Algebra is a general method of computation by certain signs and symbols, which have been contrived for this purpose, and found convenient. It is called an Universal Arithmetic, and proceeds by operations and rules similar to those in common arithmetic, founded upon the same principles. But as a number of symbols are admitted into this science, being necessary for giving it that extent and generality which is its greatest excellence, the import of those symbols must be clearly stated.

In geometry, lines are represented by a line, triangles by a triangle, and other figures by a figure of the same kind: But, in algebra, quantities are represented by the same letters of the alphabet; and various signs have been been imagined for representing their affections, relations, and dependencies.

The relation of equality is expressed by the sign =; thus, to express that the quantity represented by \(a\) is equal to that which is represented by \(b\), we write \(a = b\). But if we would express that \(a\) is greater than \(b\), we write \(a > b\); and if we would express algebraically that \(a\) is less than \(b\), we write \(a < b\).

Quantity is what is made up of parts, or is capable of being greater or less. It is increased by addition, and diminished by subtraction; which are therefore the two primary operations that relate to quantity. Hence it is, that any quantity may be supposed to enter into algebraic computations two different ways, which have contrary effects; either as an increment, or as a decrement; that is, as a quantity to be added, or as a quantity to be subtracted. The sign \(+\) (plus) is the mark of addition, and the sign \(-\) (minus) of subtraction. Thus the quantity being represented by \(a\), \(+a\) imports that \(a\) is to be added, or represents an increment; but, \(-a\) imports that \(a\) is to be subtracted, and represents a decrement.

When several such quantities are joined, the signs serve to show which are to be added and which are to be subtracted. Thus \(+a + b\) denotes the quantity that arises when \(a\) and \(b\) are both considered as increments, and therefore expresses the sum of \(a\) and \(b\). But \(+a - b\) denotes the quantity that arises, when from the quantity \(a\) the quantity \(b\) is subtracted; and expresses the excess of \(a\) above \(b\). When \(a\) is greater than \(b\), then \(a - b\) is itself an increment; when \(a = b\), then \(a - b = 0\); and when \(a\) is less than \(b\), then \(a - b\) is itself a decrement.

As addition and subtraction are opposite, or an increment is opposite to a decrement, there is an analogous opposition between the affections of quantities that are considered in the mathematical sciences; as, between excess and defect; between the value of effects or money due to a man, and money due by him. When two quantities, equal in respect of magnitude, but of those opposite kinds, are joined together, and conceived to take place in the same subject, they destroy each other's effect, and their amount is nothing. Thus, 100l. due to a man and 100l. due by him balance each other, and in estimating his stock may be both neglected. When two unequal quantities of those opposite qualities are joined in the same subject, the greater prevails by their difference. And, when a greater quantity is taken from a lesser of the same kind, the remainder becomes of the opposite kind.

A quantity that is to be added is likewise called a positive quantity; and a quantity to be subtracted is said to be negative: They are equally real, but opposite to each other, so as to take away each other's effect, in any operation, when they are equal as to quantity. Thus, \(3 - 3 = 0\), and \(a - a = 0\). But though \(+a\) and \(-a\) are equal as to quantity, we do not suppose in algebra that \(+a = -a\); because, to infer equality in this science, they must not only be equal as to quantity, but of the same quality, that in every operation the one may have the same effect as the other. A decrement may be equal to an increment, but it has in all operations a contrary effect; a motion downwards may be equal to a motion upwards; and the depression of a star below the horizon may be equal to the elevation of a star above it: But these positions are opposite, and the distance of the stars is greater than if one of them was at the horizon, so as to have no elevation above it, or depression below it. It is on account of this contrariety, that a negative quantity is said to be less than nothing, because it is opposite to the positive, and diminishes it when joined to it; whereas the addition of 0 has no effect. But a negative is to be considered no less as a real quantity than the positive. Quantities that have no sign prefixed to them are understood to be positive.

The number prefixed to a letter is called the numeral coefficient, and shows how often the quantity represented by the letter is to be taken. Thus \(2a\) imports that the quantity represented by \(a\) is to be taken twice; \(3a\) that it is to be taken thrice; and so on. When no number is prefixed, unit is understood to be the coefficient. Thus 1 is the coefficient of \(a\) or of \(b\).

Quantities are said to be like or similar, that are represented by the same letter or letters equally repeated. Thus \(+3a\) and \(-5a\) are like; but \(a\) and \(b\), or \(a\) and \(aa\) are unlike.

A quantity is said to consist of as many terms as there are parts joined by the signs \(+\) or \(-\); thus \(a + b\) consists of two terms, and is called a binomial; \(a + b + c\) consists of three terms, and is called a trinomial. These are called compound quantities: A simple quantity consists of one term only, as \(+a\), or \(+ab\), or \(+abc\).

**Chap. I. Of Addition.**

**Case I. To add quantities that are like, and have like signs.**

**Rule.** Add together the coefficients, to their sum prefix the common sign, and subjoin the common letter or letters.

**Example.** To \(+5a\) to \(-6b\) to \(a + b\)

Add \(+4a\) add \(-2b\) add \(3a + 5b\)

Sum \(+9a\) Sum \(-8b\) Sum \(4a + 6b\)

**Case II. To add quantities that are like, but have unlike signs.**

**Rule.** Subtract the lesser coefficient from the greater, prefix the sign of the greater to the remainder, and subjoin the common letter or letters.

**Example.** To \(-4a\) \(+5b - 6c\)

Add \(+7a\) \(-3b + 8c\)

Sum \(+3a\) \(2b + 2c\)

This rule is easily deduced from the nature of positive and negative quantities.

If there are more than two quantities to be added together, first add the positive together into one sum, and then the negative (by Case I.); then add these two sums together (by Case II.). CASE III. To add quantities that are unlike.

RULE. Set them all down one after another, with their signs and coefficients prefixed.

EXAMPLE. To \(+2a\) Add \(+3b\)

Sum \(2a + 3b\)

\(3a - 4x\)

CHAP. II. Of Subtraction.

GENERAL RULE. "Change the signs of the quantity to be subtracted into their contrary signs, and then add it so changed to the quantity from which it was to be subtracted, (by the rules of the last chapter): the sum arising by this addition is the remainder." For, to subtract any quantity, either positive or negative, is the same as to add the opposite kind.

EXAMPLE. From \(+5a\) Subtract \(+3a\)

Remaind. \(5a - 3a\), or \(2a\)

\(8a - 7b\)

\(3a + 4b\)

\(5a - 11b\)

It is evident, that to subtract or take away a decrement is the same as adding an equal increment. If we take away \(-b\) from \(a - b\), there remains \(a\); and if we add \(+b\) to \(a - b\), the sum is likewise \(a\). In general, the subtraction of a negative quantity is equivalent to adding its positive value.

CHAP. III. Of Multiplication.

In Multiplication, the general rule for the signs is, That when the signs of the factors are like, (i.e., both \(+\) or both \(-\)), the sign of the product is \(+\); but when the signs of the factors are unlike, the sign of the product is \(-\).

CASE I. When any positive quantity, \(+a\), is multiplied by any positive number, \(+n\), the meaning is, That \(+a\) is to be taken as many times as there are units in \(n\); and the product is evidently \(na\).

CASE II. When \(-a\) is multiplied by \(n\), then \(-a\) is to be taken as often as there are units in \(n\), and the product must be \(-na\).

CASE III. Multiplication by a positive number implies a repeated addition: But multiplication by a negative implies a repeated subtraction. And when \(+a\) is to be multiplied by \(-n\), the meaning is, That \(+a\) is to be subtracted as often as there are units in \(n\): Therefore the product is negative, being \(-na\).

CASE IV. When \(-a\) is to be multiplied by \(-n\), then \(-a\) is to be subtracted as often as there are units in \(n\); but (by chap. II,) to subtract \(-a\) is equivalent to adding \(+a\), consequently the product is \(+na\).

The IIth and IVth Cases may be illustrated in the following manner.

By the definitions, \(+a - a = 0\); therefore if we multiply \(+a - a\) by \(n\), the product must vanish, or be 0, because the factor \(a - a\) is 0. The first term of the product is \(+na\) (by Case I.) Therefore the second term of the product must be \(-na\), which destroys \(+na\); so that the whole product must be \(+na - na = 0\). Therefore \(-a\) multiplied by \(+n\) gives \(-na\).

In like manner, if we multiply \(+a - a\) by \(-n\), the first term of the product being \(-na\), the latter term of the product must be \(+na\); because the two together must destroy each other, or their amount be 0, since one of the factors (viz. \(a - a\)) is 0. Therefore \(-a\) multiplied by \(-n\), must give \(-na\).

In this general doctrine, the multiplicator is always considered as a number. A quantity of any kind may be multiplied by a number.

If the quantities to be multiplied are simple quantities, "find the sign of the product by the last rule; after it place the product of the coefficients, and then set down all the letters after one another as in one word."

EXAMPLE. Mult. \(+a\) By \(+b\)

Prod. \(+ab\)

Mult. \(-8x\) By \(-4a\)

Prod. \(-32ax\)

To multiply compound quantities, you must "multiply every part of the multiplicand by all the parts of the multiplier, taken one after another, and then collect all the products into one sum: That sum shall be the product required."

EXAMPLE. Mult. \(a + b\) By \(a + b\)

Prod. \(\{aa + ab + ab + bb\}\)

Sum \(aa + 2ab + bb\)

Mult. \(aa + ab + bb\) By \(a - b\)

Prod. \(\{aaa + aab + abb\}\)

Sum \(aaa...o...o...bbb\)

Products that arise from the multiplication of two, three, or more quantities, as \(abc\), are said to be of two, three, or more dimensions; and those quantities are called factors or roots.

If all the factors are equal, then these products are called powers; as \(aa\), or \(aaa\), are powers of \(a\). Powers are expressed sometimes by placing above the root, root, to the right hand, a figure expressing the number of factors that produce them. Thus,

\[ \begin{align*} a & \quad \text{1st Power of the} \\ a^2 & \quad \text{root } a, \text{ and } a^3 \\ a^3 & \quad \text{is shortly } a^3 \\ a^4 & \quad \text{expressed } a^4 \\ a^5 & \quad \text{thus, } a^5 \end{align*} \]

These figures which express the number of factors that produce powers, are called their indices or exponents; thus 2 is the index of \(a^2\). And powers of the same root are multiplied by adding their exponents. Thus \(a^2 \times a^3 = a^5\), \(a^4 \times a^3 = a^7\), \(a^3 \times a = a^4\).

Sometimes it is useful not actually to multiply compound quantities, but to set them down with the sign of multiplication (\(\times\)) between them, drawing a line over each of the compound factors. Thus \(a + b \times a - b\) expresses the product of \(a + b\), multiplied by \(a - b\).

**Chap. IV. Of Division.**

The same rule for the signs is to be observed in division as in multiplication; that is, "If the signs of the dividend and divisor are like, the sign of the quotient must be +; if they are unlike, the sign of the quotient must be —." This will be easily deduced from the rule in multiplication, if you consider, that the quotient must be such a quantity as, multiplied by the divisor, shall give the dividend.

The general rule in division is, "to place the dividend above a small line, and the divisor under it, expunging any letters that may be found in all the quantities of the dividend and divisor, and dividing the coefficients of all the terms by any common measure."

Thus, when you divide \(10ab + 15ac\) by \(20ad\), expunging \(a\) out of all the terms, and dividing all the coefficients by 5, the quotient is \(\frac{2b + 3c}{4d}\); and

\[ 2b) ab + bb (a + b) \]

"Powers of the same root are divided by subtracting their exponents, as they are multiplied by adding them." Thus, if you divide \(a^5\) by \(a^2\), the quotient is \(a^{5-2}\) or \(a^3\). And \(b^6\) divided by \(b^4\) gives \(b^{6-4}\) or \(b^2\); and \(a^7b^3\) divided by \(a^3b^3\) gives \(a^7b^3\) for the quotient.

"If the quantity to be divided is compound, then you must range its parts according to the dimensions of some one of its letters, as in the following example."

In the dividend \(a^2 + 2ab + b^2\), they are ranged according to the dimensions of \(a\), the quantity \(a^2\), where \(a\) is of two dimensions, being placed first, \(2ab\), where it is of one dimension, next, and \(b^2\), where \(a\) is not at all, being placed last. "The divisor must be ranged according to the dimensions of the same letters; then you are to divide the first term of the dividend by the first term of the divisor, and to set down the quotient, which, in this example, is \(a\); then multiply this quotient by the whole divisor, and subtract the product from the dividend, and the remainder shall give a new dividend, which, in this example, is \(ab + b^2\)."

\[ \begin{align*} (a + b) & \quad a^2 + 2ab + b^2 (a + b) \\ & \quad a^2 + ab \\ & \quad ab + b^2 \\ & \quad ab + b^2 \\ & \quad o. o. \end{align*} \]

"Divide the first term of this new dividend by the first term of the divisor, and set down the quotient (which in this example is \(b\)), with its proper sign. Then multiply the whole divisor by this part of the quotient, and subtract the product from the new dividend; and if there is no remainder, the division is finished." If there is a remainder, you are to proceed after the same manner, till no remainder is left; or till it appear that there will be always some remainder.

Some examples will illustrate this operation.

