ALGEBRA (1778) — Encyclopaedia Britannica Historical Editions

ALGEBRA

Volume 1 · 16,037 words · 1778 Edition

A General method of computation, wherein signs and symbols, commonly the letters of the alphabet, are made use of to represent numbers, or any other quantities.

This science, properly speaking, is no other than a kind of short-hand, or ready way of writing down a chain of mathematical reasoning on any subject whatever; so that it is applicable to arithmetic, geometry, astronomy, mensuration of all kinds of solids, &c., and the great advantages derived from it appear manifestly to arise from the conciseness and perspicuity with which every proposition on mathematical subjects can be written down in algebraic characters, greatly superior to the tedious circumlocutions which would be necessary were the reasoning to be written in words at length.

With regard to the etymology of the word algebra, it is much contested by the critics. Menage derives it from the Arabic algebrat, which signifies the restitution of any thing broken; supposing that the principal part of algebra is the consideration of broken numbers. Others rather borrow it from the Spanish, algebrista, a person who replaces dislocated bones; adding, that algebra has nothing to do with fraction. Some, with M. d'Herbelot, are of opinion, that algebra takes its name from Gebar, a celebrated philosopher, chemist, and mathematician, whom the Arabs call... call Giaher, and who is supposed to have been the inventor. Others from gesfr, a kind of parchment made of the skin of a camel, whereon Ali and Giafer Sadek wrote, in mystic characters, the fate of Mahometanism, and the grand events that were to happen, till the end of the world. But others, with more probability, derive it from gaber; a word whence, by prefixing the article al-, we have formed algebra; which is pure Arabic, and properly signifies the reduction of fractions to a whole number. However, the Arabs, it is to be observed, never use the word algebra alone, to express what we mean by it; but always add to it the word macabelah, which signifies opposition and comparison; thus algebra-almacabelah, is what we properly call algebra.

Some authors define algebra, The art of solving mathematical problems; but this is rather the idea of analysis, or the analytic art. The Arabs call it, The art of resolution and comparison; or, The art of resolution and equation. Lucas de Burgo, the first European who wrote of algebra, calls it, Regula rei et censur; that is, the rule of the root and its square; the root with them being called rei, and the square censur. Others call it Specious Arithmetic; and some, Universal Arithmetic.

It is highly probable that the Indians or Arabians first invented this noble art: for it may be reasonably supposed, that the ancient Greeks were ignorant of it; because Pappus, in his mathematical collections, where he enumerates their analysts, makes no mention of any thing like it; and, besides, speaks of a local problem, begun by Euclid, and continued by Apollonius, which none of them could fully resolve; which doubtless they might easily have done, had they known any thing of algebra.

Diophantus was the first Greek writer of algebra, who published thirteen books about the year 800, tho' only five of them were translated into Latin, by Xylander, in 1575; and afterwards, viz. anno 1621, in Greek and Latin, by M. Bachet and Fermat, with additions of their own. This algebra of Diophantus's only extends to the solution of arithmetical indeterminate problems.

Before this translation of Diophantus came out, Lucas Paciolus, or Lucas de Burgo, a Minorite friar, published at Venice, in the year 1494, an Italian treatise of algebra. This author makes mention of Leonardus Pifanus, and some others, of whom he had learned the art; but we have none of their writings. He adds, that algebra came originally from the Arabs, and never mentions Diophantus; which makes it probable, that that author was not then known in Europe. His algebra goes no farther than simple and quadratic equations.

After Paciolus, appeared Stifelius, a good author; but neither did he advance any farther.

After him, came Scipio Ferreus, Cardan, Tartaglia, and some others, who reached as far as the solution of some cubic equations. Bombelli followed these, and went a little farther. At last came Nurnius, Ramus, Schoner, Salignac, Clavius, &c. who all of them took different courses, but none of them went beyond quadratics.

In 1590, Vieta introduced what he called his Specious Arithmetics, which consists in denoting the quantities, both known and unknown, by symbols or letters. He also introduced an ingenious method of extracting the roots of equations, by approximations; since greatly improved and facilitated by Raphson, Halley, Simpson, and others.

Vieta was followed by Oughtred, who, in his Clavis Mathematicae, printed in 1631, improved Vieta's method, and invented several compendious characters, to shew the sums, differences, rectangles, squares, cubes, &c.

Harriot, another Englishman, contemporary with Oughtred, left several treatises at his death; and among the rest, an Analysis, or Algebra, which was printed in 1631, where Vieta's method is brought into a still more commodious form, and is much esteemed to this day.

In 1657, Des Cartes published his geometry, wherein he made use of the literal calculus and the algebraic rules of Harriot; and as Oughtred in his Clavis, and Marin Ghetaldus in his books of mathematical composition and resolution published in 1630, applied Vieta's arithmetic to elementary geometry, and gave the construction of simple and quadratic equations; so Des Cartes applied Harriot's method to the higher geometry, explaining the nature of curves by equations, and adding the constructions of cubic, biquadratic, and other higher equations.

Des Cartes's rule for constructing cubic and biquadratic equations, was farther improved by Thomas Baker, in his Clavis Geometrica Catholicæ, published in 1684; and the foundation of such contractions, with the application of algebra to the quadratures of curves, questions de maximis et minimis, the centrobary method of Guldinus, &c., was given by R. Slufins, in 1668; as also by Fermat in his Opera Mathematica, Roberval in the Mem. de Mathema. et de Physique, and Barrow in his Lectiones Geometricæ. In 1708, algebra was applied to the laws of chance and gaming, by R. de Montmort; and since by de Moivre and James Bernoulli.

The elements of the art were compiled and published by Kersey, in 1671; wherein the specious arithmetic, and the nature of equations, are largely explained, and illustrated by a variety of examples: the whole substance of Diophantus is here delivered, and many things added concerning mathematical composition and resolution from Ghetaldus. The like has been since done by Preflet in 1694, and by Ozanam in 1703; but these authors omit the application of algebra to geometry; which defect is supplied by Guifnee in a French treatise expressly on the subject published in 1704, and l'Hôpital in his analytical treatise of the conic sections in 1707. The rules of algebra are also compendiously delivered by Sir Isaac Newton, in his Arithmetica Universalis, first published in 1707, which abounds in select examples, and contains several rules and methods invented by the author.

Algebra has also been applied to the consideration and calculus of infinites; from whence a new and extensive branch of knowledge has arisen, called the Doctrine of Fluxions, or Analysis of Infinites, or the Calculus Differentialis.

Sect. I. Elementary Rules.

In algebra, a letter of the alphabet may stand for any quantity whatever; whether length, breadth, thickness, ness, solidity, &c., but when once a letter is appropriated to one particular kind of quantity, it cannot stand for any other, in that demonstration, or piece of reasoning. Thus, though the letter $a$ may represent any quantity of water, earth, &c., yet if it is once appropriated to any of these, water, for instance, it cannot likewise represent earth; as this would produce confusion. Each species of quantity, therefore, must be represented by a different letter.—As all quantities, concerning which we speak, must be either known or unknown; and both these are frequently represented by letters in algebraic operations; it will be proper to use the first letters of the alphabet, $a$, $b$, $c$, &c., to represent one kind of quantities; and the last letters, $x$, $y$, $z$, to represent the others; that there may be as little danger of mistake as possible.

Besides this obvious division of quantity, into known and unknown; algebraists consider quantities as positive or negative, simple or compound, roots or powers, rational or irrational.—Positive quantities are such as, by their presence, always denote an increase, or addition of something which was not there before; and therefore they have always $+$ (or $+$), the sign of addition, prefixed to them; but as quantity is generally spoken of in a positive sense, the sign is omitted before a single letter, or before the first term of any series of quantities expressed by letters. Thus, if $a$ simply is wrote down, $+a$ is supposed to be meant; in like manner, in the series $a + b + c$, &c., the first term or letter is always supposed to be positive. Negative quantities are intended to express the difference between one positive quantity and another. By themselves they cannot have any existence, as they would be less than nothing, which is absurd. These quantities have always the sign of subtraction, $-$ (or $-$), prefixed to them; whether they stand first or last. If a single letter is marked with the sign of subtraction, it is always supposed to have a respect to some other quantity which is not expressed. Thus, $a$ by itself represents a positive quantity of any kind; $-a$ does not by itself represent any thing, but only the difference between the former $a$, or $+a$, and some other quantity which at that time is not expressed; but if another quantity, expressed by $b$, is wrote down before it, as $b - a$, this denotes the difference between $b$ and $a$. The same thing would be denoted though the order of the terms were inverted; $b - a$ is the same with $-a + b$; but in writing the terms of an algebraic series, positive quantities ought to precede negative ones; and those which have like signs, whether $+$ or $-$, ought always to be placed together.

By attending to this distinction between positive and negative quantities, addition and subtraction of algebraic characters will be very easy. Every letter in algebra is supposed to represent something real; and the letter is only put for it; because it is either expressed than the name of the thing itself. Thus, suppose $a$ to represent a gallon of water; if I want to add another gallon, or another $a$, to the first one, the sum is two gallons; or, in algebraic short-hand, $2a$. In like manner, if we want to add another $a$, the sum will be $3a$. But if we want to add one species of quantity to another, as a pound of earth to a gallon of water, we must take one letter for the one species, and another for the other. Thus, let $a$ represent the earth, and $b$ the water; when these two are added together, the sum is neither two pounds of earth, nor two gallons of water: the sum of their literal representatives, therefore, can neither be $2a$, nor $2b$; but $a + b$. Here it will be observed, that, where quantities of the same kind, expressed by the same letter, are added together, some arithmetical figures must be prefixed to the algebraic ones; and these numbers, called coefficients, or inches, are to be managed exactly in the same way, as in common arithmetic. Thus $a + a$ is $2a$, and $2a + 3a$ is $5a$: added to $b$ can only be represented by $a + b$; in like manner, $7a$ added to $5b$ will neither make $12a$, nor $12b$; therefore, the sum of these two can only be represented by $7a + 5b$.

