead of supposing $b$ positive, or equal to $0$, we suppose it negative, the equation will be $ay^2\; -\; xy^2\; =\; x^3\; -\; bx^2$, 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.