**Examp. I.** \(a + b) a^2 - b^2 (a - b)\)

\[ \begin{align*} & \quad a^2 + ab \\ & \quad ab - b^2 \\ & \quad ab - b^2 \\ & \quad o. o. \end{align*} \]

**Examp. II.** \(a-b) aaa - 3aab + 3abb - bbb (aa - 2ab + bb)\)

\[ \begin{align*} & \quad aaa - aab \\ & \quad -2aab + 3abb - bbb \\ & \quad -2aab + 2ab \\ & \quad abb - bbb \\ & \quad abb - bbb \\ & \quad o. o. \end{align*} \]

It often happens, that the operation may be continued without end, and then you have an infinite series for the quotient; and by comparing the first three or four terms you may find what law the terms observe: by which means, without any more division, you may continue the quotient as far as you please. Thus, in dividing \(1\) by \(1-a\), you find the quotient to be \(1 + a + aa + aaa + aaaa + \ldots\), which series can be continued as far as you please, by adding the powers of \(a\).

The operation is thus:

\[ \begin{align*} 1 - a) & \quad 1 (1 + a + aa + aaa + \ldots) \\ & \quad 1 - a \\ & \quad + a \\ & \quad + a - aa \\ & \quad + aa \\ & \quad + aa - aaa \\ & \quad + aaa \\ & \quad + aaa - aaaa \\ & \quad + aaaa, \ldots \end{align*} \]

Note, The sign \(÷\) placed between any two quantities, ties, expresses the quotient of the former divided by the latter. Thus, \( \frac{a + b}{a - x} \) is the quotient of \( a + b \), divided by \( a - x \).

**Chap. V. Of Fractions.**

In the last chapter it was said, that the quotient of any quantity \( a \), divided by \( b \), is expressed by placing \( a \) above a small line, and \( b \) under it, thus, \( \frac{a}{b} \). These quotients are also called fractions; and the dividend, or quantity placed above the line, is called the numerator of the fraction, and the divisor, or quantity placed under the line, is called the denominator.

"If the numerator of a fraction be equal to the denominator, then the fraction is equal to unity. Thus, \( \frac{a}{a} \) and \( \frac{b}{b} \) are equal to unit. If the numerator is greater than the denominator, then the fraction is greater than unit." In both these cases, the fraction is called improper. But "if the numerator is less than the denominator, then the fraction is less than unit," and is called proper. Thus, \( \frac{5}{3} \) is an improper fraction; but \( \frac{2}{4} \) and \( \frac{2}{3} \) are proper fractions. A mixt quantity is that whereof one part is an integer, and the other a fraction. As \( 3 \frac{4}{5} \) and \( 5 \frac{2}{3} \) and \( a + \frac{a^2}{b} \).

**Problem I.**

To reduce a Mixt quantity to an Improper Fraction.

**Rule.** Multiply the part that is an integer by the denominator of the fractional part; and to the product add the numerator; under their sum place the former denominator.

Thus \( 2 \frac{1}{2} \) reduced to an improper fraction gives \( \frac{5}{2} \); \( a + \frac{a^2}{b} = \frac{ab + a^2}{b} \); and \( a - x + \frac{ax}{x} = \frac{a^2 - x^2}{x} \).

**Problem II.**

To reduce an Improper Fraction to a Mixt Quantity.

**Rule.** Divide the numerator of the fraction by the denominator, and the quotient shall give the integral part; the remainder set over the denominator shall be the fractional part.

Thus \( \frac{12}{5} = 2 \frac{2}{5} \); \( \frac{ab + a^2}{b} = a + \frac{a^2}{b} \).

**Problem III.**

To reduce fractions of different denominations to fractions of equal value that shall have the same denominator.

**Rule.** Multiply each numerator, separately taken; into all the denominators but its own, and the products shall give the new numerators. Then multiply all the denominators into one another, and the product shall give the common denominator. Thus,

The fractions \( \frac{a}{b}, \frac{b}{c}, \frac{c}{d} \), are respectively equal to these fractions \( \frac{a c d}{b c d}, \frac{b b d}{b c d}, \frac{c c b}{b c d} \), which have the same denominator \( b c d \). And the fractions \( \frac{1}{2}, \frac{1}{4}, \frac{1}{8} \), are respectively equal to these \( \frac{4}{8}, \frac{2}{8}, \frac{1}{8} \).

**Problem IV.**

To Add and Subtract fractions.

**Rule.** Reduce them to a common denominator, and add or subtract the numerators; the sum or difference set over the common denominator, is the sum or remainder required.

\[ \frac{a}{b} + \frac{c}{d} + \frac{d}{e} = \frac{a de + b ce + d^2 b}{b de}; \quad \frac{a}{b} - \frac{c}{d} = \frac{ad - bc}{bd}; \]

\[ \frac{2}{3} + \frac{3}{4} = \frac{8 + 9}{12} = \frac{17}{12} = 1 \frac{5}{12}; \quad \frac{3}{4} - \frac{2}{3} = \frac{9 - 8}{12} = \frac{1}{12}; \]

\[ \frac{4}{5} - \frac{3}{4} = \frac{16 - 15}{20} = \frac{1}{20}; \quad \frac{x}{3} - \frac{2x}{6} = \frac{x}{6}. \]

**Problem V.**

To Multiply fractions.

**Rule.** Multiply their numerators one into another to obtain the numerator of the product; and their denominators multiplied into one another shall give the denominator of the product. Thus,

\[ \frac{a}{b} \times \frac{c}{d} = \frac{a c}{b d}; \quad \frac{2}{3} \times \frac{4}{5} = \frac{8}{15}. \]

If a mixt quantity is to be multiplied, first reduce it to the form of a fraction (by Prob. I.). And, if an integer is to be multiplied by a fraction, you may reduce it to the form of a fraction by placing unit under it.

**Example.** \( 5 \frac{2}{3} \times \frac{3}{4} = \frac{17}{3} \times \frac{3}{4} = \frac{51}{12} \).

**Problem VI.**

To Divide Fractions.

**Rule.** Multiply the numerator of the dividend by the denominator of the divisor, their product shall give the numerator of the quotient. Then multiply the denominator of the dividend by the numerator of the divisor, and their product shall give the denominator.

Thus, \( \frac{4}{5} \times \frac{2}{3} = \frac{10}{12} = \frac{5}{6} \); \( \frac{5}{8} \times \frac{3}{7} = \frac{35}{24} = \frac{5}{6} \); \( \frac{a}{b} \times \frac{ad}{cb} = \frac{a^2}{bc} \). PROBLEM VII.

To find the greatest common measure of two numbers; that is, the greatest number that can divide them both without a remainder.

RULE. First divide the greater number by the lesser, and if there is no remainder, the lesser number is the greatest common divisor required. If there is a remainder, divide your last divisor by it; and thus proceed continually, dividing the last divisor by its remainder, till there is no remainder left, and then the last divisor is the greatest common measure required.

Thus, the greatest common measure of 45 and 63 is 9; and the greatest common measure of 256 and 48 is 16.

Much after the same manner the greatest common measure of algebraic quantities is discovered; only the remainders that arise in the operation are to be divided by their simple divisors, and the quantities are always to be ranged according to the dimensions of the same letter.

Thus to find the greatest common measure of \(a^2 - b^2\) and \(a^2 - 2ab + b^2\);

\[ (a^2 - b^2) \div (a^2 - 2ab + b^2) = \frac{a^2 - b^2}{a^2 - 2ab + b^2} \text{ Remainder,} \]

which divided by \(-2b\) is reduced to

\[ \frac{a^2 - b^2}{a^2 - 2ab + b^2} \text{ Remainder,} \]

Therefore \(a - b\) is the greatest common measure required.

The ground of this operation is, That any quantity that measures the divisor and the remainder (if there is any) must also measure the dividend; because the dividend is equal to the sum of the divisor multiplied into the quotient, and of the remainder added together. Thus, in the last example, \(a - b\) measures the divisor \(a^2 - b^2\), and the remainder \(-2ab + b^2\); it must therefore likewise measure their sum \(a^2 - 2ab + b^2\). You must observe in this operation to make that the dividend which has the highest powers of the letter, according to which the quantities are ranged.

PROBLEM VIII.

To reduce any fraction to its lowest terms.

RULE. Find the greatest common measure of the numerator and denominator; divide them by that common measure, and place the quotients in their room, and you shall have a fraction equivalent to the given fraction expressed in the least terms.

Thus,

\[ \frac{25abc}{25abc} \cdot \frac{75abc}{125b^2x} = \frac{3a}{5x}; \quad \frac{156aa + 156ab}{572aa - 572ab} = \frac{3a + 3b}{11a - 11b}. \]

When unit is the greatest common measure of the numbers and quantities, then the fraction is already in its lowest terms. Thus \(\frac{3a}{5d}\) cannot be reduced lower.

And, numbers whose greatest common measure is unit, are said to be prime to one another.

If a vulgar fraction is to be reduced to a decimal (that is, a fraction whose denomination is 10, or any of its powers,) "annex as many cyphers as you please to the numerator, and then divide it by the denominator, the quotient shall give a decimal equal to the vulgar fraction proposed." Thus,

\[ \frac{2}{3} = .66666, \text{ &c.} \quad \frac{3}{5} = .6; \]

\[ \frac{2}{7} = .2857142, \text{ &c.} \]

These fractions are added and subtracted like whole numbers; only care must be taken to set similar places above one another, as units above units, and tenths above tenths, &c. They are multiplied and divided as integer numbers; only there must be as many decimal places in the product as in both the multiplicand and multiplier; and in the quotient as many as there are in the dividend more than in the divisor. And in division the quotient may be continued to any degree of exactness you please, by adding cyphers to the dividend. The ground of these operations is easily understood from the general rules for adding, multiplying, and dividing fractions.

CHAP. VI. Of the Involution of Quantities.

The products arising from the continual multiplication of the same quantity were called (in Chap. III.) the powers of that quantity. Thus, \(a, a^2, a^3, \text{ &c.}\) are the powers of \(a\); and \(a^2, a^3, b^2, a^3, b^3, \text{ &c.}\) are the powers of \(ab\). In the same chapter, the rule for the multiplication of powers of the same quantity is, "To add the exponents, and make their sum the exponent of the product." Thus \(a^4 \times a^3 = a^7\); and \(a^3 b^3 \times a^6 b^2 = a^9 b^5\). In Chap. IV. you have the rule for dividing powers of the same quantity, which is "To subtract the exponents, and make the difference the exponent of the quotient."

Thus,

\[ \frac{a^6}{a^2} = a^{6-2} = a^4; \quad \frac{a^5 b^3}{a^4 b} = a^{5-4} b^{3-1} = ab^2. \]

If you divide a lesser power by a greater, the exponent of the quotient must, by this rule, be negative.

Thus, Thus, \( \frac{a^4}{a^5} = a^{4-5} = a^{-1} \). But, \( \frac{a^4}{a^5} = \frac{1}{a^1} \); and hence \( \frac{1}{a^2} \) is expressed also by \( a^2 \) with a negative exponent.

It is also obvious, that \( \frac{a}{a} = a^{1-1} = a^0 \); but \( \frac{a}{a} = 1 \), and therefore \( a^0 = 1 \). After the same manner \( \frac{1}{a} = a^0 = a^0 - 1 = a^{-1} \); \( \frac{1}{a^2} = a^0 - 2 = a^{-2} \); \( \frac{1}{a^3} = a^0 - 3 = a^{-3} \); so that the quantities \( a, 1, \frac{1}{a}, \frac{1}{a^2}, \frac{1}{a^3}, \ldots \) may be expressed thus, \( a^1, a^0, a^{-1}, a^{-2}, a^{-3}, a^{-4}, \ldots \). Those are called the negative powers of \( a \) which have negative exponents; but they are at the same time positive powers of \( \frac{1}{a} \) or \( a^{-1} \).

Negative powers (as well as positive) are multiplied by adding, and divided by subtracting their exponents.

Thus the product of \( a^{-2} \) (or \( \frac{1}{a^2} \)) multiplied by \( a^{-3} \) (or \( \frac{1}{a^3} \)) is \( a^{-2-3} = a^{-5} \) (or \( \frac{1}{a^5} \)); also \( a^{-6} \times a^6 = a^{-6+6} = a^0 \) (or \( \frac{1}{a^0} \)); and \( a^{-3} \times a^3 = a^0 = 1 \). And, in general, any positive power of a multiplied by a negative power of \( a \) of an equal exponent gives unit for the product; for the positive and negative destroy each other, and the product gives \( a^0 \), which is equal to unit.

Likewise \( \frac{a^{-3}}{a^{-2}} = a^{-3+2} = a^{-1} \); and \( \frac{a^{-2}}{a^{-3}} = a^{-2+3} = a^1 \). But also, \( \frac{a^{-3}}{a^{-5}} = \frac{a^{-3}}{a^{-2} \times a^{-3}} = \frac{1}{a^{-3}} \); therefore \( \frac{1}{a^{-3}} = a^3 \): And, in general, "Any quantity placed in the denominator of a fraction may be transposed to the numerator, if the sign of its exponent be changed." Thus \( \frac{1}{a^3} = a^{-3} \), and \( \frac{1}{a^{-3}} = a^3 \).

The quantity \( a^m \) expresses any power of \( a \) in general, the exponent (\( m \)) being undetermined; and \( a^{-m} \) expresses \( \frac{1}{a^m} \), or a negative power of \( a \) of an equal exponent; and \( a^m \times a^{-m} = a^{m-m} = a^0 = 1 \) is their product. \( a^n \) expresses any other power of \( a \); \( a^m \times a^n = a^{m+n} \) is the product of the powers \( a^m \) and \( a^n \), and \( a^{m-n} \) is their quotient.