When quantities occur which have contrary signs, there is a necessity for subtracting the one from the other, in order to come at the true sum. Thus if a man has $£10$ of stock in hand, and $£5$ of debt; in order to come at his real worth, we must subtract the debt from the goods. If the $£10$ of goods is represented by $10a$, and the $£5$ of debt by $5a$; it is as plain, that the sum must be only $10a - 5a$, or $5a$. If dissimilar letters occur, having contrary signs, they must be wrote down with the signs prefixed that are proper to each. Thus, the sum of $2a$, $3a$, and $-7b$, is $2a + 3a - 7b$, or $5a - 7b$; of $2a$, $5a$, $6b$, and $-7b$, is $2a + 5a + 6b - 7b$, or $7a - b$, &c.

Subtraction of algebraic characters consists only in changing the sign of the quantity to be subtracted, and then following the above rules for addition. Thus, if I am to take $2a$ from $5a$, I change the sign of the $2a$, and write it thus, $5a - 2a$; adding these, I find the sum to be $3a$, as already mentioned. If the letters are dissimilar, they must be wrote down with the sign of subtraction — between them: as, if I subtract $b$ from $a$, the remainder will be $a - b$; but if I take $-b$ from $a$, I must change its sign to $+$, and then the remainder is $a + b$. The reason is evident, from the former example. If a man has $£10$ in goods, and owes $£5$; if I want to take away his debt, I must add to his stock, or prevent the debt from affecting it, which is the same thing. If I represent the goods by $a$, and the debt by $b$, the true state of his affairs will be represented by $a - b$. If I want to take away $-b$ from this, I must change its sign to $+$; and then the $+b$ and $-b$ destroy one another so that the remainder, after taking away the debt, is $a$, or $£10$; which is agreeable to truth.

Quantities are considered by algebraists as simple or compound. The simple quantities are such as are represented by single letters, as $a$, $b$, $c$, &c. Compound quantities arise only from the addition or subtraction of dissimilar simple ones; thus, $a + b$, $b + c$, and all others connected by the signs $+$ or $-$, are called compound quantities. By multiplication of simple quantities, compound ones are not produced: for letters are multiplied into one another by writing them down in connection, without any sign, or with $X$, the sign of multiplication, between them; as $a \times b$, or $ab$, denotes the product of $a$ multiplied into $b$. In algebra, the signs prefixed to the quantities, are objects of multiplication, as well as the letters or coefficients of the letters themselves: thus, $+$ multiplied into $+$, always gives $+$ for the product; and $-$ multiplied into $-$ gives the same; but $-$ into $+$, or $+$ into $-$, give $-$ for the product.

That $+$ multiplied into $+$ should give $+$, or that $+$ into $-$ should give $-$ for the product, will readily be comprehended; but why $-$ multiplied into $-$ should give $+$, is not so easily understood. Different methods have Elementary have been used to illustrate the reason of this; but all Rules of them seem involved in some degree of obscurity, from which we hope the following will be altogether free.

We have already observed, that no quantity is itself negative, but only as it stands in relation to another. Positive and negative quantities, therefore, arise only from addition and subtraction, but not from multiplication. Four inches in measure are a positive quantity in themselves, and are positive or negative in algebraic writing according as they are added to or taken away from anything. Negative quantities, therefore, are capable of being added or subtracted, but not of being multiplied, as negatives. Suppose one merchant owes £100, another £50, and a third buys the stock, and becomes liable for the debts of both. His capital will then be negatively affected by both debts; and if we call it $a$, the debt of the first merchant $b$, and of the second $c$, his real worth will be expressed by $a - b - c$, and may be found by subtracting the sum of the debts from his stock; but it is impossible to multiply the two debts together in any manner of way, so as to affect him by the product of the numbers; the reason is, because we change the relation by multiplying them. In like manner, if we cut four inches from a ruler, these with respect to the whole ruler will be $-4$; but if we multiply the $-4$, or the part cut off, by itself, we produce $+16$ square inches, which have not, nor can have, any relation to the ruler itself, but will become positive or negative with regard to another quantity, just as we please to add or subtract them. The case is different when a negative quantity is multiplied by a positive one; because then the relation is not changed. Thus, in the former example, if we cut off four inches from a ruler, the quantity cut off is $-4$; if we multiply this $-4$ by $+2$, or, which is the same thing, want to add other four inches to those already cut off, we must take them from the ruler, and thus the product will be $-8$.

In multiplication of algebraic characters, there is not the least difficulty. The signs are multiplied as we have already mentioned; the coefficients, as in common arithmetic; and the letters, by writing them down without any sign between them: thus, $2a$ multiplied into $3b$, produces $6ab$, or $6ba$; for the order of the letters is of no consequence. If the multiplier and multiplicand are both compound quantities, each term of the first must be multiplied into all those of the second, and all the products added together: thus, if $a-b$ is to be multiplied by $a-b$, I first multiply by $a$, which produces $aa-ab$; I then multiply by $-b$, and the produce is $-ab+bb$; and, adding these two products together, we have $aa-zab+bb$ for the total produce.

Division being the converse of multiplication, what has been said concerning the latter, will also serve to make the former easily understood. When the same letters are contained in the divisor and dividend, there division may properly take place: thus, if I am required to divide $abc$ by $a$, the quotient will be $bc$; because $bc$ multiplied into $a$, produces $abc$ the dividend. If I am to divide it by $b$, the quotient will be $ac$; because $ac\times b$ is $abc$, or $abc$. With regard to the signs, they are to be managed so, that the sign of the divisor multiplied into that of the quotient may produce the sign of the dividend; and it must always be carefully observed to change the sign of that quantity which is subtracted from the dividend, whether the subtraction can properly take place or not. The coefficients, or pure numbers, are to be divided exactly as in common arithmetic. Suppose now it is required to divide $aa-zab+bb$ by $a-b$, I begin with considering what sign multiplied into that of the divisor will give that of the dividend for a product: as they are both positive quantities, this must be $+$. I next consider what letter multiplied into the first term of the divisor will give the first term of the dividend for a product. This I find to be $a$; for $a\times a$ gives $aa$ for the product. I then multiply this first term of the quotient into both terms of the divisor; and behoved to do so, though there were three, four, or more terms in it. The product is $aa-ab$. Subtracting this product from the dividend, there remains $-ab+bb$ for a new dividend. I must now again consider what sign multiplied into that of the first term of the divisor will give the sign of the first term of the dividend; which I here find to be $-$. By again considering what letter multiplied into the first term of the divisor will give the first term of the dividend for a product, I find it to be $b$; which multiplied into both terms of the divisor, produces $-ab+bb$; which, subtracted from the new dividend, leaves no remainder.

If the letters are totally different, or the first term of the divisor cannot be found in the dividend, there division cannot take place; the quantities must in this case be wrote down with $-$ the sign of division between them, or placed the one over the other like fractions, as $\frac{a}{b}, \frac{bd}{cf}, \frac{bd}{cf}$ &c., but as long as the first term of the divisor will divide the first term of the dividend, the operation may be continued; and sometimes the quotient will run out to an infinite series of terms, as in the following example:

\[ \begin{align*} 1 + x & \quad (1 - x + xx - xxx, &c.) \\ 1 + x & \\ -x & \\ -x - xx & \\ +xx & \\ +xx + xxx & \\ -xxx & \\ -xxx - xxxx & \\ +xxxx, &c. \end{align*} \]

If a quantity is multiplied into itself any number of times, the products are said to be the powers of that quantity, which is called the root, with respect to them. The powers are distinguished by the names of square; cube, or third power; biquadratic, or fourth power; solid, or fifth power; cube squared, or sixth power, &c.; and are thus wrote: $a^1$, or simply $a$, the radical quantity; $a^2$, or $a$ squared, or multiplied into itself; $a^3$, $a$ cubed, or the square of $a$ multiplied by $a$; $a^4$, signifying the square of $a$ multiplied by itself, &c. The multiplying a quantity by itself any number of times is called involving that quantity to a certain height, the sign of which is $\circ$; and if the root of an involved quantity is required, the operation by which it is found is called evolution, and is expressed by the sign $\circ$.

Involution of a simple quantity is performed merely by writing it down with a figure above; as $a^2, a^6, a^7, &c.$, expressing the height of the power to which it is involved. involved. These figures are named the indices, or exponents of the powers. Involution of compound quantities is performed by continual multiplication; but any root, consisting of only two terms, such as $a + b$, or $a - b$, (the first of which is called a binomial, and the second a residual root) may be involved to any height, by the following rule.