To raise any simple quantity to its second, third, or fourth power, is to add its exponent twice, thrice, or four times to itself; therefore the second power of any quantity is had by doubling its exponent, and the third by trebling its exponent; and, in general, the power expressed by \( m \) of any quantity is had by multiplying the exponent by \( m \), as is obvious from the multiplication of powers. Thus the second power or square of \( a \) is \( a^2 \times a^1 = a^3 \); its third power or cube is \( a^3 \times a^2 = a^5 \); and the \( m \)th power of \( a \) is \( a^m \times a^m = a^{2m} \). Also, the square of \( a^4 \) is \( a^4 \times a^4 = a^8 \); the cube of \( a^4 \) is \( a^4 \times a^4 = a^{12} \); and the \( m \)th power of \( a^4 \) is \( a^4 \times a^m \). The square of \( a b c \) is \( a^2 b^2 c^2 \), the cube is \( a^3 b^3 c^3 \), the \( m \)th power \( a^m b^m c^m \).

The raising of quantities to any power is called involution, and any simple quantity is involved by multiplying the exponent by that of the power required, as in the preceding examples.

The coefficient must also be raised to the same power by a continual multiplication of itself by itself, as often as unit is contained in the exponent of the power required. Thus the cube of \( 3 a b \) is \( 3 \times 3 \times 3 \times a^3 b^3 = 27 a^3 b^3 \).

As to the signs, When the quantity to be involved is positive, it is obvious that all its powers must be positive. And, when the quantity to be involved is negative, yet all its powers, whose exponents are even numbers, must be positive: for any number of multiplications of a negative, if the number be even, gives a positive; since \( -x = + \), therefore \( -x - x - x - x = +x + = + \); and \( -x - x - x - x - x - x = +x + x + = + \).

The power then only can be negative when its exponent is an odd number, though the quantity to be involved be negative. The powers of \( -a \) are \( -a, +a^2, -a^3, +a^4, -a^5, \ldots \). Those whose exponents are 2, 4, 6, etc., are positive; but those whose exponents are 1, 3, 5, etc., are negative.

The involution of compound quantities is a more difficult operation. The powers of any binomial \( a + b \) are found by a continual multiplication of it by itself, as follows:

\[ \begin{align*} a + b & = \text{Root.} \\ \times a + b & \\ a^2 + ab & + ab + b^2 \\ a^2 + 2ab + b^2 & = \text{the square or 2d power.} \\ \times a + b & \\ a^3 + 2a^2b + ab^2 & + a^2b + 2ab^2 + b^3 \\ a^3 + 3a^2b + 3ab^2 + b^3 & = \text{cube or 3d power, etc.} \end{align*} \]

If the powers of \( a - b \) are required, they will be found the same as the preceding, only the terms in which the exponent of \( b \) is an odd number, will be found negative; "because an odd number of multiplications of a negative produces a negative." Thus, the cube of \( a - b \) will be found to be \( a^3 - 3a^2b + 3ab^2 - b^3 \). Where the 2d and 4th terms are negative, the exponent of \( b \) being an odd number in these terms. In general, "The terms of any power of \( a - b \) are positive and negative by turns."

It is to be observed, That "in the first term of any power of \( a - b \), the quantity \( a \) has the exponent of the power required; that in the following terms, the exponents of \( a \) decrease gradually by the same difference..." ence (viz: unit), and that in the last terms it is never found. The powers of \( b \) are in the contrary order; it is not found in the first term, but its exponent in the second term is unit, in the third term its exponent is 2; and thus its exponent increases, till in the last term it becomes equal to the exponent of the power required."

As the exponents of \( a \) thus decrease, and at the same time those of \( b \) increase, "the sum of their exponents is always the same, and is equal to the exponent of the power required." Thus, in the 6th power of \( a + b \), viz. \( a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + b^6 \), the exponents of \( a \) decrease in this order, 6, 5, 4, 3, 2, 1, 0; and those of \( b \) increase in the contrary order, 0, 1, 2, 3, 4, 5, 6. And the sum of their exponents in any term is always 6.

To find the coefficient of any term, the coefficient of the preceding term being known, you are to "divide the coefficient of the preceding term by the exponent of \( b \) in the given term, and to multiply the quotient by the exponent of \( a \) in the same term, increased by unit." Thus to find the coefficients of the terms of the 6th power of \( a + b \), you find the terms are

\[ a^6, a^5b, a^4b^2, a^3b^3, a^2b^4, ab^5, b^6; \]

and you know the coefficient of the first term is unit; therefore, according to the rule, the coefficient of the second term will be \( \frac{1}{1} \times 5 + 1 = 6 \); that of the third term will be \( \frac{6}{2} \times 4 + 1 = 3 \times 5 = 15 \); that of the fourth term will be \( \frac{15}{3} \times 3 + 1 = 5 \times 4 = 20 \); and those of the following will be 15, 6, 1, agreeable to the preceding table.

In general, if \( a + b \) is to be raised to any power \( m \), the terms, without their coefficients will be \( a^m, a^{m-1}b, a^{m-2}b^2, a^{m-3}b^3, a^{m-4}b^4, \ldots \), &c., continued till the exponent of \( b \) becomes equal to \( m \).

The coefficients of the respective terms, according to the last rule, will be

\[ 1, m, m \times \frac{m-1}{2}, m \times \frac{m-1}{2}, \times \frac{m-2}{3}, m \times \frac{m-1}{2} \times \frac{m-2}{3} \times \frac{m-3}{4}, \ldots \]

&c., continued until you have one coefficient more than there are units in \( m \).

It follows therefore by these last rules, that \( a + b^m = a^m + m \times a^{m-1}b + m \times \frac{m-1}{2} \times a^{m-2}b^2 + m \times \frac{m-1}{2} \times \frac{m-2}{3} \times a^{m-3}b^3 + m \times \frac{m-1}{2} \times \frac{m-2}{3} \times \frac{m-3}{4} \times a^{m-4}b^4 + \ldots \), which is the general theorem for raising a quantity consisting of two terms to any power \( m \).

If quantity consisting of three or more terms is to be involved, "you may distinguish it into two parts, considering it as a binomial, and raise it to any power by the preceding rules; and then, by the same rules, you may substitute, instead of the powers of these compound parts, their values." Thus,

\[ a + b + c^2 = a + b + c^2 = a + b^2 + 2c \times a + b + c^2 = a^2 + 2a + b^2 + 2ac + 2bc + c^2. \]

In these examples, \( a + b + c \) is considered as composed of the compound part \( a + b \) and the simple part \( c \); and then the powers of \( a + b \) are formed by the preceding rules, and substituted for \( a + b^3 \) and \( a + b^2 \).

**Chap. VII. Of Evolution.**

The reverse of involution, or the resolving of powers into their roots, is called evolution. The roots of single quantities are easily extracted "by dividing their exponents by the number that denominates the root required." Thus, the square root of \( a^8 \) is \( a^{\frac{8}{2}} = a^4 \); and the square root of \( a^6b^3c^2 \) is \( a^{\frac{6}{2}}b^{\frac{3}{2}}c = a^3b^{\frac{3}{2}}c \). The cube root of \( x^3y^2z^4 \) is \( x^{\frac{3}{3}}y^{\frac{2}{3}}z^{\frac{4}{3}} \). The ground of this rule is obvious from the rule for involution. The powers of any root are found by multiplying its exponent by the index that denominates the power; and therefore, when any power is given, the root must be found by dividing the exponent of the given power by the number that denominates the kind of root that is required.

It appears, from what was said of involution, that "any power that has a positive sign may have either a positive or negative root, if the root is denominated by any even number." Thus the square-root of \( +a^2 \) may be \( +a \) or \( -a \), because \( +a \times +a = -a \times -a \) gives \( +a^2 \) for the product.

But if a power have a negative sign, "no root of it denominated by an even number can be assigned," since there is no quantity that multiplied into itself an even number of times can give a negative product. Thus the square root of \( -a^2 \) cannot be assigned, and is what we call an impossible or imaginary quantity.

But if the root to be extracted is denominated by an odd number, then shall "the sign of the root be the same as the sign of the given number whose root is required." Thus the cube root of \( -a^3 \) is \( -a \), and the cube root of \( -a^6b^3 \) is \( -a^2b \).

If the number that denominates the root required is a divisor of the exponent of the given power, then shall the root be only a lower power of the same quantity. As the cube root of \( a^3 \) is \( a \), the number 3 that denominates the cube root being a divisor of 3.

But if the number that denominates what sort of root is required is not a divisor of the exponent of the given power, "then the root required shall have a fraction for its exponent." Thus the square root of \( a^3 \) is \( a^{\frac{3}{2}} \); the cube root of \( a^5 \) is \( a^{\frac{5}{3}} \), and the square root of \( a \) itself is \( a^{\frac{1}{2}} \).

These powers that have fractional exponents are called imperfect powers or surds; and are otherwise expressed by placing the given power within the radical sign \( \sqrt{} \), and placing above the radical sign the number that denominates what kind of root is required. Thus \( \frac{3}{2} = \sqrt[2]{a^3} \), \( \frac{2}{3} = \sqrt[3]{a^2} \); and \( \frac{n}{m} = \sqrt[m]{a^n} \). In numbers the square root of 2 is expressed by \( \sqrt{2} \), and the cube root of 4 by \( \sqrt[3]{4} \). The square root of any compound quantity, as \(a^2 + 2ab + b^2\) is discovered after this manner. "First," take care to dispose the terms according to the dimensions of the alphabet, as in division; then find the square root of the first term \(a\), which gives \(a\) for the first member of the root. Then subtract its square from the proposed quantity, and divide the first term of the remainder \((2ab + b^2)\) by the double of that member, viz., \(2a\), and the quotient \(b\) is the second member of the root. Add this second member to the double of the first, and multiply their sum \((2a + b)\) by the second member \(b\), and subtract the product \((2ab + b^2)\) from the foreaid remainder \((2ab + b^2)\) and if nothing remains, then the square root is obtained;" and in this example it is found to be \(a + b\).

The manner of the operation is thus,

\[ \begin{align*} & a^2 + 2ab + b^2 \\ & \quad (a + b) \\ & \quad \frac{2a + b}{2ab + b^2} \\ & \quad \frac{2ab + b^2}{0} \end{align*} \]

"But if there had been a remainder, you must have divided it by the double of the sum of the two parts already found, and the quotient would have given the third member of the root."

Thus, if the quantity proposed had been \(a^2 + 2ab + 2ac + b^2 + 2bc + c^2\); after proceeding as above, you would have found the remainder \(2ac + 2bc + c^2\), which divided by \(2a + 2b\) gives \(c\) to be annexed to \(a + b\) as the 3d member of the root. Then adding \(c\) to \(2a + 2b\), and multiplying their sum \(2a + 2b + c\) by \(c\), subtract the product \(2ac + 2bc + c^2\) from the foreaid remainder; and since nothing now remains, you conclude that \(a + b + c\) is the square root required.

The square root of any number is found out after the same manner. If it is a number under 100, its nearest square root is found by the following table; by which also its cube root is found if it be under 1000, and is biquadrate if it be under 10000.

| The root | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | |----------|---|---|---|---|---|---|---|---|---| | Square | 1 | 4 | 9 | 16| 25| 36| 49| 64| 81| | Cube | 1 | 8 | 27| 64|125|216|343|512|729| | Biquad. | 1 |16|81|256|625|1296|2401|4096|6561|

But if it is a number above 100, then its square root will consist of two or more figures, which must be found by different operations by the following

**Rule.**

"To find the square root of any number, place a point above the number that is in the place of units, pass the place of tens, and place again a point over that of hundreds, and go on towards the left hand placing a point over every 2d figure; and by these points the number will be distinguished into as many parts as there are figures in the root. Then find the square root of the first part, and it will give the first figure of the root; subtract its square from that part, and annex the second part of the given number to the remainder. Then divide this new number (neglecting its last figure) by the double of the first figure of the root, annex the quotient to that double, and multiply the number thence arising by the said quotient, and if the product is less than your dividend, or equal to it, that quotient shall be the second figure of the root. But if the product is greater than the dividend, you must take a less number for the second figure of the root than that quotient." Much after the same manner may the other figures of the quotient be found, if there are more points than two placed over the given number.

To find the square root of 99856, first point it thus, 99856; then find the square root of 9 to be 3, which therefore is the first figure of the root; subtract 9, the square of 3, from 9, and to the remainder annex the second part 98, and divide (neglecting the last figure 8) by the double of 3, or 6, and place the quotient after 6, and then multiply 61 by 1, and subtract the product 61 from 98. Then to the remainder (37) annex the last part of the proposed number (56), and dividing 3756 (neglecting the last figure 6) by the double of 31, that is by 62, place the quotient after, and multiplying 626 by the quotient 6, you will find the product to be 3756, which subtracted from the dividend, and leaving no remainder, the exact root must be 316.

**Example.**

\[ \begin{array}{c} 99856 \\ -9 \\ \hline 6198 \\ \times 61 \\ \hline 626 \times 3756 \\ \times 6 \times 3756 \\ \hline 0 \end{array} \]

In general, to extract any root out of any given quantity, "First range that quantity according to the dimensions of its letters, and extract the said root out of the first term, and that shall be the first member of the root required. Then raise this root to a dimension lower by unit than the number that denominates the root required, and multiply the power that arises by that 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."

Thus to extract the root of the 5th power out of \(a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5\), I find, that the root of the 5th power out of \(a^5\) gives \(a\), which I raise to the 4th power, and multiplying by 5, the product is \(5a^4\); then dividing the second term of the given quantity \(5a^4b\) by \(5a^4\), I find \(b\) to be the second member; and raising \(a + b\) to the 5th power, and subtracting it, there being no remainder, I conclude that \(a + b\) is the root required. If the root has three members, the third is found after the same manner from the first two considered as one member, as the second member was found from the first, which may be easily understood from what was said of extracting the square root. In extracting roots it will often happen that the exact root cannot be found in finite terms; thus the square root of $a^2 + x^2$ is found to be

$$a + \frac{x^2}{2a} + \frac{x^4}{8a^3} + \frac{x^6}{16a^5} + \frac{5x^8}{128a^7} + \ldots$$

The operation is thus;

$$a^2 + x^2 \left(a + \frac{x^2}{2a} - \frac{x^4}{8a^3} + \frac{x^6}{16a^5} - \ldots\right)$$

$$= a^2 + \frac{x^2}{2a} + \frac{x^4}{4a^2}$$

$$- \frac{x^6}{8a^3} + \frac{x^8}{16a^5} - \frac{x^{10}}{32a^7} + \ldots$$

The general theorem which we gave for the involution of binomials will serve also for their evolution; because to extract any root of a given quantity is the same thing as to raise that quantity to a power whose exponent is a fraction that has its denominator equal to the number that expresses what kind of root is to be extracted. Thus, to extract the square root of $a + b$ is to raise $a + b$ to a power whose exponent is $\frac{1}{2}$.

The roots of numbers are to be extracted as those of algebraic quantities. Place a point over the units, and then place points over every third, fourth, or fifth figure towards the left hand, according as it is the root of the cube, of the 4th or 5th power that is required; and, if there be any decimals annexed to the number, point them after the same manner, proceeding from the place of units towards the right hand. By this means the number will be divided into so many periods as there are figures in the root required. Then inquire which is the greatest cube, biquadrate, or 5th power in the first period, and the root of that power will give the first figure of the root required. Subtract the greatest cube, biquadrate, or 5th power, from the first period, and to the remainder annex the first figure of your second period, which shall give your dividend.

Raise the first figure already found to a power less by unit than the power whose root is sought, that is, to the 2d, 3d, or 4th power, according as it is the cube root, the root of the 4th, or the root of the 5th power that is required, and multiply that power by the index of the cube, 4th, or 5th power, and divide the dividend by this product, so shall the quotient be the second figure of the root required.

Raise the part already found of the root, to the power whose root is required, and if that power be found less than the two first periods of the given number, the second figure of the root is right. But, if it be found greater, you must diminish the second figure of the root till that power be found equal to or less than those periods of the given number. Subtract it, and to the remainder annex the next period; and proceed till you have gone through the whole given number, finding the third figure by means of the two first, as you found the second by the first; and afterwards finding the fourth figure (if there be a fourth period) after the same manner from the three first.

Thus to find the cube root of 13824, point it 13824; find the greatest cube in 13, viz. 8, whose cube root 2 is the first figure of the root required. Subtract 8 from 13, and to the remainder 5 annex 8, the first figure of the second period; divide 58 by triple the square of 2, viz. 12, and the quotient is 4, which is the second figure of the root required, since the cube of 24 gives 13824, the number proposed. After the same manner the cube root of 13312053 is found to be 237.

$$\begin{array}{c} 13824 \\ \text{Subtr. } 8 = 2 \times 2 \times 2 \\ 3 \times 4 = 12 \quad 58 \quad (4) \\ \text{Subtr. } 24 \times 24 \times 24 = 13824 \\ \end{array}$$

Rem. . . . 0 .

In extracting of roots, after you have gone through the number proposed, if there is a remainder, you may continue the operation by adding periods of cyphers to that remainder, and find the true root in decimals to any degree of exactness.

**Chap. VIII. Of Proportion.**

When quantities of the same kind are compared, it may be considered, either how much the one is greater than the other, and what is their difference; or, it may be considered how many times the one is contained in the other; or, more generally, what is their quotient. The first relation of quantities is expressed by their arithmetical ratio; the second by their geometrical ratio. That term whose ratio is inquired into is called the antecedent, and that with which it is compared is called the consequent.

When of four quantities the difference betwixt the first and second is equal to the difference betwixt the third and fourth, those quantities are called arithmetical proportionals; as the numbers 3, 7, 12, 16. And the quantities $a, a+b, e, c+b$. But quantities form a series in arithmetical proportion, when they increase or decrease by the same constant difference. As these, $a, a+b, a+2b, a+3b, a+4b$, &c., $x, x-b, x-2b$, &c., or the numbers 1, 2, 3, 4, 5, &c., and 10, 7, 4, 1, —2, —5, —8, &c.

In four quantities arithmetically proportional, "the sum of the extremes is equal to the sum of the mean terms." terms." Thus, \(a, a+b, e, e+b\), are arithmetical proportionals, and the sum of the extremes \((a+c+b)\) is equal to the sum of the mean terms \((a+b+e)\). Hence, to find the fourth quantity arithmetically proportional to any three given quantities; "Add the second and third, and from their sum subtract the first term, the remainder shall give the fourth arithmetical proportional required."

In a series of arithmetical proportionals, "the sum of the first and last terms is equal to the sum of any two terms equally distant from the extremes." If the first terms are \(a, a+b, a+2b, \ldots\) and the last term \(x\), the last term but one will be \(x-b\), the last but two \(x-2b\), the last but three \(x-3b, \ldots\) So that the first half of the terms, having those that are equally distant from the last term set under them, will stand thus;

\[ \begin{align*} a, & \quad a+b, \quad a+2b, \quad a+3b, \quad a+4b, \quad \ldots \\ x, & \quad x-b, \quad x-2b, \quad x-3b, \quad x-4b, \quad \ldots \\ \end{align*} \]

And it is plain, that if each term be added to the term above it, the sum will be \(a+x\), equal to the sum of the first term \(a\) and the last term \(x\). From which it is plain, that "the sum of all the terms of an arithmetical progression is equal to the sum of the first and last taken half as often as there are terms;" that is, the sum of an arithmetical progression is equal to the sum of the first and last terms multiplied by half the number of terms.

Thus, in the preceding series, if \(n\) be the number of terms, the sum of all the terms will be \(a + x \times \frac{n}{2}\).

The common difference of the terms being \(b\), and \(b\) not being found in the first term, it is plain that "its coefficient in any term will be equal to the number of terms that precede that term." Therefore in the last term \(x\) you must have \(n-1 \times b\), so that \(x\) must be equal to \(a + n-1 \times b\). And the sum of all the terms being

\[ \frac{a+x \times n}{2}, \text{ it will also be equal to } \frac{2an+n^2b-nb}{2}, \]

or

\[ \frac{a+nb-b}{2} \times n. \text{ Thus for example, the series } 1+2+3+4+5, \ldots \text{ continued to a hundred, must be equal to } \frac{2 \times 100 + 10000 - 100}{2} = 5050. \]

If a series have (o) nothing for its first term, then "its sum shall be equal to half the product of the last term multiplied by the number of terms." For then \(a\) being \(=0\), the sum of the terms, which is in general \(a + x \times \frac{n}{2}\), will in this case be \(\frac{nx}{2}\). From which it is evident, that "the sum of any number of arithmetical proportionals beginning from nothing, is equal to half the sum of as many terms equal to the greatest term."

Thus,

\[ \begin{align*} 0+1+2+3+4+5+6+7+8+9 &= \frac{9+9+9+9+9+9+9+9+9+9}{2} = \frac{10 \times 9}{2} = 45. \end{align*} \]

"If of four quantities the quotient of the first and second be equal to the quotient of the third and fourth,"

And you read them by saying, As 2 is to 6, so 4 is to 12; or, as \(a\) is to \(ar\), so \(b\) to \(br\).

In four quantities geometrically proportional, "the product of the extremes is equal to the product of the middle terms." Thus, \(axbr = ar \times xb\). And, if it is required to find a fourth proportional to any three given quantities, "multiply the second by the third, and divide their product by the first, the quotient shall give the fourth proportional required." Thus, to find a fourth proportional to \(a, ar, \text{ and } b\), multiply \(ar\) by \(b\), and divide the product \(arb\) by the first term \(a\), the quotient \(br\) is the fourth proportional required.

In calculations it sometimes requires a little care to place the terms in due order; for which you may observe the following

**Rule.** First set down the quantity that is of the same kind with the quantity sought; then consider, from the nature of the question, whether that which is given is greater or less than that which is sought; if it is greater, then place the greatest of the other two quantities on the left hand; but if it is less, place the least of the other two quantities on the left hand, and the other on the right.

Then shall the terms be in due order; and you are to proceed according to the rule, multiplying the second by the third, and dividing their product by the first.

**Exam.** "If 30 men do any piece of work in 12 days, how many men shall do it in 18 days?"

Because it is a number of men that is sought, first set down 30, the number of men that is given: you will easily see that the number that is given is greater than the number that is sought; therefore place 18 on the left hand, and 12 on the right; and find a 4th proportional to 18, 30, 12, viz. \(\frac{30 \times 12}{18} = 20\).

When a series of quantities increase by one common multiplicator, or decrease by one common divisor, they are said to be in geometrical proportion continued.

As, \(a, ar, ar^2, ar^3, ar^4, ar^5, \ldots\), or,

\[ \frac{a}{r}, \frac{a}{r^2}, \frac{a}{r^3}, \frac{a}{r^4}, \frac{a}{r^5}, \ldots \]

The common multiplier or divisor is called their common ratio.

In such a series, "the product of the first and last is always equal to the product of the second and last but one, or to the product of any two terms equally remote from the extremes." In the series, \(a, ar, ar^2, ar^3, \ldots\), if \(y\) be the last term, then shall the four last terms of the series be \(y, \frac{y}{r}, \frac{y}{r^2}, \frac{y}{r^3}\); now it is plain, that \(axy = ar \times \frac{y}{r} = ar^2 \times \frac{y}{r^2} = ar^3 \times \frac{y}{r^3}, \ldots\). The sum of a series of geometrical proportionals wanting the first term, is equal to the sum of all but the last term multiplied by the common ratio.

For \( ar + ar^2 + ar^3 + \ldots \), etc.,

\[ \frac{y}{r} + \frac{y}{r^2} + \frac{y}{r^3} + \ldots + y = r \times a + ar + ar^2 + \ldots \]

Therefore if \( s \) be the sum of the series, \( s - a \) will be equal to \( s - y \times r \); that is, \( s - a = sr - yr \), or \( sr - s = yr - a \), and \( s = \frac{yr - a}{r - 1} \).

Since the exponent of \( r \) is always increasing from the second term, if the number of terms be \( n \), in the last term its exponent will be \( n - 1 \). Therefore \( y = ar^{n-1} \); and \( yr = ar^{n-1} \times ar^n \); and \( r = \left( \frac{yr - a}{r - 1} \right) = \frac{ar^n - a}{r - 1} \). So that having the first term of the series, the number of the terms, and the common ratio, you may easily find the sum of all the terms.

If it is a decreasing series whose sum is to be found, as of \( y + \frac{y}{r} + \frac{y}{r^2} + \frac{y}{r^3} + \ldots \), etc., \( + ar^3 + ar^2 + ar + a \), and the number of the terms be supposed infinite, then shall \( a \), the last term, be equal to nothing. For, because \( n \), and consequently \( r^{n-1} \) is infinite, \( a = \frac{r}{r^n - 1} = 0 \).

The sum of such a series \( s = \frac{yr}{r - 1} \); which is a finite sum, though the number of terms be infinite. Thus,

\[ 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots = \frac{1 \times 2}{2 - 1} = 2 \]

and \( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots = \frac{1 \times 2}{3 - 1} = \frac{3}{2} \)

**Chap. IX. Of Equations that involve only one unknown Quantity.**

An equation is "a proposition asserting the equality of two quantities." It is expressed most commonly by setting down the quantities, and placing the sign (=) between them.

An equation gives the value of a quantity, when that quantity is alone on one side of the equation; and that value is known, if all those that are on the other side are known. Thus if I find that \( x = \frac{4 \times 6}{3} = 8 \), I have a known value of \( x \). These are the last conclusions we are to seek in questions to be resolved; and if there be only one unknown quantity in a given equation, and only one dimension of it, such a value may always be found by the following rules.

**Rule I.** Any quantity may be transposed from one side of the equation to the other, if you change its sign.

For to take away a quantity from one side, and to place it with a contrary sign on the other side, is to subtract it from both sides; and it is certain, that "when from equal quantities you subtract the same quantity, the remainders must be equal."

By this rule, when the known and unknown quantities are mixed in an equation, you may separate them by bringing all the unknown to one side, and the known to the other side of the equation; as in the following examples,

Suppose \( 5x + 50 = x + 56 \)

by Transposit. \( 5x - 4x = 56 - 50 \), or, \( x = 6 \)

And if \( 2x + a = b + c \)

\( 2x - x = b - a \), or, \( x = b - a \).

**Rule II.** Any quantity by which the unknown quantity is multiplied may be taken away, if you divide all the other quantities on both sides of the equation by it.

For that is to divide both sides of the equation by the same quantity, and when you divide equal quantities by the same quantity, the quotients must be equal. Thus,

If \( ax = b \)

then \( x = \frac{b}{a} \);

and if \( 3x + 12 = 27 \)

by Rule 1st. \( 3x = 27 - 12 = 15 \)

and by Rule 2d. \( x = \frac{15}{3} = 5 \)

**Rule III.** If the unknown quantity is divided by any quantity, that quantity may be taken away if you multiply all the other members of the equation by it.