The power must always consist of one term more than is expressed by its index: that is, if it is required to raise $a + b$ to the square, the power will consist of three terms; if to the cube, of four terms; to the biquadrate, of five; to the surfoldic, of six, &c. The first and last terms are both pure powers, without any coefficients, the one of the first and the other of the last term of the root, the indices of both which express the height of the power. Thus, if I am to involve $a + b$ to the fifth power, the first term must be $a^5$, and the last $b^5$. In the intermediate terms the index of $a$ decreases, and that of $b$ gradually increases, till it attains the same height that $a$ had at first. The letters of the 6th power of $a + b$, therefore, without their coefficients, will stand thus:

$$a^6 + a^5b + a^4b^2 + a^3b^3 + a^2b^4 + ab^5 + b^6.$$

To find the coefficients, multiply the index of any term into its coefficient, and divide by the number of terms; the quotient is the coefficient of the term immediately following. In the first term, the coefficient, though not expressed, is supposed to be 1. This multiplied by 6 the index, and divided by the number of terms 1, quotes 6 for the coefficient of the second term, which therefore is $6ab$: multiplying then the index 5, by this coefficient 6, and dividing by 2, the number of terms, I have 15 for the coefficient of the third, and the term is $15a^4b$. Proceeding in this manner, I find the power required, to be

$$a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + b^6.$$

The residual root, $a - b$, is involved by the very same rules; only the signs, instead of being constantly $+$, are $-$ and $+$ alternately; and thus the 6th power of $a - b$ will be

$$a^6 - 6a^5b + 15a^4b^2 - 20a^3b^3 + 15a^2b^4 - 6ab^5 + b^6.$$

If the root consists of three or more terms, no rule can be formed by which the quantity can be so easily involved to the required height, as continual multiplication; because there are such a number of terms, and the letters are so intermingled with one another, that it would be difficult to remember the numerous directions necessary in such a case: nor do such tedious multiplications often occur; but where they do, it is proper to range the product according to the number of times that a certain letter is repeated in every term, which is called the ranging it according to the dimensions of that letter. Thus, suppose I am to raise $a + b + c$ to the cube; by multiplying it twice, I find the product to be

$$a^3 + 3a^2b + 3ab^2 + b^3 + 3a^2c + 6abc + 3ac^2 + 3bc^2 + c^3 + 3a^2b^2 + 3ab^2c + 3a^2c^2 + 3b^2c^2 + 3abc^2 + 3ab^2c^2 + 3ac^2b^2 + 3bc^2a^2 + 3bc^2b^2 + 3c^2a^2b + 3c^2b^2a + 3c^2a^2c + 3c^2b^2c + 3c^2a^2c^2 + 3c^2b^2c^2.$$

This long line is exceedingly confused, and difficult to be comprehended at one view; but by ranging it according to the dimensions of any of its letters, is much more plain and intelligible: according to the dimensions of the letter $a$, it stands thus:

As Evolution, or the extraction of roots, is proper-

ly the solution of a certain kind of equations, it will be more properly treated of, after the nature of equations in general, and the methods of solving the more simple ones, are considered.

In algebra, as in common arithmetic, fractions arise from the division of quantities that are incommensurable to one another, or those of which the lesser will not divide the greater without a remainder; but as the rules for adding, subtracting, multiplying, &c. of algebraic fractions are exactly the same with those for performing the same operations on arithmetical ones, only making allowance for the difference between adding, subtracting, &c. letters, instead of figures, we refer to the article Arithmetic.

Hitherto we have only considered such quantities as Surds, or must be supposed always to have a positive or real existence, and consequently can be expressed by a certain symbol; but, besides these, there are other imaginary quantities, the existence of which it is often necessary to suppose, though in fact they have not, nor cannot have, an existence. Thus, if I am required to find a number which, multiplied into itself, will produce 16; it is easily found, and such a number may be expressed by $a$; but if I am required to find one, which, multiplied by itself, will produce 15, it cannot be found by any art, and consequently cannot be expressed by a letter. Quantities of this kind are denominated, by algebraists, Surds, or Irrational ones; and have the sign $\sqrt{}$ prefixed to them, which denotes their imaginary existence. This sign denotes the extraction of a root; and the different kinds of roots defined, are expressed by figures set over it. Thus, $\sqrt{2}$, or simply $\sqrt{}$, denotes that the square root is desired; $\sqrt[3]{2}$, the cube-root, &c. Sometimes this sign is prefixed to a number, or to an algebraic series which is capable of affording a true rational root; but it then only denotes that the root hath not been extracted, and consequently exists as yet only in idea. The prefixing this sign to any letter makes no other difference with regard to addition, subtraction, multiplication, or division, than causing the letter represent a different quantity than otherwise it would have done, and so must be added or subtracted by signs. Thus $a$ added to $a$, makes $2a$; but $a$ added to $\sqrt{a}$, is $a + \sqrt{a}$. Among themselves surds are as easily managed as other quantities: for $\sqrt{a} + \sqrt{a}$ is $2\sqrt{a}$, and $\sqrt{a} - \sqrt{a}$ is $0$; $\sqrt{a} + \sqrt{a} = \sqrt{a}$, is $2\sqrt{a}$; $\sqrt{a}\times\sqrt{a}$ is $a$; $\sqrt{a}\div\sqrt{a}$ is $1$; $\sqrt{a}\div\sqrt{a}$ is $a$, &c. When the root of any compound quantity is sought, it must, besides the radical sign, have a line drawn over it, to denote that it is only to be reckoned a simple quantity; thus $\sqrt{ab+cd}$, &c. In cases where irrational quantities of this kind occur, it will be proper to put some letter, as $x$, $y$, $z$, or any other not already used, for the surd, and let that symbol remain till the last step of the operation, when the true value may be substituted in its place.

Surds, like fractions, may be reduced to their least terms; or two unlike surd quantities may be reduced to two having the same denomination. To reduce a surd quantity to its lowest terms, a certain rational root must be found in it, multiplied by a surd; the root must be extracted according to the rules hereafter given. Equations, even for evolution, and prefixed to the other quantity with the radical sign. Thus, though no number multiplied into itself will produce 8, yet such an imaginary quantity may be expressed otherwise than by $ \sqrt{8} $; for 8 contains the number 4, which is a perfect square, and produced by multiplying 2 into itself. $ \sqrt{8} $ therefore is reduced to $ \sqrt{4} \times \sqrt{2} $; but one of these is a perfect square; and therefore $ \sqrt{4} \times \sqrt{2} = 2\sqrt{2} $, which is the fird in its lowest terms. In like manner, $ \sqrt{28} $ is $ \sqrt{4} \times \sqrt{7} $ or, $ 2\sqrt{7} $; $ \sqrt{18} $ is $ \sqrt{9} \times \sqrt{2} $, or $ 3\sqrt{2} $.

The same rule holds in algebraic quantities. $ \sqrt{a^2b^2} $ is $ \sqrt{a^2} \times \sqrt{b^2} $, or $ ab $; $ \sqrt{a^2b^2c^2} $ is $ \sqrt{a^2} \times \sqrt{b^2} \times \sqrt{c^2} $; which being both complete squares, the fird is reduced to $ 2ab $, or $ 2ab $.

This method of reducing surds is often very convenient for bringing them into less compass, so as to facilitate their addition or subtraction. Thus $ \sqrt{18} + \sqrt{32} $, being reduced to their least terms, become $ 3\sqrt{2} + 4\sqrt{2} $, or $ 7\sqrt{2} $; and $ \sqrt{8a^2} + \sqrt{50a^2} - \sqrt{72a^2} $, is reduced to $ 2a\sqrt{2} + 5a\sqrt{2} - 6a\sqrt{2} $, or $ a\sqrt{2} $; $ \sqrt{12a^2} + \sqrt{75a^2} $, becomes $ 2a\sqrt{3} + 5a\sqrt{3} $, or $ 7a\sqrt{3} $, &c.

Surds are reduced to the same denomination, by involving them to a proper height; but in order to understand this the more readily, it is proper to take notice, that in any series of powers, as $ a^1, a^2, a^3, a^4, a^5, a^6, $ &c., the addition of the indices is equivalent to the involution of the power, and the subtraction of the indices is equivalent to the division of the powers by one another. Thus, by subtracting the index 4 from 7 in the powers $ a^4 $ and $ a^7 $, there remains $ a^3 $; which is the quotient of $ a^7 $ divided by $ a^4 $; as is evident from dividing $ aaaaaaa $ by $ aaa $. In like manner, the division of the indices answers to the extraction of the root: thus, to divide the index of $ a^6 $ by 2, is the same thing as to extract its square root; to divide it by 3, is the same thing as to extract its cube root; the quotients being $ a^3 $ and $ a^2 $, answering to the powers $ aaa $ and $ aa $.

This division cannot go farther in rational quantities, than that of 2 the index of the square by itself. The quotient is 1, which is the index of $ \sqrt{a^2} $, being $ a^1 $, or simply $ a $. The square or cube root of $ a $, then, must be expressed by a division of its index 1, by 2 or 3, and may be wrote $ a^{1/2} $, as well as $ \sqrt{a} $ and $ \sqrt[3]{a} $.

When surds are to be reduced to the same denomination, it will be most proper to write them with these fractional indices; the fractions have then only to be reduced to a common denominator, according to the rules of arithmetic: and thus, $ a^{1/2} $ and $ a^{1/3} $ will become $ a^{1/6} $ and $ a^{1/6} $. This reduction is convenient when surds are to be multiplied or divided by one another. For example; suppose I was to multiply the two above-mentioned surds into one another, no more is necessary than to add the two indices together, after having reduced them to a common denominator, and the product is $ a^{1/6} $; which intimates, that the product of $ \sqrt{a} $ into $ \sqrt{3} $ is equivalent to $ \sqrt{a^3} $; $ \sqrt{2} $ and $ \sqrt{3} $ will become $ \sqrt{2^3} $ and $ \sqrt{3^2} $, or $ \sqrt{2^3} $ and $ \sqrt{3^2} $, which is $ \sqrt{8} $ and $ \sqrt{9} $; multiplied together, they become $ \sqrt{72} $, &c.