Thus,

If \( \frac{x}{b} = b + 5 \)

then shall \( x = bb + 5b \)

By this rule, an equation whereof any part is a fraction may be reduced to an equation that shall be expressed by integers. If there are more fractions than one in the given equation, you may, by reducing them to a common denominator, and then multiplying all the other terms by that denominator, abridge the calculation thus;

If \( \frac{x}{5} + \frac{k}{3} = x - 7 \)

then \( 3x + 5k = 15x - 105 \)

and by this Rule \( 3x + 5k = 15x - 105 \)

and by R. 1. and 2. \( x = \frac{15}{3} = 5 \)

**Rule IV.** If that member of the equation that involves the unknown quantity be a surd root, then the equation is to be reduced to another that shall be free from any surd, by bringing that member first to stand alone upon one side of the equation, and then taking away the radical sign from it, and raising the other side of the equation to the power denominated by the surd.

Thus if \( \sqrt{4x + 16} = 12 \)

\( 4x + 16 = 144 \)

and \( 4x = 144 - 16 = 128 \)

and \( x = \frac{128}{4} = 32 \). Rule V. If that side of the equation that contains the unknown quantity be a compleat square, cube, or other power; then extract the square root, cube root, or the root of that power, from both sides of the equation, and thus the equation shall be reduced to one of a lower degree.

If \( x^2 + 6x + 9 = 20 \) then \( x + 3 = \sqrt{20} \) and \( x = \sqrt{20} - 3 \).

Rule VI. A proportion may be converted into an equation asserting the product of the extreme terms equal to the product of the mean terms; or any one of the extremes equal to the product of the means divided by the other extreme.

If \( 12 - x : \frac{x}{2} :: 4 : 1 \) then \( 12 - x = 2x \) ... \( 3x = 12 \) ... and \( x = 4 \).

Or if \( 20 - x : x :: 7 : 3 \) then \( 60 - 3x = 7x \) ... \( 10x = 60 \) ... and \( x = 6 \).

Rule VII. If any quantity be found on both sides of the equation with the same sign prefixed, it may be taken away from both: Also, if all the quantities in the equation are multiplied or divided by the same quantity, it may be struck out of them all. Thus,

If \( 3x + b = a + b \) ... \( 3x = a \) ... and \( x = \frac{a}{3} \).

Rule VIII. Instead of any quantity in an equation, you may substitute another equal to it.

Thus if \( 3x + y = 24 \) and \( y = 9 \) then \( 3x + 9 = 24 \) ... \( x = \frac{24 - 9}{3} = 5 \).

The further improvements of this rule shall be taught in the following chapter.

Chap. X. Of the Solution of Questions that produce Simple Equations.

Simple equations are those "wherein the unknown quantity is only of one dimension." In the solution of which, we are to observe the following directions.

Direct. I. "After forming a distinct idea of the question proposed, the unknown quantities are to be expressed by letters, and the particulars to be translated from the common language into the algebraic manner of expressing them, that is, into such equations as shall express the relations or properties that are given of such quantities."

Thus, if the sum of two quantities must be 60, that condition is expressed thus, \( x + y = 60 \). If their difference must be 24, that condition gives \( x - y = 24 \).

If their product must be 1640, then \( xy = 1640 \). If their quotient must be 6, then \( \frac{x}{y} = 6 \). If their proportion is as 3 to 2, then \( x : y :: 3 : 2 \) or \( 2x = 3y \); because the product of the extremes is equal to the product of the mean terms.

Direct. II. "After an equation is formed, if you have one unknown quantity only, then, by the rules of the preceding chapter, bring it to stand alone on one side, so as to have only known quantities on the other side:" thus you shall discover its value.

Examp. "A person being asked what was his age, answered that \( \frac{1}{4} \) of his age multiplied by \( \frac{1}{2} \) of his age gives a product equal to his age. Qu. What was his age?"

It appears from the question, that if you call his age \( x \), then shall \( \frac{3x}{4} \times \frac{x}{12} = x \) that is \( \frac{3x^2}{48} = x \) and by Rule 3... \( 3x^2 = 48x \) and by R. 7... \( 3x = 48 \) whence by R. 2... \( x = 16 \).

Direct. III. "If there are two unknown quantities, then there must be two equations arising from the conditions of the question: Suppose the quantities \( x \) and \( y \); find a value of \( x \) or \( y \) from each of the equations, and then, by putting these two values equal to each other, there will arise a new equation involving one unknown quantity; which must be reduced by the rules of the former chapter."

Examp. I. "Let the sum of two quantities be \( s \) and their difference \( d \). Let \( s \) and \( d \) be given, and let it be required to find the quantities themselves." Suppose them to be \( x \) and \( y \), then, by the supposition,

\[ \begin{align*} x + y &= s \\ x - y &= d \\ \end{align*} \]

whence \( \begin{cases} x = s - y \\ x = d + y \end{cases} \)

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

Examp. II. "A privateer running at the rate of 10 miles an hour, discovers a ship 18 miles off making way at the rate of 8 miles an hour: It is demanded how many miles the ship can run before she be overtaken?"

Let the number of miles the ship can run before she be overtaken be called \( x \), and the number of miles the privateer must run before she come up with the ship be \( y \); then shall (by supp.) \( y = x + 18 \) ... and \( x : y :: 8 : 10 \) whence $10x = 8y \quad \text{and} \quad x = y - 18$. whence

$y - 18 = \frac{4y}{5}$ and $y = 90 \quad \text{and} \quad x = y - 18 = 72$.

To find the time, say, If 8 miles give 1 hour, 72 miles will give 9 hours . . thus, $8 : 1 :: 72 : 9$.

**Examp. III.** "Suppose the distance between London and Edinburgh to be 360 miles; and that a courier sets out from Edinburgh, running at the rate of 10 miles an hour; another sets out at the same time from London, and runs 8 miles an hour: It is required to know where they will meet?"

Suppose the courier that sets out from Edinburgh runs $x$ miles, and the other $y$ miles, before they meet, then shall,

by suppos. $\begin{cases} x + y = 360 \\ x : y = 5 : 4 \end{cases}$

$x = \frac{5y}{4}$

$x = 360 - y$

$\frac{5y}{4} = 360 - y$

$5y + 4y = 360$

$9y = 1440$

$y = \frac{1440}{9} = 160$

$x = 360 - y = 200$

**Examp. IV.** "Two merchants were copartners; the sum of their stock was 300l. One of their stocks continued in company 11 months, but the other drew out his stock in 9 months; when they made up their accounts they divided the gain equally. Qu. What was each man's stock?" Suppose the stock of the first to be $x$, and the stock of the other to be $y$; then,

by suppos. $\begin{cases} x + y = 300 \\ 11x = 9y \end{cases}$

$x = \frac{9y}{11} = 300 - y$

$11y + 9y = 3300$

$20y = 3300$

$y = \frac{3300}{20} = 165 \quad \text{so} \quad x = 300 - y = 135.$

**Direct. IV.** "When in one of the given equations the unknown quantity is of one dimension, and in the other of a higher dimension; you must find a value of the unknown quantity from that equation where it is of one dimension, and then raise that value to the power of the unknown quantity in the other equation; and by comparing it, so involved, with the value you deduce from that other equation, you shall obtain an equation that will have only one unknown quantity, and its powers."

That is, when you have two equations of different dimensions, if you cannot reduce the higher to the same dimension with the lower, you must raise the lower to the same dimension with the higher.

**Examp. V.** "The sum of two quantities, and the difference of their squares, being given, to find the quantities." Suppose them to be $x$ and $y$, their sum $s$, and the difference of their squares $d$. Then,

$\begin{cases} x + y = s \\ x^2 - y^2 = d \end{cases}$

$x = s - y$

$x^2 = s^2 - 2sy + y^2$

$x^2 = d + y^2$, whence $x = \frac{s^2 + d}{2s}$.

**Direct. V.** "If there are three unknown quantities, there must be three equations in order to determine them, by comparing which, you may, in all cases, find two equations involving only two unknown quantities; and then, by Direct. 3d, from these two you may deduce an equation involving only one unknown quantity; which may be resolved by the rules of the last chapter."

From three equations involving any three unknown quantities, $x$, $y$, and $z$, to deduce two equations, involving only two unknown quantities, the following rule will always serve.

**Rule.** "Find three values of $x$ from the three given equations; then, by comparing the first and second value, you will find an equation involving only $y$ and $z$; again, by comparing the first and third, you will find another equation involving only $y$ and $z$; and, lastly, those equations are to be resolved by Dir. 3."

**Examp. VI.**

Suppose $\begin{cases} x + y + z = 12 \\ x + 2y + 3z = 20 \end{cases}$ then, $\begin{cases} 12 - y - z = 1st \\ 20 - 2y - 3z = 2nd \\ 18 - 3y - 3z = 3rd \end{cases}$

These two last equations involve only $y$ and $z$, and are to be resolved by Direct. 3d, as follows,

$\begin{cases} 2y + 3z - z = 20 - 12 = 8 \\ y + 2z = 8 \end{cases}$

$36 - 3y - 6z = 24 - 2y - 2z$

$12 - y + 4z$

whence $y = \begin{cases} 8 - 2z \quad \text{1st value} \\ 12 - 4z \quad \text{2d value} \end{cases}$

$8 - 2z = 12 - 4z$

$2z = 12 - 8 = 4$

and $z = 2$

$y = (8 - 2z) = 4$

$x = (12 - y - z) = 6.$

This method is general, and will extend to all equations that involve 3 unknown quantities; but there are often often easier and shorter methods to deduce an equation involving one unknown quantity only; which will be best learned by practice.

**Examp. VII.**

Supp. \( \begin{align*} x + y &= a \\ x + z &= b \\ y + z &= c \end{align*} \)

\[ \begin{align*} x &= a - y \\ a - y + z &= b \\ y + z &= c \\ a + 0 + 2z &= b + c \\ 2z &= b + c - a \\ z &= \frac{b + c - a}{2} \\ y &= \frac{c + a - b}{2} \\ x &= \frac{a + b - c}{2}. \end{align*} \]

It is obvious from the 3d and 5th directions, in what manner you are to work if there are four, or more, unknown quantities, and four, or more, equations given. By comparing the given equations, you may always at length discover an equation involving only one unknown quantity; which, if it is a simple equation, may always be resolved by the rules of the last chapter. We may conclude then, that "When there are as many simple equations given as quantities required, these quantities may be discovered by the application of the preceding rules."

If indeed there are more quantities required than equations given, then the question is not limited to determinate quantities; but is capable of an infinite number of solutions. And, if there are more equations given than there are quantities required, it may be impossible to find the quantities that will answer the conditions of the question; because some of these conditions may be inconsistent with others.

**Chap. XI. Containing some general Theorems for the exterminating unknown Quantities in given Equations.**

In the following Theorems, we call those coefficients of the same order that are prefixed to the same unknown quantities in the different equations. Thus in Theor. 2d, \(a, d, g\) are of the same order, being the coefficients of \(x\); also \(b, e, h\) are of the same order, being the coefficients of \(y\); and those are of the same order that affect no unknown quantity.

But those are called opposite coefficients that are taken each from a different equation, and from a different order of coefficients: As, \(a, e, k\) in the first theorem; and \(a, e, k\) in the second; also, \(a, b, f\) and \(d, b, k\), &c.

**Theorem I.**

Suppose that two equations are given, involving two unknown quantities, as,

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

then shall \(y = \frac{af - dc}{ae - db}\),

where the numerator is the difference of the products of the opposite coefficients in the orders in which \(y\) is not found, and the denominator is the difference of the products of the opposite coefficients taken from the orders that involve the two unknown quantities.

For, from the first equation, it is plain, that

\[ \begin{align*} ax &= c - by \\ x &= \frac{c - by}{a} \end{align*} \]

from the 2d, \(dx = f - ey\), and \(x = \frac{f - ey}{d}\),

therefore \(\frac{c - by}{a} = \frac{f - ey}{d}\), and \(cd - db = af - acy\),

whence \(acy - db = af - cd\),

and \(y = \frac{af - cd}{ae - db}\),

after the same manner, \(x = \frac{ct - bf}{ae - db}\).

**Examp. Supp.**

\[ \begin{align*} 5x + 7y &= 100 \\ 3x + 8y &= 80 \end{align*} \]

then \(y = \frac{5 \times 80 - 3 \times 100}{5 \times 8 - 3 \times 7} = \frac{100}{19} = 5 \frac{5}{19}\),

and \(x = \frac{240}{19} = 12 \frac{12}{19}\).

**Theorem II.**

Suppose now that there are three unknown quantities and three equations, then call the unknown quantities \(x, y, z\).

Thus,

\[ \begin{align*} ax + by + cz &= m \\ dx + ey + fz &= n \\ gx + hy + kz &= p \end{align*} \]

Then shall \(z = \frac{ace - ahn + dhm - dbp + gbm - gem}{ask - ahf + dhc - dbk + gbf - gec}\),

where the numerator consists of all the different products that can be made of three opposite coefficients taken from the orders in which \(z\) is not found, and the denominator consists of all the products that can be made of the three opposite coefficients taken from the orders that involve the three unknown quantities.

**Chap. XII. Of Quadratic Equations.**

In the solution of any question, where you have got an equation that involves one unknown quantity, but involves volves at the same time the square of that quantity, and the product of it multiplied by some known quantity, then you have what is called a Quadratic equation; which may be resolved by the following

RULE.

1. "Transport all the terms that involve the unknown quantity to one side, and the known terms to the o- ther side of the equation.

2. "If the square of the unknown quantity is multi- plied by any coefficient, you are to divide all the terms by that coefficient, that the coefficient of the square of the unknown quantity may be unit.

3. "Add to both sides the square of half the coeffici- ent prefixed to the unknown quantity itself, and the side of the equation that involves the unknown quan- tity will then be a compleat square.