Sect. II. Equations, or the application of the foregoing general rules to the solution of various kinds of problems.

The word equation implies no more than simply the making one thing equal to another, or asserting it to be Equation. So, if the assertion is really true; and, in fact, it is by this very simple operation that the most abstruse and difficult algebraic problems are resolved. The method of noting down equations, or making the affirmation of equality, is by writing down the two quantities, with $ = $, the sign of equality, between them; and the quantities are then called the two different sides of the equation. Thus, $ a + b = c $; that is, the sum of $ a $ and $ b $ is equal to the third quantity $ c $, where $ a + b $ are one side of the equation, and $ c $ is the other: $ 4 + 5 = 9 $. Here, $ 4 + 5 = 9 $ are one side, and $ 9 $ is the other side, of the equation.

It is needless to observe, that no problem can be resolved by making false equations, or affirming a thing to be equal to what it is not: but tho' this will never be done intentionally, it is very often done by mistake; and to prevent mistakes of this kind, it will be always necessary to keep in view the following self-evident axioms.

1. If equal quantities are added to equal quantities, the sums will be equal. Thus, if a bottle contains a gallon of water, and a cask contains another gallon; if a third gallon is poured into the bottle, and a fourth one into the cask, there will be equal quantities of water in the bottle and the cask.

2. If equal quantities are subtracted from, multiplied into, or divided by, equal quantities; the remainders, products, or quotients, will be equal.

In conformity to these axioms, it is plain, that an algebraist may do what he pleases with his equations, provided he does the same thing with both sides of them; thus, if $ a = 4 $, I may then say $ 2a = 8 $, $ 7a = 28 $, $ a - 4 = 4 - 4 = 0 $; or $ a + 2 = 2 + 2 = 4 $, $ a + 8 = 8 + 8 = 16 $, &c.; where every one of these equations is as true as the first; because what is done to one side of the equation is likewise done to the other: but if I either add, subtract, multiply, or divide, one side, without doing so to the other, I evidently affirm a falsehood; for if $ a = 4 $, then it is plain that if I multiply one side by 2, and only add 2 to the other, I make $ 2a = 6 $, or say that twice four is six.

As there is no science whatever wherein people are more liable to mistake, and to perplex themselves, than algebraic operations, it will be very proper for young algebraists to number the steps of their operation, and on the left-hand margin to mark what is done in each step, that a more full and distinct view of the whole may be at once obtained, and any mistake more easily corrected, as in the following example:

| Step | Equation | |------|----------| | 1 | $ a = 6 $ | | 2 | $ 2a = 12 $ | | 3 | $ 2a + b = 12 + b $ | | 4 | $ 2a = \frac{12}{6} $ | | 5 | $ 4a^2 = 144 $, &c. |

Here the figures on the margin denote what is done with each preceding step, or equation; $ 1 \times 2 $ denotes that the first equation is multiplied, not by the second equation, but by the number 2; which, for this reason, has a line drawn over it; $ 2 + b $ signifies, that $ b $ is added to both sides of the second equation: $ 2 + 1 $ signifies, that both sides of the second equation are divided by both sides of the first: $ 2a^2 $, that both sides of the second equation are involved to the second power or square. In all equations there are some quantities supposed to be known, and others unknown; the design of the equation is to discover the value of the unknown quantities; in order to which they must be compared with those quantities which are known; for if the equation consists only of unknown quantities, it is impossible to know anything about them.

The end proposed in every equation is to place the unknown quantities all by themselves on one side of the equation, and the known ones by themselves on the other: when this is done, the equation is said to be reduced, and the operation is at an end.

Equations may be reduced, (1.) By addition and subtraction; or, as it is commonly called, by transposition. This is performed by adding to, or subtracting from, both sides of the equation, a quantity with which it is encumbered, and which tends to obscure the true meaning. Thus, $x + 6 = 7$; here the unknown quantity $x$ is combined, by addition, with 6 a known one; which I want to get clear of, that I may know the precise value of $x$. For this purpose I make an equation of $6 - x = 6$, which I subtract from the former, and the work stands thus:

\[ \begin{array}{c|c} 1 & x + 6 = 7 \\ 2 & 6 = 6 \\ 3 & x = 7 - 6 = 1 \end{array} \]

Here I find the true value of $x$, because it stands alone upon one side, and a known quantity stands alone on the other. It is evident also, that if, instead of writing down the equation $6 = 6$, I only change the sign of the known quantity, and carry it over to the other side of the equation with the sign so changed, the event will be the same; for, if $x + 6 = 7$, then undoubtedly $x = 7 - 6$, or $6 = 7 - x$. It is a rule, therefore, in algebra, That whatever quantity is carried over from one side of an equation to another, must have its sign changed, whether it was + or −, and whether the quantity be known or unknown; it will then produce the effects of a positive or negative quantity, among those to which it is carried, according as the sign is changed from − to +, or from + to −. Suppose the following equation given,

\[ \begin{array}{c|c} 1 & 3x + 5 = 2x + 6 \\ 2 & 3x + 5 = 6 \\ 3 & x = 1 \end{array} \]

The reason of this operation is obvious: for carrying over $2x$ with its sign changed, it meets with $3x$, which it destroys as far as it can; the remainder is then only $x$, which being still combined with 5, makes the transposition again necessary, as in the former example.

(2.) When the unknown quantity is combined with any known one by multiplication, it is necessary to divide both sides of the equation by that quantity into which the unknown one is multiplied. Thus, suppose $4x = 20$, I cannot make $x$ stand alone upon one side of the equation, unless I divide $4x$ by 4; the quotient is $x$; and dividing the other side also by 4, we have $x = 5$. In like manner, if $4x - 2 = 8$, then, by transposition, $4x = 8 + 2 = 10$, and, by division, $x = \frac{10}{4} = 2.5$, &c.

(3.) If the unknown quantity is divided by any known one, both sides of the equation must be multiplied by that quantity which divides the unknown one, in order to take away the fraction, without which the equation could not be conveniently reduced. With regard to fractional quantities, according to the rules of arithmetic, it is the same thing to multiply a fraction by its denominator, and merely to throw away that denominator; hence, if one side of an equation is divided by any quantity, and not another, it will be sufficient to multiply by the dividing quantity that side of the equation which is not affected by it. This is equally evident with any of the former methods of reduction; for if $\frac{x}{2} = 4$, then it is plain, that $x = 8$, and so of others. If several fractional quantities occur in one, or both sides of the equation, the same operation must be repeated with every one of them, as

\[ \frac{x}{2} + \frac{x}{3} + \frac{x}{4} = 6; \quad \text{then } x + \frac{2x}{3} + \frac{2x}{4} = 12; \quad \text{and } 3x + 2x, \text{ or } 5x + \frac{6x}{4} = 36, \text{ and } 20x + 6x = 144; \quad \text{whence, by division, } x = \frac{144}{26} = 5.5385, \text{ nearly.} \]

(4.) Reduction by involution takes place when the unknown quantity is under the radical sign. In this case, in order to come at its value, both sides of the equation must be involved to the power expressed by the index of the surd quantity, as $\sqrt{x} = 4$; then, $x = 16$, by involving both sides of the equation to the square; if $\sqrt[3]{x} = 3$, then $x = 27$, &c.

There are all the methods of reduction that are applicable to simple equations, or those where the unknown quantity is not multiplied by itself; in which case, very different methods are to be used, which shall be explained under quadratic, cubic, &c. equations: we must now take notice of the preliminary steps necessary to be taken in order to the solution of an algebraic problem.

The first thing to be done is to state the question, as Method of stating or writing down an algebraic problem.

It is required to find a number, which being multiplied by 5, and 8 subtracted from the product, the remainder shall be 52. As the thing here sought is only one number, I put $x$, or any letter at pleasure, for it; then, as the question intimates that the number sought is multiplied by 5, and 8 subtracted from the product, I do the same with the letter taken to represent it; and find the remainder to be 52−8: this therefore, by the question, being equal to 52, I write it down in algebraic characters, thus, $5x - 8 = 52$. By transposition $5x = 52 + 8 = 60$; and by division, $x = \frac{60}{5} = 12$, the number sought.

When only one thing is sought, generally the solution of algebraic problems is not difficult; but when two or more things are required to be discovered, the difficulty becomes proportionably greater. It is necessary, however, that where there are two or more unknown quantities, there should be data sufficient to find them all out; because questions proposed without sufficient data, cannot be resolved but in an indeterminate manner. Thus, if it be required to find two numbers $x$ and $y$, with this single condition, that their sum shall be 100; it is evident, that the question is capable of 99 different answers, each of which shall ful- fil the condition required; for $ x $ may be 1, and $ y = 99 $; or $ x $ may be 2, and $ y = 98 $, &c., but if to the foregoing condition I add another, namely, that the difference of the two numbers required is 50, the question is then properly limited, and capable only of one direct answer.

If a third condition is required, suppose, that their product should be 740; this condition is either superfluous, because the values of $ x $ and $ y $ may be found without it; or absurd, as being inconsistent with the rest. It is therefore a general rule in algebra, That where there are two unknown quantities, the problem must be laid down in such a manner as to admit of two equations being formed from it, which shall neither be inconsistent with, nor consequences of, one another; for if this last is the case, it is the same thing as tho' only one equation were given; for instance, if am required to find two numbers whose sum is 100, and double their sum 200, this last equation is only the first one doubled; and consequently the question is still as unlimited as before.

For the solution of problems where two or more quantities are concerned, there is one general rule which will certainly hold in all cases, namely, to find a value of each of the unknown quantities from each of the equations, treating the other unknown quantity exactly as a known one. By this means we have two sides of a new equation, where only one unknown quantity is concerned, the other being exterminated, as it is called, by the preceding operation; and it is evident, that if the equations are consistent with one another, the value of the unknown quantity found by one equation, will be precisely equal to that found by the other. We shall illustrate this by the preceding example, which, being stated, will be $ x + y = 100 $, and $ x - y = 50 $. By transposing the first equation, we have $ x = 100 - y $; and by transposing the second, $ x = 50 + y $; it is plain, that $ x $, though an unknown quantity, must always be equal to itself; and therefore the values of it obtained from both these equations will be equal to one another; of these therefore I form the new equation $ 100 - y = 50 + y $; by transposition, we have first $ 100 - 50 = 2y $, and then $ 50 = 2y $; whence, by division, $ 25 = y $, and $ 100 - y = 75 $.

The same method is to be followed when there are three, or four unknown quantities; but the operation will then be much more tedious; because, having formed a new equation in which one quantity is exterminated, we must still continue to form new ones in order to exterminate the others, as in the following example.

It is required to find three numbers whose sum is 130; if the third is multiplied by 3, and that product is subtracted from the sum of the first and second, the remainder will be 10; if the first is multiplied by 2, the second by 3, these two products are added together, and 15 subtracted from the sum, the remainder will be 7 times the third number.

Having put $ x, y, z $, for the three numbers, the question resolves itself into the following equations.

\[ \begin{align*} 1. & \quad x + y + z = 130 \\ 2. & \quad 2x + 3y - 15z = 7z \\ 3. & \quad 3x + 2y - 15z = 7z \\ \end{align*} \]

By transposing the first equation, we have $ x = 130 - y - z $; by transposing the second, $ x = 10 + 3z - y $; on transposing the third, and dividing by two, we have $ x = \frac{7z + 15 - 3y}{2} $. These three values of $ x $ must necessarily be equal to one another; I therefore form a new equation from the first and third; then $ 130 - y = \frac{7z + 15 - 3y}{2} $. Reducing this equation by multiplication and transposition, it becomes $ y = 92 - 245 $. To have another value of $ y $, I form a new equation from the second and third values of $ x $, or I might for the same purpose make an equation of the first and second values of $ x $; this will be $ 10 + 3z - y = \frac{7z + 15 - 3y}{2} $. Reducing this equation in the same manner as before, we have $ y = 5 $. We must now form a third equation from the two values of $ y $ already found; and thus we will have $ 92 - 245 = 5 $; from whence, by transposition and division, we have $ z = 30 $.

In the same manner we might now proceed to find the values of the other unknown quantities: but it is evident, that though this method must infallibly answer, a great deal of needless trouble is occasioned by it in the present case; for, if, instead of finding the three values of $ x $, I only find one from the first equation, and substitute that in place of the letter $ x $ in the second, the quantity $ y $ will be exterminated at once. The value of $ x $ from the first equation is $ x = 130 - y - z $, the second equation is $ x + y - 3z = 10 $; writing therefore into this equation, $ 130 - y - z $, in place of $ x $, we have $ 130 - y - z + y - 3z = 10 $, where the positive and negative destroy one another, and the equation becomes $ 130 - 4z = 10 $; whence, $ 4z = 120 $, and $ z = 30 $. But it is plain, that the remarkable success of this substitution depends entirely upon the circumstance of a single $ y $ in the second equation; for had there been 2$ y $ there, the same advantage would not have been derived from following this method. There can therefore be no rules laid down for obtaining the solution of algebraic problems in the most easy manner possible; these must depend on the particular circumstances of each problem; and hence there is no science where the rational faculties and ingenuity are put to a greater stretch than in algebra, and no branch of education is more proper for producing a quickness of understanding, provided the algebraist does not lose himself in the depths of his science, in which case he will be quick-lighted only to algebra itself.

As so much difficulty is occasioned by a number of unknown quantities, it will seldom be proper to state a question with two unknown quantities, where one will answer the purpose, though sometimes the unknown quantities may be made to disappear surprisingly, by proper management. On some occasions, instead of chusing a single letter to represent an unknown quantity, it will be proper to express it by a sum, or a difference; as $ x + y $, or $ x - y $. As an example, we shall give three methods of solving the former problem, "Required to find two numbers whose sum is 100, and difference 50." With one unknown quantity, the question is stated in the following manner.

\[ \begin{align*} 1. & \quad x = \text{the least number sought} \\ 2. & \quad 100 - x = \text{the greatest} \\ 3. & \quad 100 - x = 50 = \text{their difference by question} \\ 4. & \quad 3x + 2y = 100 = \text{the least number sought} \\ 5. & \quad 100 - x = 50 = \text{the greatest number} \\ \end{align*} \]

With two unknown quantities this may be solved otherwise. Equations, therwise than by forming a new equation, thus:

1. $ x = \text{greatest number} $ 2. $ y = \text{least} $ 3. $ x + y = 100 $ \{ by question 4. $ x - y = 50 $

$ \begin{array}{c} 3+4 \\ 5+2 \\ 5-2 \\ 3-4 \\ 7+2 \\ 8-2 \\ \end{array} $

$ \begin{array}{c} x = 150 \\ x = 75 \\ y = 50 \\ y = 25 \\ \end{array} $

Representing one of the numbers by a sum, and the other by a difference, the work will stand thus:

1. $ x + y = \text{greatest number} $ 2. $ x - y = \text{least} $ 3. $ x + y = 100 $ \{ by question 4. $ x - y = 50 $

$ \begin{array}{c} 1+2 \\ 3+2 \\ 1-2 \\ 4-2 \\ 6-2 \\ 7+2 \\ 8-2 \\ \end{array} $

$ \begin{array}{c} x = 150 \\ x = 75 \\ y = 50 \\ y = 25 \\ \end{array} $

Though this problem is so easily resolved by all the three methods, that it is difficult to say which has the advantage; yet it is sufficient to show the prodigious diversity of operation that must occur in the solution of algebraic problems, according as we use different methods. The last method is exceedingly proper, where equations have to be multiplied into one another, which is the origin of quadratic, cubic, and other high equations, of which we are now to treat.

If an equation is multiplied into itself once, the produce is another equation, which is as strictly just as the former: but after having reduced it by all the methods proposed for the reducing simple equations, and having brought the unknown quantities to one side, and the known ones to the other, we are still at a loss; because the unknown quantity being multiplied into itself, we know not what relation it bears to the known one. Thus, if the equation $ ax^2 + bx + c = 0 $ is multiplied once into itself, the produce is $ a^2x^4 + 2abx^3 + b^2x^2 + 2acx + 2bcx + c^2 = 0 $; where the unknown quantity cannot be discovered till we know what number multiplied into itself will produce 144. The above equation is one of that kind called quadratic equations; and, from its containing only of the literal quantity multiplied into itself, is called a simple quadratic: but if we multiply the equation $ ax^2 + bx + c = 0 $ once into itself, the produce is $ a^2x^4 + 2abx^3 + b^2x^2 + 2acx + 2bcx + c^2 = 0 $; and reducing this by transposition, we have $ a^2x^4 + 2abx^3 + b^2x^2 + 2acx + 2bcx + c^2 = 0 $, where the literal quantity is not only multiplied by itself, but by the number 6. This addition is called the affeetion of an equation, and the last mentioned one is of that kind called quadratic affeeted equations.

It is not to be supposed that any person would produce equations of this kind by multiplying such simple ones as those above mentioned; but very often the circumstances of the question oblige him to state them in this manner, or they are unavoidably multiplied in the course of the operation. Thus, suppose it is required to find two numbers whose sum is 100, and product 1875; by the common method, we have $ x + y = 100 $, and $ xy = 1875 $. From the first equation $ x = 100 - y $, and from the second $ x = \frac{1875}{y} $; whence $ 100 - y = \frac{1875}{y} $. Reducing this, we have $ 100y - y^2 = 1875 $. We do not get clear of this difficulty by using only one unknown quantity; for putting $ x $ for the one, and $ 100 - x $ for the other, we come at once to the equation $ 100x - x^2 = 1875 $.

Neither is it to be totally avoided by making $ x + y = $ one of the numbers, and $ x - y = $ the other; thus indeed, by the question, we have $ 2x = 100 $, and $ x^2 - y^2 = 1875 $; whence, by substituting the value of $ x $, we have $ y = 625 $; so that, though the equation is now only a simple quadratic, we must still remain ignorant of the value of $ y $, till we know what number multiplied into itself will produce 625. Here, however, we see the utility of sometimes representing unknown quantities by a sum and a difference.