4. "Extract the square root from both sides of the equation; which you will find, on one side, always to be the unknown quantity, with half the foresaid coef- ficient subjoined to it; so that, by transposing this half, you may obtain the value of the unknown quan- tity expressed in known terms." Thus,

Suppose \( y^2 + ay = b \)

Add the square of \( \frac{a}{2} \) to both sides

\[ y^2 + ay + \frac{a^2}{4} = b + \frac{a^2}{4} \]

Extract the root, \( y + \frac{a}{2} = \sqrt{b + \frac{a^2}{4}} \)

Transpose \( \frac{a}{2} \), \( y = \sqrt{b + \frac{a^2}{4}} - \frac{a}{2} \).

The square root of any quantity, as \( +aa \), may be \( +a \), or \( -a \); and hence, "All quadratic equations ad- mit of two solutions." In the last example, after finding that \( y^2 + ay + \frac{a^2}{4} = b + \frac{a^2}{4} \), it may be inferred that \( y + \frac{a}{2} \) \( = \pm \sqrt{b + \frac{a^2}{4}} \) or to \( \sqrt{b + \frac{a^2}{4}} \); since \( \sqrt{b + \frac{a^2}{4}} \times \) \( \sqrt{b + \frac{a^2}{4}} \) gives \( b + \frac{a^2}{4} \), as well as \( + \sqrt{b + \frac{a^2}{4}} \times \) \( \sqrt{b + \frac{a^2}{4}} \).

There are therefore two values of \( y \); the one gives \( y = \) \( + \sqrt{b + \frac{a^2}{4}} - \frac{a}{2} \), the other, \( y = -\sqrt{b + \frac{a^2}{4}} - \frac{a}{2} \).

Since the squares of all quantities are positive, it is plain that "the square root of a negative quantity is imaginary, and cannot be assigned." Therefore there are some quadratic equations that cannot have any solution. For example, suppose

\[ y^2 - ay + 3a^2 = 0 \]

then \( y^2 - ay = -3a^2 \)

add \( \frac{a^2}{4} \) to both, \( y^2 - ay + \frac{a^2}{4} = 3a^2 + \frac{a^2}{4} \)

\[ = \frac{11a^2}{4} \]

extract the root, \( y - \frac{a}{2} = \sqrt{\frac{11a^2}{4}} \)

and \( y = \frac{a}{2} \pm \sqrt{\frac{11a^2}{4}} \).

whence the two values of \( y \) must be imaginary or impossi- ble, because the root of \( \frac{11a^2}{4} \) cannot possibly be af- signed.

Suppose that the quadratic equation proposed to be resolved is \( y^2 - ay = b \)

then \( y^2 - ay + \frac{a^2}{4} = b + \frac{a^2}{4} \)

\( y - \frac{a}{2} = \sqrt{b + \frac{a^2}{4}} \)

and \( y = \frac{a}{2} \pm \sqrt{b + \frac{a^2}{4}} \). If the square root of \( b + \frac{a^2}{4} \) cannot be extracted exactly, you must, in order to determine the value of \( y \), nearly approximate to the value of \( \sqrt{b + \frac{a^2}{4}} \), by the rules in chap. 7. The fol- lowing examples will illustrate the rule for quadratic equations.

EXAMP. I. "To find that number which if you multiply by the product shall be equal to the square of the same number having 12 added to it." Call the number \( j \), then

\[ y^2 + 12 = 8y \]

transp. \( y^2 - 8y = -12 \)

add the Sq. of 4, \( y^2 - 8y + 16 = -12 + 16 = 4 \)

extract the R. \( y - 4 = \pm 2 \)

transp. \( y = 4 \pm 2 = 6 \) or 2.

EXAMP. II. "To find a number such, that if you subtract from 105, and multiply the remainder by the number itself, the product shall give 21."

Call it \( y \). Then

\[ 105 - y = 21 \]

that is, \( 105 - y = 21 \)

transp. \( y^2 - 105 = -21 \)

add the sq. of 5, \( y^2 - 105 + 25 = -21 + 25 = 4 \)

extract, \( y - 5 = \pm \sqrt{4} = \pm 2 \)

and \( y = 5 \pm 2 = 7 \) or 3.

EXAMP. III. "A company dining together in an inn, find their bill amounts to 175 shillings; two of them were not allowed to pay, and the rest found that their shares amounted to 10 s. a man more than if all had paid. Qu. How many were in company?"

Suppose their number \( x \); then if all had paid, each man's share would have been \( \frac{175}{x} \): but now the share of each person is \( \frac{175}{x-2} \), seeing \( x-2 \) is the number of those that pay. It is therefore, by the question,

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

and \( 175x - 175x + 350 = 10x^2 - 20x \)

that is, \( 10x^2 - 20x = 350 \)

and \( x^2 - 2x = 35 \)

add 1 \( x^2 - 2x + 1 = 35 + 1 = 36 \)

extr. \( \sqrt{x^2 - 2x + 1} = \pm 6 \)

\( x = 1 \pm 6 = 7 \), or, \(-5\). It is obvious, that the positive value \( y \) gives the solution of the question; the negative value \(-5\) being, in the present case, useless.

Any equation of this form \( y^2 + ay = b \), where the greatest index of the unknown quantity \( y \) is double to the index of \( y \) in the other term, may be reduced to a quadratic \( z^2 + az = b \), by putting \( y = z \), and consequently \( y^2 = z^2 \). And this quadratic resolved as above gives

\[ z = -\frac{a}{2} \pm \sqrt{\frac{b + a^2}{4}}. \]

And seeing \( y = z = -\frac{a}{2} \pm \sqrt{\frac{b + a^2}{4}}, \quad y = \]

\[ = \sqrt{-\frac{a}{2} \pm \sqrt{\frac{b + a^2}{4}}}. \]

**Examp.** "The product of two quantities is \( a \), and the sum of their squares \( b \). Q. The quantities?"

Supp.

\[ \begin{align*} xy &= a \\ x^2 + y^2 &= b \\ \end{align*} \]

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

mult. by \( y^2 \)

\[ \begin{align*} by^2 &= a^2 \\ \end{align*} \]

tranf. \( y^4 - by^2 = a^2 \).

Put now \( y^2 = z \) and conseq. \( y^4 = z^2 \), and it is

\[ \begin{align*} z^2 - bz + \frac{b^2}{4} &= a^2 \\ \end{align*} \]

add \( \frac{b^2}{4} \)

\[ \begin{align*} z^2 - bz + \frac{b^2}{4} &= a^2 \\ \end{align*} \]

ext. \( \sqrt{\cdot} \)

\[ \begin{align*} z - \frac{b}{2} &= \pm \sqrt{\frac{b^2}{4} - a^2} \\ \end{align*} \]

and \( z = \frac{b}{2} \pm \sqrt{\frac{b^2}{4} - a^2} \)

and seeing \( y = \sqrt{z} \),

\[ y = \pm \sqrt{\frac{b}{2} \pm \sqrt{\frac{b^2}{4} - a^2}}. \]

**Chap. XIII. Of Surds.**

If a lesser quantity measures a greater so as to leave no remainder, as \( 2a \) measures \( 10a \), being found in it five times, it is said to be an aliquot part of it, and the greater is said to be a multiple of the lesser. The lesser quantity in this case is the greatest common measure of the two quantities: for as it measures the greatest, so it also measures itself, and no quantity can measure it that is greater than itself.

When a third quantity measures any two proposed quantities, as \( 2a \) measures \( 6a \) and \( 10a \), it is said to be a common measure of these quantities; and if no greater quantity measure them both, it is called their greatest common measure.

Those quantities are said to be commensurable which have any common measure; but if there can be no quantity found that measures them both, they are said to be incommensurable; and if any one quantity be called rational, all others that have any common measure with it, are also called rational: But those that have no common measure with it, are called irrational quantities.

If any two quantities \( a \) and \( b \) have any common measure \( x \), this quantity \( x \) shall also measure their sum or difference \( a-b \). Let \( x \) be found in \( a \) as many times as unit is found in \( m \), so that \( a = mx \); and in \( b \) as many times as unit is found in \( n \), so that \( b = nx \); then shall \( a-b = mx-nx = (m-n)x \); so that \( x \) shall be found in \( a-b \), as often as unit is found in \( m-n \): now since \( m \) and \( n \) are integer numbers, \( m-n \) must be an integer number or unit, and therefore \( x \) must measure \( a-b \).

It is also evident, that if \( x \) measure any number as \( a \), it must measure any multiple of that number. If it be found in \( a \) as many times as unit is found in \( m \) so that \( a = mx \), then it will be found in any multiple of \( a \), as \( na \), as many times as unit is found in \( mn \); for \( na = nmx \).

If two quantities \( a \) and \( b \) are proposed, and \( b \) measure \( a \) by the units that are in \( m \) (that is, be found in \( a \) as many times as unit is found in \( m \)) and there be a remainder \( c \), and if \( x \) be supposed to be a common measure of \( a \) and \( b \), it shall be also a measure of \( c \). For by the supposition \( a = mb + c \), since it contains \( b \) as many times as there are units in \( m \), and there is \( c \) besides of remainder. Therefore \( a = mb + c \). Now \( x \) is supposed to measure \( a \) and \( b \), and therefore it measures \( mb \), and consequently \( a - mb \), which is equal to \( c \).

If \( c \) measure \( b \) by the units in \( n \), and there be a remainder \( d \), so that \( b = nc + d \), and \( b = nc - d \), then shall \( x \) also measure \( d \); because it is supposed to measure \( b \), and it has been proved that it measures \( c \), and consequently \( nc \), and \( b - nc \) which is equal to \( d \). Whence, as, after subtracting \( b \) as often as possible from \( a \), the remainder \( c \) is measured by \( x \); and, after subtracting \( c \) as often as possible from \( b \), the remainder \( d \) is also measured by \( x \); so, for the same reason, if you subtract \( d \) as often as possible from \( c \), the remainder (if there be any) must still be measured by \( x \); and if you proceed, still subtracting every remainder from the preceding remainder, till you find some remainder, which, subtracted from the preceding, leaves no further remainder, but exactly measures it, this last remainder will still be measured by \( x \), any common measure of \( a \) and \( b \).

The last of these remainders, viz., that which exactly measures the preceding remainder, must be a common measure of \( a \) and \( b \): suppose that \( d \) was this last remainder, and that it measured \( c \) by the units in \( r \), then shall \( c = rd \), and we shall have these equations,

\[ \begin{align*} a &= mb + c \\ b &= nc + d \\ c &= rd. \end{align*} \]

Now it is plain that since \( d \) measures \( c \), it must also measure \( nc \), and therefore must measure \( nc + d \), or \( b \). And since it measures \( b \) and \( c \), it must measure \( mb + c \), or \( a \); so that it must be a common measure of \( a \) and \( b \). But further, it must be their greatest common measure; for every common measure of \( a \) and \( b \) must measure \( d \), by the last article; and the greatest number that measures \( d \), is itself, which therefore is the greatest common measure of \( a \) and \( b \).

But if, by continually subtracting every remainder from the preceding remainder, you can never find one that measures that which precedes it exactly, no quantity can be be found that will measure both $a$ and $b$: and therefore they will be incommensurable to each other.

For if there was any common measure of these quantities, as $x$, it would necessarily measure all the remainders $c$, $d$, &c. For it would measure $a - mb$, or $c$, and consequently $b - nc$, or $d$, and so on; now these remainders decrease in such a manner, that they will necessarily become at length less than $x$, or any assignable quantity. For $c$ must be less than $\frac{1}{2}x$; because $c$ is less than $b$, and therefore less than $mb$, and consequently less than $\frac{1}{2}c + \frac{1}{2}mb$, or $\frac{1}{2}a$. In like manner $d$ must be less than $\frac{1}{2}b$; for $d$ is less than $c$, and consequently less than $\frac{1}{2}d + \frac{1}{2}nc$, or $\frac{1}{2}b$. The third remainder, in the same manner, must be less than $\frac{1}{2}e$, which is itself less than $\frac{1}{2}a$: Thus these remainders decrease, so that every one is less than the half of that which preceded it next but one. Now if from any quantity you take away more than its half, and from the remainder more than its half, and proceed in this manner, you will come at a remainder less than any assignable quantity. It appears therefore, that if the remainders $c$, $d$, &c. never end, they will become less than any assignable quantity, as $x$, which therefore cannot possibly measure them, and therefore cannot be a common measure of $a$ and $b$.

In the same way the greatest common measure of two numbers is discovered. Unit is a common measure of all integer numbers, and two numbers are said to be prime to each other when they have no greater common measure than unit; such as 9 and 25. Such always are the least numbers that can be assumed in any given proportion; for if these had any common measure, then the quotients that would arise by dividing them by that common measure would be in the same proportion, and, being less than the numbers themselves, these numbers would not be the least in the same proportion; against the supposition.

The least numbers in any proportion always measure any other numbers that are in the same proportion. Suppose $a$ and $b$ to be the least of all integer numbers in the same proportion, and that $c$ and $d$ are other numbers in that proportion, then will $a$ measure $c$, and $b$ measure $d$.

For if $a$ and $b$ are not aliquot parts of $c$ and $d$, then they must contain the same number of the same kind of parts of $c$ and $d$; and therefore dividing $a$ into parts of $c$, and $b$ into an equal number of like parts of $d$, and calling one of the first $m$, and one of the latter $n$; then as $m$ is to $n$, so will the sum of all the $m$'s be to the sum of all the $n$'s; that is, $m : n :: a : b$, therefore $a$ and $b$ will not be the least in the same proportion; against the supposition. Therefore $a$ and $b$ must be aliquot parts of $c$ and $d$. Hence we see that numbers which are prime to each other are the least in the same proportion; for if there were others in the same proportion less than them, these would measure them by the same number, which therefore would be their common measure against the supposition, for we supposed them to be prime to each other.