We have already observed, that, when literal powers are to be divided by one another, the division is performed by subtracting their indices. The extraction of their roots, in like manner, is performed by dividing their indices by 2, 3, 4, &c. according as we want the square, cube, or biquadratic root; so, if required to find the square root of $ a^4 $, I divide its index 8 by 2, the quotient 4 is the root required. If the root of any series of terms is required, as of $ x^2 + 6x + 9 $, we must proceed to find it by supposing it to be $ a + b $. This root we involve to the square, and then make the following equation $ a^2 + 2ab + b^2 = x^2 + 6x + 9 $. From this it is evident, that if $ a^2 $ corresponds with $ x^2 $, $ 2ab $ must correspond with $ 6x $, and $ b^2 $ to 9; therefore, as $ x $ is the first term of the root, and corresponds with $ a $, the coefficient of $ x $ must correspond with $ 2b $ the coefficient of $ a $. Dividing, therefore, the coefficient of $ x $ in the second term by 2, the quotient 3 is the second term of the root, and the square root of $ x^2 + 6x + 9 $ is $ x + 3 $. Hence we have an easy rule for completing an imperfect square, viz. to take half the coefficient of the unknown quantity, multiply it by itself, and add it to both sides of the equation, which will then be exact squares. Thus the affected quadratics are easily reduced to simple ones, as in the following example. Suppose, $ x^2 + 14x = 32 $, then taking the half of 14 or 7, multiplying it by itself, and adding it to both sides of the equation, we have $ x^2 + 14x + 49 = 81 $. From the foregoing example we are sure that the root of the literal part is $ x + 7 $, and from the multiplication table we know that 9 multiplied into itself produces 81. Extracting the root on both sides, therefore, we have $ x + 7 = 9 $; whence $ x = 2 $.

As long as the root of the number sought does not exceed some of the 9 digits, there is no difficulty; but supposing it to consist of many places of figures, a tedious operation is required, which will be best understood by an example. Suppose the following equation is given; $ x^2 = 2985984 $, I take $ x = a + b $; whence $ x^2 = a^2 + 2ab + b^2 $, which consequently must be equal to the number given. The extraction of the root is now facilitated by the following consideration, that no digit multiplied into itself can produce more than two places of figures. To ascertain the number of places therefore in the root of the abovementioned number, I place a point over every third figure, beginning at the right hand, and the equation will stand thus:

\[ a^2 + 2ab + b^2 = 2985984. \]

Hence I conclude, that the root required must consist of 4 places of figures, or be above 1000. I next consider what digit multiplied into itself will produce the nearest square under 2, the first figure of the power. Had the point been placed over the second figure, I must Equations must have considered what digit multiplied into itself would have produced the nearest square under the first two figures. In the present case, I find it to be 1. I therefore suppose $a = 1000$; multiply it by itself, and subtract it from the power, in the following manner:

\[ a^2 + 2ab + b^2 = 2985984 \quad (1000 = a) \]

\[ a^2 = 1000000 \]

\[ 2ab + b^2 = 1985984 \]

It now appears, that if this remainder was divided by $2a + b$, the quotient must be $b$ for $2a + b = 2ab + b^2$. But as $b$ is still unknown, I must first proceed with $2a$, as in common division; but as it has something to be added, I must have regard to this in choosing the quotient figure; therefore, though in common division I might choose 8 for the quotient, I only choose 7, $a$ being $= 1000$, $2a = 2000$; and to find the other term $b$, the work will stand thus:

\[ \begin{array}{c} 2a = 2000 \\ b = 700 \\ 2a + b = 2700 \end{array} \]

\[ \begin{array}{c} 1985984 \\ 1890000 \\ \hline 95984 \end{array} \]

To find the other figures of the root, I must now suppose $a = 1700$; in which case, the former $a^2 + 2ab + b^2$ will now only be equivalent in value to $a^3$, and 95984 $= 2ab + b^2$. The operation is now to be repeated; $a$ being 1700, $2a = 3400$; which I set as a new divisor, and proceed as follows:

\[ \begin{array}{c} 2a = 3400 \\ b = 20 \\ 2a + b = 3420 \end{array} \]

\[ \begin{array}{c} 95984 \\ 68400 \\ \hline 27584 \end{array} \]

I now make a third supposition, of $a = 1728$, and proceed as before; thus,

\[ \begin{array}{c} 2a = 3448 \\ b = 8 \\ 2a + b = 3456 \end{array} \]

Here there being no remainder, we find 1728 to be the true root of the number required; and if the above example is attended to, the reasons of the arithmetical rules given for extracting roots will be sufficiently understood. See Arithmetic.

If the equations are multiplied into themselves twice, the produce is called a cubic equation; and, like the quadratic, is either simple, or affected; thus, $x^3 + 1728$ is a simple cubic; $x^3 + 10x^2 + 3x = 997474$, $x^3 + 10x^2 + 104x$, &c., are cubic affected equations.

The solution of simple cubic equations, or the method of extracting the cube root, will easily be understood by an example of the same kind with that by which we illustrated the extraction of the square root.

If the cube root of any simple algebraic power is required, it is found by dividing the index of that power by 3, as already observed. If of any series, the root must be supposed $= a + b$, as before; then, this involved to the cube, or $a^3 + 3a^2b + 3ab^2 + b^3$, will be equal to the cube proposed. Let it be required to find the cuberoot of $x^3 + 18x^2 + 108x + 216$. Here, taking $a + b =$ the root required, and involving it to the cube, we have $a^3 + 3a^2b + 3ab^2 + b^3 = x^3 + 18x^2 + 108x + 216$. From inspection, it is evident, that if $a^3$ corresponds to $x^3$, $3a^2b$ must correspond to $18x^2$, and consequently that $b = 18 - 3 = 6$.

By attentively considering this, we may easily see how an algebraic cube can be completed. Let us suppose the equation $x^3 + 6x^2 + 32$ given, and it is required to complete the cube. Here it is plain that $b = 2$, and consequently that the cube which wants the terms equivalent to $3ab^2$ and $b^3$ will be completed by adding them. As $b = 2$, they are easily found to be $12x + 8$; and adding these to both sides of the equation, we have $x^3 + 6x^2 + 12x + 8 = 40 + 12x$. Both sides of this equation are complete cubes; but it is impossible to reduce an affected cubic equation by completing its cube, as we reduce a quadratic equation by completing its square: the reason is, because the square consists of but three terms; if it wants the third, that can always be made up from the known quantity with which the unknown one is multiplied in the second; if it wants both the second and the third, it is a complete square; but in a cube which consists of four terms, the unknown quantity enters into them all except the last; and therefore, if any other than the last is wanting, the unknown quantity must again be added to both sides of the equation, as in the last example. Some cases may indeed occur, as the following, where the cube can be advantageously completed. Suppose the following equation is given; $x^3 + 12x^2 + 48x = 448$. As these terms are equivalent to $a^3 + 3a^2b + 3ab^2 + b^3$, and only want $b^3$ to make it complete, we need only take the third of the coefficient of the second term, and, involving it to the cube, add it to both sides of the equation, which will then be $x^3 + 12x^2 + 48x + 64 = 512$. By extracting the root, we have $x + 4 = 8$, and $x = 4$. Instances of this kind, however, occur so rarely, that we should not have mentioned this had it not been to show the reason why cubic equations cannot be solved on the same principles with quadratics.

If the cube root of a large number is to be extracted, the principles are the same with those on which the extraction of the square root depends, but the operation is more tedious. Let it be required to find the cube root of 5832. Taking $a + b =$ the root required, as before; we have then $a^3 + 3a^2b + 3ab^2 + b^3 = 5832$. The number of places in the root must be determined by points, as in the extraction of the square root; but for the cube they must be placed at the interval of two figures from one another, because the cube of some of the digits extends to three places of figures. I then choose the digit which produces the cube next less than that of the first one, two, or three figures of the resolved, according as the point happens to fall, for the significant figure of $a$, annexing to it as many cyphers as there are places of figures in the root; then having cubed this and subtracted it, I take $3a^2$ for a divisor, multiplying it by $b$, and adding $3ab^2$, and $b^3$, thus:

\[ \begin{array}{c} a^3 + 3a^2b + 3ab^2 + b^3 = 5832 \\ a^3 = 1000 \quad (8 = b) \end{array} \]

\[ \begin{array}{c} 3a^2 = 300 \quad 4832 \\ 3ab^2 = 2400 \\ 3ab^2 = 1920 \\ b^3 = 512 \end{array} \]

\[ 3a^2b + 3ab^2 + b^3 = 4832 \] Here is the root required; but had there been a remainder, a mult have been taken =180, and the operation repeated. The finding of b is attended with much more difficulty in the cube than in the square, on account of the great additions to be made; and the higher the powers, the greater is this difficulty: but as it is evident that an algebraic theorem will be sufficient direction for the extraction of every root, however high the power may be involved, we shall take no farther notice of the evolution of simple powers; only that all powers whose indices are multiples of 2 and 3, may be evolved by repeated extractions of the square or cube roots: thus, if I want the biquadratic root of any power, it is obtained by extracting the square twice; if the root of the fifth power, it may be had by extracting the square thrice, or the cube twice; of the 8th power, by extracting the square four times; of the 9th power, by extracting the cube root thrice, &c.; but the roots of the 5th, 7th, and 11th powers, can only be had by following an algebraic theorem constructed on purpose for themselves.

Hitherto we have considered cubic and other high equations as originating from a continued multiplication of one simple equation into itself; but their most common origin is from the multiplication of three or more different equations into one another. Thus, if we multiply the equations $x+1=5$, $x-3=1$, $x+2=6$, into one another, the cubic equation $x^3-7x-6=30$, and by transposition $x^3-7x=30$, will be produced. Here it is observable, that this equation wants the second term, because some of the numbers combined with x are negative, and others positive, and the negative and positive ones are exactly equal to one another. Had the negative quantity been either greater or less than the two positive ones, all the three terms would have remained in the product; and hence, when we see a cubic equation without the second term, we may know that the positive and negative quantities combined with x in the simple equations, or roots, from which it is formed, have been exactly equal to one another.

Cubic equations in which the third term is wanting arise from the multiplication of a simple quadratic by $x^2+1$, $x^2-2$, $x^2-3$, &c. thus $x^4-x^2=x^2-x^2$; $x^4+x^2=x^2+2x^2$, &c.

We have already observed, that the higher equations are produced by the terms of a question which secretly oblige us either to state it in equations already involved, or to involve them when we attempt their reduction. An example or two, we apprehend, will here be proper. Let it be required to find two numbers, of which, if the second is subtracted from 220, and the remainder divided by the unknown number, the quotient will be the first number; also, if the second is multiplied by itself, and the original number subtracted from the product, it will be 38 times the first.

\[ \begin{align*} 1 & \quad x = \text{the one number}, \\ 2 & \quad y = \text{the other}, \\ 3 & \quad 220-y = x \\ 4 & \quad y^2-y = 38x \\ 5 & \quad 38 = x \end{align*} \]

By a little variation in the terms of this question, a cubic equation, in which the third term is wanting, may be produced. Suppose two numbers, x and y, are required, of which 200 divided by the second may equal the first; and the square of the second may be equal to 38 times the first + the second. Here,

\[ \begin{align*} 1 & \quad x = \frac{200}{y} \\ 2 & \quad y = 38x+y \\ 3 & \quad y^2-y = 38x \\ 4 & \quad y^2-y = 38x \\ 5 & \quad y^2-y = 38x \\ 6 & \quad y^2-y = 38x \\ 7 & \quad y^2-y = 38x \end{align*} \]

If the simple equations, or roots, of which a cubic or other high equation is composed, are of such a nature that one of them destroys itself and becomes =0, a new species of cubic will arise, which is capable of three different solutions, and consequently a kind of indeterminate problem. If the equation is a biquadratic, it will have four solutions of this kind, of the fifth power five, and so on, the number of solutions always being expressed by the index of the power. This does not hold, however, in any other kind of equations than those where one of the original ones destroys itself; as will appear from the following examples.

If we multiply the equations $x+1=5$, $x-1=4$, and $x-4=0$, into one another, we will produce the cubic equation $x^3-4x^2-x+4=0$; or, by transposition, $x^3-4x^2-x=4$. Here, x may either be -1, -1, or 4; for if either of these are substituted in place of x, it answers the terms of the question. If $x=1$, then $x^3=1$; $-4x^2=-4$, and $-x=-1$; and $x^3-4x^2-x=1-4-1=-4$. If $x=-1$, then $x^3=-1$; $-4x^2=4$, and $-x=1$, according to the rules of subtraction; consequently $x^3-4x^2-x=-1-4+1=-4$ as the equation imports. Lastly, if $x=4$; then $x^3=64$; $-4x^2=-64$, and $-x=-4$, as in the other cases. In like manner, in the equation $x^3-9x^2+26x=24$, the value of x may be either 2, 3, or 4; for if $x=2$, then $x^3-9x^2+26x=8-36+52=24$; if $x=3$, then $x^3-9x^2+26x=27-81+78=24$; and if $x=4$, then $x^3-9x^2+26x=64-144+104=24$; and so of others.

But, when cubics are formed from the multiplication of equations into one another, all of which have some positive value, it is evident, that then they can only have one true solution: and the reason is plain; because, when any of the equations destroys itself, it likewise destroys the value of all the rest, and the whole becomes =0; and were it not that algebra can represent imaginary beings as well as real ones, there could be nothing to work upon in such a case. In such equations, the absolute number which constitutes their value is obtained from the continual multiplication of the known quantities combined with x into one another; or the last term. Equations term transposed. Thus in the first example, $ x^3 - 4x^2 + 4x = 0 $, the number $ +4 $ is formed by the multiplication of $ +1 \times -1 $, and $ -4 $, wherewith $ x $ was combined, into one another; for $ +1 \times -1 = -1 $; and $ -1 \times -4 = +4 $, according to the rules of multiplication. It is not possible, therefore, but that what has multiplied, must also divide; and as the taking $ x = 0 $ destroys all the product on the other side which alone could have truly limited the value of $ x $, it is the same thing as though we had taken $ x = 0 $, $ x + 1 = 0 $, and $ x - 4 = 0 $, and multiplied them all into one another, or given $ x $ three different values originally.

We shall evidently see the difference between the two species of cubics just now mentioned, by another example. The equations, $ x^3 - 2x^2 + 2x = 0 $, and $ x^3 + 2x^2 = 0 $, produce the following; $ x^3 - x^2 - 4x + 6 = 0 $. Here, as the number $ 6 $ is not produced by the multiplication of $ -1 $, $ -2 $, and $ +2 $, into one another, the value of $ x $ must be different from any one of them; and it is found to be so upon trial; for supposing $ x = -1 $, then $ x^3 - x^2 - 4x = -1 - 1 + 4 = 2 $. If $ x = -2 $, then $ x^3 - x^2 - 4x = -8 - 4 + 12 = 0 $. If $ x = +2 $, then $ x^3 - x^2 - 4x = 8 - 4 - 12 = -8 $; but neither of these are agreeable to the terms of the question; therefore $ x $ is neither $ -1 $, $ -2 $, nor $ +2 $. But if we take $ x = 3 $, then $ x^3 - x^2 - 4x = 27 - 9 - 12 = 6 $, according to the question; and this is therefore the only true value of $ x $.

Having thus explained at large the origin of all the different kinds of high equations that can possibly occur (for what is said of cubics, applies equally to Biquadratics, or those of any dimension whatever), we must now give some account of the different methods of obtaining an exact solution of them with as little trouble as possible. A ready method of doing this hath always been reckoned a desideratum in algebra, and indeed is likely to continue so. — From what we have already said, we hope it will be evident why a cube cannot be completed in a manner similar to that of completing the square in quadratic equations; another method hath therefore been chosen, namely, of destroying the second and third terms, and thus reducing the affected cube to a simple one.

The destruction of the second term is easily effected, and may be understood from the following considerations. (1.) In every cube whose root is a binomial, or expressible by $ a + b $, the signs are all $ + $; thus the cube of $ a + b = a^3 + 3a^2b + 3ab^2 + b^3 $. (2.) In a residual root, or $ a - b $, the signs of the cube are $ + $ and $ - $ alternately; thus the cube of $ a - b = a^3 - 3a^2b + 3ab^2 - b^3 $. (3.) By adding the cube of a binomial to the cube of a residual, the second and fourth terms always destroy one another, because they have contrary signs; but the first and third remain, because their signs are like, and they can only be destroyed by subtracting the equation from one another: thus the sum of the two cubes of $ a + b $, and $ a - b $, is $ 2a^3 + 6ab^2 $; their difference is $ 6a^2b + 2b^3 $.

It hath already been observed, that the coefficient of the second term of any cube is always equal to three times the known quantity forming one part of the root; as, if the root is $ a + b $, the coefficient of $ a^2 $ in the second term will be $ 3b $; if the root is $ x + 3 $, the coefficient of the second term will be $ +9 $; if it is $ x - 3 $, the coefficient will be $ -9 $, &c. Let it now be required to destroy the second term of the equation $ x^3 - 12x^2 + 47x = -60 $. Here, because the sign is negative, I suppose Equations $ x = a + 4 $, the third part of the coefficient of the second term, and substitute this instead of $ x $ into all the terms of the equation, in the following manner:

\[ \begin{align*} 1 & : x^3 - 12x^2 + 47x = -60 \\ 2 & : x = a + 4 \text{ by supposition} \\ 3 & : x^3 = a^3 + 12a^2 + 48a + 64 \\ 2 \times 7 & : 4x^2 = 12a^2 - 96a - 192 \\ 2 \times 7 & : 47x = 47a + 188 \\ 2 + 4 + 5 & : x^3 - 12x^2 + 47x = a^3 - a + 60 \\ 1 = 6 & : a^3 - a + 60 = -60 \\ 7 - 60 & : a^3 - a = -120 \end{align*} \]

From this example it will easily appear when the assumed value of $ x $ ought to be a binomial, and when a residual, and the destroying the second term of any equation can never be a matter of difficulty; but the destruction of the third term, it is plain, must depend upon quite other principles; for as its sign remains always $ + $ whether the root is binomial or residual, it cannot be destroyed by any addition of a positive; and as it is also generated from all the three steps of the new substitution, it is impossible to calculate matters so as to make the positive and negative terms at all times to destroy one another. In the last example, indeed, they have done so very nearly; and if the equation had been $ a^3 - 12x^2 + 48x = -55 $, they would have done so altogether, and the equation would have become $ a^3 = -125 $; but this is evidently a mere accident.

A method of destroying the third term of cubics as well as the second, has been invented by Cardan. It is very laborious; however, it shows in an eminent manner the powers of algebra, and how much a dextrous management of literal quantities may conduce to the resolution of problems utterly impossible to be solved without them.