If two numbers $a$ and $b$ are prime to one another, and a third number $c$ measures one of them $a$, it will be prime to the other $b$. For if $c$ and $b$ were not prime to each other, they would have a common measure, which, because it would measure $c$, would also measure $a$, which is measured by $c$, therefore $a$ and $b$ would have a common measure, against the supposition.

If two numbers $a$ and $b$ are prime to $c$, then shall their product $ab$ be also prime to $c$: For if you suppose them to have any common measure as $d$, and suppose that $d$ measures $ab$ by the units in $c$, so that $de = ab$, then shall $d : a :: b : e$. But since $d$ measures $c$, and $c$ is supposed to be prime to $a$, it follows that $d$ and $a$ are prime to each other; and therefore $d$ must measure $b$; and yet, since $d$ is supposed to measure $c$ which is prime to $b$, it follows that $d$ is also prime to $b$; that is, $d$ is prime to a number which it measures, which is absurd.

It follows from the last article, that if $a$ and $c$ are prime to each other, then $a^2$ will be prime to $c$: For by supposing that $a$ is equal to $b$, then $ab$ will be equal to $a^2$; and consequently $a^2$ will be prime to $c$. In the same manner $c^2$ will be prime to $a$.

If two numbers $a$ and $b$ are both prime to other two $c$, $d$, then shall the product $ab$ be prime to the product $cd$; for $ab$ will be prime to $c$ and also to $d$, and therefore, by the same article, $cd$ will be prime to $ab$.

From this it follows, that if $a$ and $c$ are prime to each other, then $a^2$ be prime to $c^2$, by supposing, in the last, that $a = b$, and $c = d$. It is also evident that $a^3$ will be prime to $c^3$, and in general any power of $a$ to any power of $c$ whatsoever.

Any two numbers, $a$ and $b$, being given, to find the least numbers that are in the same proportion with them, divide them by their greatest common measure $x$, and the quotients $c$ and $d$ shall be the least numbers in the same proportion with $a$ and $b$.

For if there could be any other numbers in that proportion less than $c$ and $d$, suppose them to be $e$ and $f$, and these being in the same proportion as $a$ and $b$ would measure them: And the number by which they would measure them, would be greater than $x$, because $e$ and $f$ are supposed less than $c$ and $d$, so that $x$ would not be the greatest common measure of $a$ and $b$; against the supposition.

Let it be required to find the least number that any two given numbers, as $a$ and $b$, can measure. First, "If they are prime to each other, then their product $ab$ is the least number which they can both measure."

For if they could measure a less number than $ab$ as $c$, suppose that $c$ is equal to $ma$, and to $nb$; and since $c$ is less than $ab$, therefore $ma$ will be less than $ab$, and $m$ less than $b$; and $nb$ being less than $ab$, it follows that $n$ must be less than $a$; but since $ma = nb$, and consequently $a : b :: n : m$, and $a$ and $b$ are prime to each other, it would follow that $a$ would measure $n$, and $b$ measure $m$, that is, a greater number would measure a less, which is absurd.

But if the numbers $a$ and $b$ are not prime to each other, and their greatest common measure is $x$, which measures $a$ by the units in $m$, and measures $b$ by the units in $n$, so that $a = mx$, and $b = nx$, then shall $an$ (which is equal to $bm$, because $a : b :: mx : nx :: m : n$, and therefore $an = bm$) be the least number that $a$ and $b$ can both measure. For if they could measure any number $c$ less than than \( na \), so that \( a = kb \), then \( a : b : : m : n : : k : l \); and because \( x \) is supposed to be the greatest common measure of \( a \) and \( b \), it follows that \( m \) and \( n \) are the least of all numbers in the same proportion; and therefore \( m \) measures \( k \), and \( n \) measures \( l \). But as \( c \) is supposed to be less than \( na \), that is, \( la \) less than \( na \), therefore \( l \) is less than \( n \), so that a greater would measure a lesser, which is absurd. Therefore \( a \) and \( b \) cannot measure any number less than \( an \); which they both measure, because \( na = mb \).

It follows from this reasoning, that if \( a \) and \( b \) measure any quantity \( c \), the least quantity \( na \), which is measured by \( a \) and \( b \), will also measure \( c \). For if you suppose, as before, that \( c = la \), you will find, that \( n \) must measure \( l \), and \( na \) must measure \( la \) or \( c \).

Let \( a \) express any integer number, and \( \frac{m}{n} \) any fraction reduced to its lowest terms, so that \( m \) and \( n \) may be prime to each other, and consequently \( an + m \) also prime to \( n \), it will follow that \( an + m \) will be prime to \( n \), and consequently \( \frac{an + m}{n^2} \) will be a fraction in its least terms, and can never be equal to an integer number. Therefore the square of the mixt number \( a + \frac{m}{n} \) is still a mixt number, and never an integer. In the same manner, the cube, biquadrate, or any power of a mixt number, is still a mixt number, and never an integer. It follows from this, that the square root of an integer must be an integer or an incommensurable. Suppose that the integer proposed is \( B \), and that the square root of it is less than \( a + 1 \), but greater than \( a \), then it must be an incommensurable; for if it is a commensurable, let it be \( a + \frac{m}{n} \) where \( \frac{m}{n} \) represents any fraction reduced to its least terms; it would follow, that \( a + \frac{m}{n} \) squared would give an integer number \( B \), the contrary of which we have demonstrated.

It follows from the last article, that the square roots of all numbers but of \( 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, \ldots \) (which are the squares of the integer numbers \( 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, \ldots \)) are incommensurables: After the same manner, the cube roots of all numbers but of the cubes of \( 1, 2, 3, 4, 5, 6, 7, 8, 9, \ldots \) are incommensurables; and quantities that are to one another in the proportion of such numbers must also have their square roots or cube roots incommensurable.

The roots of such numbers being incommensurable are expressed therefore by placing the proper radical sign over them; thus, \( \sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{10}, \ldots \) express numbers incommensurable with unit. These numbers, though they are incommensurable themselves with unit, are commensurable in power with it, because their powers are integers, that is, multiples of unit. They may also be commensurable sometimes with one another, as the \( \sqrt{8} \), and the \( \sqrt{2} \), because they are to one another as 2 to 1: And when they have a common measure, as \( \sqrt{2} \) is the common measure of both, then their ratio is reduced to an expression in the least terms, as that of commensurable quantities, by dividing them by their greatest common measure. This common measure is found as in commensurable quantities, only the root of the common measure is to be made their common divisor. Thus, \( \frac{\sqrt{12}}{\sqrt{3}} = \sqrt{4} = 2 \), and \( \frac{\sqrt{18a}}{\sqrt{2}} = 3\sqrt{a} \).

A rational quantity may be reduced to the form of any given surd, by raising the quantity to the power that is denominated by the name of the surd, and then setting the radical sign over it thus, \( a = \sqrt{a^2} = \sqrt{a^3} = \sqrt{a^4} = \sqrt{a^n} = \sqrt{a^n} \), and \( 4 = \sqrt{16} = \sqrt{64} = \sqrt{256} = \sqrt{1024} = \sqrt{4^n} \).

As surds may be considered as powers with fractional exponents, "they are reduced to others of the same value that shall have the same radical sign, by reducing these fractional exponents to fractions having the same value and a common denominator." Thus, \( \sqrt{a} = \frac{a}{\sqrt{m}} \), \( \sqrt{a} = \frac{m}{\sqrt{n}} \), \( \sqrt{a} = \frac{n}{\sqrt{m}} \), and therefore \( \sqrt{a} \) and \( \sqrt{a} \), reduced to the same radical sign, become \( \sqrt{a^m} \) and \( \sqrt{a^n} \). If you are to reduce \( \sqrt{3} \) and \( \sqrt{2} \) to the same denominator, consider \( \sqrt{3} \) as equal to \( 3^{\frac{1}{2}} \), the \( \sqrt{2} \) as equal to \( 2^{\frac{1}{2}} \), whose indices reduced to a common denominator, you have \( 3^{\frac{1}{2}} = 3^{\frac{1}{2}} \) and \( 2^{\frac{1}{2}} = 2^{\frac{1}{2}} \), and consequently \( \sqrt{3} = \sqrt{3^{\frac{1}{2}}} = \sqrt{27} \), and \( \sqrt{2} = \sqrt{2^{\frac{1}{2}}} = \sqrt{4} \); so that the proposed surds \( \sqrt{3} \) and \( \sqrt{2} \) are reduced to other equal surds \( \sqrt{27} \) and \( \sqrt{4} \), having a common radical sign.

Surds of the same rational quantity are multiplied by adding their exponents, and divided by subtracting them; thus \( \sqrt{a} \times \sqrt{a} = a^{\frac{1}{2}} \times a^{\frac{1}{2}} = a^{\frac{1}{2}} = a^{\frac{1}{2}} \); and \( \frac{\sqrt{a}}{\sqrt{a}} = \frac{a^{\frac{1}{2}}}{a^{\frac{1}{2}}} = a^{\frac{1}{2}} = a^{\frac{1}{2}} \); \( \frac{m}{\sqrt{a}} \times \frac{n}{\sqrt{a}} = \frac{m+n}{mn} \); \( \frac{n}{\sqrt{a}} = \frac{n-m}{mn} \); \( \sqrt{2} \times \sqrt{2} = \sqrt{2^2} = \sqrt{4} \); \( \sqrt{32} = \sqrt{2^5} = \sqrt{2} \).

If the surds are of different rational quantities, as \( \sqrt{a^2} \) and \( \sqrt{b^2} \), and have the same sign, "multiply these rational quantities into one another, or divide them..." them by one another, and set the common radical sign over their product or quotient." Thus, \( \sqrt[n]{a^m} \times \sqrt[n]{b^m} = \sqrt[n]{a^m b^m} \); \( \sqrt{2 \times 5} = \sqrt{10} \); \( \frac{\sqrt{a^4}}{\sqrt{b^3 a}} = \frac{\sqrt{a^4}}{\sqrt{b^3 a}} = \frac{\sqrt{a^4}}{\sqrt{b^3 a}} \).

If the surds have not the same radical sign, "reduce them to such as shall have the same radical sign, and proceed as before," \( \sqrt{a} \times \sqrt{b} = \sqrt{a b} \); \( \frac{\sqrt{a}}{\sqrt{x}} = \frac{\sqrt{a}}{\sqrt{x}} \); \( \sqrt{2 \times 4} = \sqrt{8} \); \( \sqrt{2^3 \times 4^2} = \sqrt{64} \); \( \sqrt{8 \times 16} = \sqrt{128} \). If the surds have any rational coefficients, their product or quotient must be prefixed; thus, \( 2\sqrt{5} \times 5\sqrt{6} = 10\sqrt{18} \).

The powers of surds are found as the powers of other quantities, "by multiplying their exponents by the index of the power required;" thus the square of \( \sqrt{2} \) is \( \sqrt{2^2} = \sqrt{4} \); the cube of \( \sqrt{5} = \sqrt{5^3} = \sqrt{125} \).

Or you need only, in involving surds, "raise the quantity under the radical sign to the power required, continuing the same radical sign; unless the index of that power is equal to the name of the surd, or a multiple of it, and in that case the power of the surd becomes rational." Evolution is performed "by dividing the fraction which is the exponent of the surd by the name of the root required." Thus the square root of \( \sqrt{a^4} \) is \( \sqrt{a^2} \) or \( \sqrt{a^4} \).

The surd \( \sqrt{a^m x} = a \sqrt{x} \); and in like manner, if a power of any quantity of the same name with the surd divides the quantity under the radical sign without a remainder, as here \( a^m \) divides \( a^m x \), and \( 2^5 \) the square of \( 5 \) divides \( 75 \) the quantity under the sign in \( \sqrt{75} \) without a remainder, then place the root of that power rationally before the sign, and the quotient under the sign, and thus the surd will be reduced to a more simple expression.

Thus, \( \sqrt{75} = \sqrt{3 \times 25} = \sqrt{3} \times \sqrt{25} = 5\sqrt{3} \); \( \sqrt{48} = \sqrt{3 \times 16} = 4\sqrt{3} \); \( \sqrt{81} = \sqrt{9 \times 9} = 9\sqrt{3} \); \( \sqrt{27 \times 3} = 3\sqrt{3} \).

When surds by the last article are reduced to their least expressions, if they have the same irrational part, they are added or subtracted, "by adding or subtracting their rational coefficients, and prefixing the sum or difference to the common irrational part." Thus, \( \sqrt{75} + \sqrt{48} = 5\sqrt{3} + 4\sqrt{3} = 9\sqrt{3} \); \( \sqrt{31} + \sqrt{24} = 3\sqrt{3} + 2\sqrt{3} = 5\sqrt{3} \).

Compound surds are such as consist of two or more joined together. The simple surds are commensurable in power, and by being multiplied into themselves give at length rational quantities; yet compound surds multiplied into themselves commonly give still irrational products. But when any compound surd is proposed, there is another compound surd which multiplied into itself gives a rational product. Thus, \( \sqrt{a} + \sqrt{b} \) multiplied by \( \sqrt{a} - \sqrt{b} \) gives \( a - b \), and "the investigation of that surd which multiplied into the proposed surd will give a rational product," is made easy by the following theorems.

**THEOREM I.**

Generally, if you multiply \( a^n - b^n \) by \( a^n - m + a^{n-2} b^m + a^{n-3} b^m + a^{n-4} b^m \), &c., continued till the terms be in number equal to \( n \), the product shall be \( a^n - b^n \); for \( a^n - m + a^{n-2} b^m + a^{n-3} b^m + a^{n-4} b^m \), &c., \( b^n \).