Before this method can be followed, the second term must be destroyed as we have shewn above; then $ x $ must be supposed $ y + z $, and we proceed as in the following example:

\[ \begin{align*} 1 & : x^3 + 7x = 92 \text{ by question} \\ 2 & : x = y + z \text{ by supposition} \\ 3 & : y^3 + z^3 + 3yz = y + z + 1 \\ 2 \times 7 & : 4y^2 + 3z^2 + 3yz = 7 \\ 5 & : 3yz = -7 \text{ by supposition} \\ 6 & : y + z = -7 \\ 6 \text{ substituted into } 1 & : x^3 + 7x = -343 + 27 \\ 4 + 7 & : x^3 + 7x = -343 + 27 \\ 5 + 3 & : y^3 + z^3 = -1372 + 27 \\ 9 & : y^3 + z^3 = -1372 + 27 \\ 10 & : y^3 + z^3 = -1372 + 27 \\ 11 & : y^3 + z^3 = -1372 + 27 \\ 12 & : y^3 + z^3 = -1372 + 27 \\ 13 & : y^3 + z^3 = -1372 + 27 \\ 14 & : y^3 + z^3 = -1372 + 27 \\ 15 & : y^3 + z^3 = -1372 + 27 \\ 16 & : y^3 + z^3 = -1372 + 27 \\ 17 & : y^3 + z^3 = -1372 + 27 \\ 18 & : y^3 + z^3 = -1372 + 27 \\ 19 & : y^3 + z^3 = -1372 + 27 \\ 20 & : y^3 + z^3 = -1372 + 27 \\ 21 & : y^3 + z^3 = -1372 + 27 \\ 22 & : y^3 + z^3 = -1372 + 27 \end{align*} \]

In the above operation there is no difficulty, except in the assuming $ 3yz = -7 $, after having determined their sum. Equations.

\[ y + z \text{ to be } = x; \text{ but it must be considered, that the product of two numbers is by no means determined by their sum; for by making one of the numbers a fraction and the other an integer, by making one of them positive, and the other negative, we may fix their product, or any number of times their product, at what we please, without affecting their sum in the least. But we must be careful, if we have once assumed a sum, not to assume a difference also; for that would determine the unknown quantities. Thus, having assumed $ y + z = x $, we cannot assume $ y - z = $ any known quantity, because it might alter the value of $ y $ and $ z $ with regard to $ x $; but though we assume any imaginable product, we only alter the value of $ y $ and $ z $ with regard to one another, which is of no consequence.}

From the above operation may be deduced a general rule for the solution of all cubics to which this method is applicable; which, as corrected by Mr Simson, may be expressed in the following words. "Multiply the whole value of the equation by itself; divide the product by four; to the quotient add the cube of the coefficient of $ x $ in the third term (the second being destroyed) divided by 27; extract the square root of this sum, to which add half the value of the equation, and extract the cube root of the whole. Divide, now, one third of the coefficient of $ x $ by the root just found; subtract the quotient from the divisor, and the remainder is the value of $ x $." For the better understanding this theorem, in the foregoing example, $ x^3 + 7x = 9z $, let $ a = 7 $, and $ b = 9z $; then, the rule we have just now mentioned will stand thus in algebraic characters:

\[ \frac{x}{2} + \sqrt{\frac{b^2 + a^3}{4}} - \frac{b}{2} + \sqrt{\frac{b^2 + a^3}{4}} = 0 \]

Though this theorem seems capable of resolving every kind of cubic equation, yet one unlucky circumstance destroys its utility in a great many cases. For instance; let the equation $ x^3 - 12x = -9 $ be proposed. Here, according to the theorem, I multiply -9 by itself; the product is -81; this, divided by 4, quotes, 20.25. I now divide the cube of -12, or -1728, by 27; and the quotient -64 added to +20.25, destroys it entirely, and leaves a remainder of -44.25. From this the square root ought to be extracted; but this is impossible, because it is a negative quantity, and is formed neither from the multiplication of a positive into itself, nor of a negative into itself, but of a positive into a negative. Here, therefore, the operation must stop; and it is easy from this example to see when Cardan's method will succeed, and when it will not.

Other methods have been invented of solving the higher equations; but all of them are excessively laborious, and even precarious. A very ingenious method was invented by Sir Isaac Newton from finding the divisors of the absolute number by which the value of the equation is expressed; each of these was to be substituted in place of the unknown quantity, till some of them was found to answer the terms of the question.

It is easily shown, indeed, that $ x $ must always be a divisor of this number, and thus equations may be solved which could not be solved by Cardan's method; of which the last-mentioned one $ x^3 - 12x = -9 $, is an instance: for here, the only divisors of -9 are, +1, -1, +3, -3, and +9, -9; and substituting these successively in place of $ x $, 3 will be found to answer, and is the true value of $ x $. Notwithstanding this advantage, however, when the number is large, it is excessively tedious to substitute all the divisors; and indeed, as we may easily know within a figure or two of the true value, perhaps we might succeed as well by random trials as any other way. The last term, and consequently the number of divisors, however, may be lessened by changing the equation into another, wherein a binomial or residual root is put for the unknown quantity; thus, in the equation $ x^3 - 4x^2 - 8x + 32 = 0 $, if $ x + 1 $ be substituted for $ y $, it will become $ x^3 - 16x^2 - 16x + 21 = 0 $.

Another very curious method is, instead of substituting all the divisors of the last term, to substitute successively the terms of the arithmetical progression 2, 0, -1, -2, &c. with the numbers thence resulting; and find all the divisors of each of these numbers, and write them down over against the number they divide. This being done, search for one or more arithmetical progressions, either ascending or descending, whose common difference is either unity, or some divisor of the index of the highest power of $ x $; that term of such progression which stands over against 0, if divided by the common difference, and substituted into the equation with the sign + or -, according as the progression from whence it was taken was ascending or descending, will be one of the roots of the equation. If $ x $ has more values than one, there will be more arithmetical progressions. Sometimes indeed there will be deceptions by this method, and progressions will appear, which do not point out the true root; but these would fail if the substitution was continued two or three steps further: an example or two will sufficiently illustrate this.

Let the equation given be $ x^3 + x = 68 $. By transposition it becomes $ x^3 + x = 68 = 0 $. Here I first suppose $ x = 2 $; which being substituted, produces -58; then I suppose $ x = 1 $, which produces 66; if $ x = 0 $, then -68 is produced; with -1, then -70 is produced; with -2, 78 is produced; and so on. Having thus made the requisite substitutions, they are wrote down with the terms of the arithmetical progression from which they are produced, on one hand, and their divisors on the other, thus:

| Divisors | Progression ascending | |----------|-----------------------| | 2 | 1, 2, 29, &c. | | 1 | 66 | | 0 | 68 | | 1 | 70 | | 2 | 78 | | 3 | 98 |

Among these divisors only one progression is discovered; and the number 4, pointing over against 0, shows 4 to be the only true root of the equation.

Let now the equation $ x^3 + x^2 - 29x^2 - 9x + 180 = 0 $ be proposed, and the work will stand as follows.

| Divisors | Progressions | |----------|-------------| | 2 | 70 | | 1 | 144 | | 0 | 180 | | -1 | 160 | | -2 | 90 |

In this example there are four progressions, two ascending, and two descending; which show the four roots of the equation to be +3, +4, -3, and -5. The reason Equations. reason of our success in this method is, that all the va- lues of $ x $ must necessarily be divisors of the absolute number by which the value of the whole equation is expressed. When $ x $ is supposed = 0, then that number stands alone; because it cannot be affected by any of the values of $ x $. The true roots of the equation must therefore lie in that line of divisors opposite to 0. The progressions serve to point them out; because, as +1, +2, or -1, -2, are successively substituted in place of $ x $, there is a proportionable alteration in the value of the equation, and consequently in the divisors of the number by which it is expressed; and as long as the substitution is continued, using quantities that differ by one certain increase or decrease, the same progres- sion must continue among the divisors.

A method of depriving biquadratic equations into cubic ones was invented by Des Cartes, which is pub- lished in Simpson's algebra, together with an improve- ment: but as the difficulty of solving cubic equations is very little inferior to that of solving biquadratics, we think it unnecessary to take farther notice of this, or any other method that is applicable to particular cases; and shall therefore explain the method of solving equa- tions by approximation, or by the converging se- ries; which, though sufficiently laborious, will cer- tainly answer in all cases, and for every kind of equa- tion.

Let the proposed equation be $ x^3 + 10x^2 + 50x = 2600 $. Here it is plain, that $ x $ cannot much exceed 10; mak- ing trial of 11, therefore, I find it too much, so that the true value of $ x $ must lie between 10 and 11. The difference between 10 and the true root, I call $ e $, which is an unknown quantity; and for the more easy Equations, finding its value, I put $ r $ for 10, and say $ x = r + e $. Then,

\[ \begin{align*} 1 & : x^3 + 3r^2 + 3re + e^3 \\ 2 & : 10x^2 + 10r^2 + 20re + 10e^2 \\ 3 & : 50x = 50r + 50e \\ 4 & : x^3 + 10x^2 + 50x = 10r^3 + 30r^2 + 30re + 10e^2 \\ & : 10r^2 + 20re + 10e^2 + 50r + 50e \end{align*} \]

Because $ e $ is of small value in comparison of $ r $, and to avoid being involved in high equations, I reject all the powers of $ e $ above the first; and having thrown them out, the equation becomes $ x^3 + 10x^2 + 50x $, or $ 2600 = r^3 + 30r^2 + 30re + 10e^2 $; whence, by transposition, $ 2600 = r^3 + 10r^2 + 50r + 50e $; and, by division, $ \frac{2600}{3r^2 + 20r + 50} = e $.

As the value of $ r $ is known, I substitute that value into this new equation; and having made the division, $ e $ is found to be 0.18 nearly. Having then assumed $ r = 10.18 $, and substituted this value into the equation in- stead of $ r $, in order to find the value of $ e $ more ex- actly, it will come out $ -0.0005247 $; which added to 10.18, gives 10.1794753; and if this value is again substituted, we will have another value of $ e $, which will determine the root still more exactly; and so on, to as many places of decimals as we please.

It is not necessary, in the solution of equations by this method, to take $ r $ always the nearest root less than just; the same purpose will be answered by taking it more than just, making $ r = e $, and proceeding ac- cordingly.