**THEOREM II.**

\( a^n - m + a^{n-2} b^m + a^{n-3} b^m + a^{n-4} b^m \), &c., multiplied by \( a^n + b^n \) gives \( a^n + b^n \), which is demonstrated as the other. Here the sign of \( b^n \) is positive, when \( n \) is an odd number.

When any binomial surd is proposed, "suppose the index of each number equal to \( m \), and let \( n \) be the least integer number that is measured by \( m \), then shall \( a^n - m + a^{n-2} b^m + a^{n-3} b^m \), &c., give a compound surd, which multiplied into the proposed surd \( a^n + b^n \) will give a rational product." Thus to find the surd which multiplied by \( \sqrt{a} - \sqrt{b} \), will give a rational quantity. Here \( m = 2 \), and the least number which is measured by \( \frac{1}{2} \) is unit; let \( n = 1 \), then shall \( a^n - m + a^{n-2} b^m \), &c., \( = a^{1-\frac{1}{2}} + a^{1-\frac{3}{2}} b^{\frac{1}{2}} + a^{1-\frac{5}{2}} b^{\frac{3}{2}} = a^{\frac{1}{2}} + a^{\frac{3}{2}} b^{\frac{1}{2}} + a^{\frac{5}{2}} b^{\frac{3}{2}} \), which multiplied by \( \sqrt{a} - \sqrt{b} \) gives \( a - b \).

To find the surd which multiplied by \( \sqrt{a} + \sqrt{b} = a^{\frac{1}{2}} + b^{\frac{1}{2}} \), gives a rational product. Here \( m = 2 \) and \( n = 3 \), and \( a^n - m + a^{n-2} b^m + a^{n-3} b^m \), &c., \( = a^{\frac{3}{2}} - a^{\frac{5}{2}} b^{\frac{1}{2}} + a^{\frac{7}{2}} b^{\frac{3}{2}} - a^{\frac{9}{2}} b^{\frac{5}{2}} + a^{\frac{11}{2}} b^{\frac{7}{2}} - a^{\frac{13}{2}} b^{\frac{9}{2}} + \cdots \).

**THEOREM III.**

Let \( a^n + b^n \) be multiplied by \( a^n - m + a^{n-2} b^m + a^{n-3} b^m \), and the product shall give \( a^n + b^n \); therefore \( n \) must be taken the least integer that shall give \( \frac{n}{m} \) also an integer.

Dem. Dem. \(a^n - m = an - 2mb^1 + an - 3mb^2 + \ldots\)

\[a^n - m = an - mb^1 + an - 2mb^2 + \ldots\]

The sign of \(b^m\) is positive only when \(m\) is an odd number, and the binomial proposed is \(an + b\).

If any binomial surd is proposed whose two numbers have different indices, let these be \(m\) and \(l\), and take \(n\) equal to the least integer number that is measured by \(m\) and by \(l\); and \(an - m = an - mb^1 + an - 3mb^2 + \ldots\)

\[a^n - m = an - mb^1 + an - 2mb^2 + \ldots\]

shall give a compound surd, which multiplied by the proposed \(an + b\) shall give a rational product. Thus \(\sqrt{a} - \sqrt{b}\) being given, suppose \(m = \frac{1}{2}, l = \frac{1}{2}\), and \(n = 3\), therefore you have \(n = 3\), and

\[a^n - m + 2an - mb^1 + an - 3mb^2 + \ldots = a^3 - \frac{1}{4} + a^3 - \frac{1}{2}b^1 + a^3 - \frac{1}{2}b^2 + \ldots\]

\[+ a^3 - \frac{1}{2}b^3 + a^3 - \frac{1}{2}b^4 + \ldots = a^3 - \frac{1}{4} + a^3 - \frac{1}{2}b^1 + a^3 - \frac{1}{2}b^2 + \ldots\]

\[+ ab + \sqrt{a} \times \sqrt{b} + \sqrt{b}^2 = a^3 - \frac{1}{4} + a^3 - \frac{1}{2}b^1 + a^3 - \frac{1}{2}b^2 + \ldots\]

which multiplied by the \(\sqrt{a} - \sqrt{b}\) gives \(a^n - m = a^3 - b^2\).

By these theorems any binomial surd whatsoever being given, you may find a surd which multiplied by it shall give a rational product.

Suppose that a binomial surd was to be divided by another, as \(\sqrt{20} + \sqrt{12}\), by \(\sqrt{5} - \sqrt{3}\), the quotient may be expressed by \(\frac{\sqrt{20} + \sqrt{12}}{\sqrt{5} - \sqrt{3}}\). But it may be expressed in a more simple form by multiplying both numerator and denominator by that surd which, multiplied into the denominator, gives a rational product: Thus

\[\frac{\sqrt{20} + \sqrt{12}}{\sqrt{5} - \sqrt{3}} = \frac{\sqrt{20} + \sqrt{12}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} = \frac{16 + 2\sqrt{60}}{8 + 2\sqrt{15}}.\]

In general, when any quantity is divided by a binomial surd, as \(a^n - b\), where \(m\) and \(l\) represent any fractions whatsoever, take \(n\) the least integer number that is measured by \(m\) and \(l\), multiply both numerator and denominator by \(a^n - m + an - 2mb^1 + an - 3mb^2 + \ldots\), and the denominator of the product will become rational, and equal to \(a^n - b\); then divide all the members of the numerator by this rational quantity, and the quote arising will be that of the proposed quantity divided by the binomial surd, expressed in its least terms. Thus,

\[\frac{3}{\sqrt{3} - \sqrt{2}} = \frac{3\sqrt{5} + 3\sqrt{2}}{3} = \sqrt{5} + \sqrt{2};\]

\[\frac{\sqrt{6}}{\sqrt{7} - \sqrt{3}} = \frac{\sqrt{42} + \sqrt{18}}{4}.\]

When the square root of a surd is required, it may be found nearly by extracting the root of a rational quantity that approximates to its value. Thus to find the square root of \(3 + 2\sqrt{2}\), we first calculate \(\sqrt{2} = 1, 41421\); and therefore \(3 + 2\sqrt{2} = 5, 82842\), whose root is found to be nearly \(2, 41421\): So that \(\sqrt{3 + 2\sqrt{2}}\) is nearly \(2, 41421\). But sometimes we may be able to express the roots of surds exactly by other surds; as in this example the square root of \(3 + 2\sqrt{2}\) is \(1 + \sqrt{2}\), for \(1 + \sqrt{2} \times 1 + \sqrt{2} = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}\).

In order to know when and how this may be found, let us suppose that \(x + y\) is a binomial surd, whose square will be \(x^2 + y^2 + 2xy\): If \(x\) and \(y\) are quadratic surds, then \(x^2 + y^2\) will be rational, and \(2xy\) irrational; so that \(2xy\) shall always be less than \(x^2 + y^2\), because the difference is \(x^2 + y^2 - 2xy = x^2 - y^2\) which is always positive. Suppose that a proposed surd consisting of a rational part \(A\), and an irrational part \(B\), coincides with this, then \(x^2 + y^2 = A\) and \(xy = \frac{B}{2}\): Therefore by what was said of equations, Chap. 12th,

\[y^2 = A - x^2 = \frac{B^2}{4x^2},\] and therefore,

\[Ax^2 - x^4 = \frac{B^2}{4} \quad \text{and} \quad x^4 - Ax^2 + \frac{B^2}{4} = 0;\]

from whence we have \(x^2 = \frac{A + \sqrt{A^2 - B^2}}{2}\) and \(y^2 = \frac{A - \sqrt{A^2 - B^2}}{2}\).

Therefore when a quantity partly rational, partly irrational, is proposed to have its root extracted, call the rational part \(A\), the irrational \(B\), and the square of the greatest member of the root shall be \(A + \sqrt{A^2 - B^2}\), and the square of the lesser part shall be \(A - \sqrt{A^2 - B^2}\). And as often as the square root of \(A^2 - B^2\) can be extracted, the square root of the proposed binomial surd may be expressed itself as a binomial surd. For example, if \(3 + 2\sqrt{2}\) is proposed, then \(A = 3, B = 2\sqrt{2}\) and \(A^2 - B^2 = 9 - 8 = 1\). Therefore \(x^2 = \frac{A + \sqrt{A^2 - B^2}}{2} = 2\), and \(y^2 = \frac{A - \sqrt{A^2 - B^2}}{2} = 1\). Therefore \(x + y = 1 + \sqrt{2}\).

To find the square root of \(-1 + \sqrt{-8}\), suppose \(A = -1, B = \sqrt{-8}\), so that \(A^2 - B^2 = 9\) and \(\frac{A + \sqrt{A^2 - B^2}}{2} = \frac{-1 + 3}{2} = 1\), and \(\frac{A - \sqrt{A^2 - B^2}}{2} = \frac{-1 - 3}{2} = -2\), therefore the root required is \(1 + \sqrt{-2}\). But though \( x \) and \( y \) are not quadratic surds or roots of integers, if they are the roots of like surds, as if they are equal to \( \sqrt{m}z \) and \( \sqrt{n}z \), where \( m \) and \( n \) are integers, then \( A = m + n \times z \) and \( B = \sqrt{mnz} \);

\[ A^2 - B^2 = (m + n)^2 \times z \quad \text{and} \quad x^2 = \frac{A + \sqrt{A^2 - B^2}}{2} = \frac{m + n \times z + mn}{2} \times z = \frac{m + n}{z}, \quad y^2 = \frac{A - \sqrt{A^2 - B^2}}{2} = \frac{m + n}{z}, \quad \text{and} \quad x + y = \sqrt{\frac{m + n}{z}} + \sqrt{\frac{m + n}{z}}. \]

The part \( A \) here easily distinguishes itself from \( B \) by its being greater.

If \( x \) and \( y \) are equal to \( \sqrt{m}z \) and \( \sqrt{n}z \), then \( x^2 + 2xy + y^2 = m + n + 1 + 2\sqrt{mnz} \). So that if \( z \) or \( t \) be not multiples of one of the other, or of some number that measures them both by a square number, then will \( A \) itself be a binomial.

Let \( x + y + z \) express any trinomial furd, its square \( x^2 + y^2 + z^2 + 2xy + 2xz + 2yz \) may be supposed equal to \( A + B \) as before. But rather multiply any two radicals as \( 2xy \) by \( 2xz \), and divide by the third \( 2yz \) which gives the quotient \( 2x \) rational, and double the square of the furd \( x \) required. The same rule serves when there are four quantities \( x^2 + y^2 + z^2 + r^2 + 2xy + 2xz + 2yz + 2rz + 2zx \), multiply \( 2xy \) by \( 2zx \), and the product \( 4x^2y^2 \) divided by \( 2yz \) gives another rational quotient, half the square of \( 2x \). In like manner \( 2xy \times 2yz = 4y^2xz \), which divided by \( 2xz \) another member gives \( 2y^2 \), a rational quote, the half of the square of \( 2y \). In the same manner \( z \) and \( r \) may be found; and their sum \( x + y + z + r \), the square root of the septinomial \( x^2 + y^2 + z^2 + r^2 + 2xy + 2xz + 2yz + 2zx + 2ry + 2xz \), discovered.

For example, to find the square root of \( 10 + \sqrt{40} + \sqrt{60} \); I try \( \frac{\sqrt{40} \times \sqrt{60}}{\sqrt{60}} \), which I find to be \( \frac{4}{16} = 4 \), the half of the square root of the double of which, viz., \( \frac{1}{2} \times \sqrt{8} = \sqrt{2} \), is one member of the square root required; next \( \frac{\sqrt{40} \times \sqrt{60}}{\sqrt{40}} = 6 \), the half of the square root of the double of which is \( \sqrt{3} \) another member of the root required; lastly, \( \frac{\sqrt{40} \times \sqrt{60}}{\sqrt{24}} = 10 \), which gives \( \sqrt{5} \) for the third member of the root required; from which we conclude, that the square root of \( 10 + \sqrt{40} + \sqrt{60} \), is \( \sqrt{2} + \sqrt{3} + \sqrt{5} \); and trying, you find it succeeds, since multiplied by itself it gives the proposed quadrinomial.

For extracting the higher roots of a binomial, whose two members being squared are commensurable numbers, there is the following.

**Rule.** "Let the quantity be \( A + B \), whereof \( A \) is the greater part, and \( c \) the exponent of the root required. Seek the least number \( n \) whose power \( n^c \) is divisible by \( AA - BB \), the quotient being \( Q \).

Compute \( \sqrt{A + B} \times \sqrt{Q} \) in the nearest integer number, which suppose to be \( r \). Divide \( A + B \) by its greatest rational divisor, and let the quotient be \( s \), and let \( \frac{r + \frac{n}{2}}{z} \) in the nearest integer number, be

" \( r \), so shall the root required be \( \frac{r + \frac{n}{2}}{\sqrt{Q}} \), if the \( c \) root of \( A + B \) can be extracted.

**Exampl. I.** Thus to find the cube root of \( \sqrt{968} + 25 \), we have \( A^2 - B^2 = 343 \), whose divisors are \( 7, 7, 7 \), whence \( n = 7 \), and \( Q = 1 \). Further, \( A + B \times \sqrt{Q} \), that is, \( \sqrt{968} + 25 \) is a little more than \( 56 \), whose nearest cube root is \( 4 \). Therefore \( r = 4 \). Again, dividing \( \sqrt{968} \) by its greatest rational divisor, we have \( A \sqrt{Q} = 22 \sqrt{2} \), and the radical part \( \sqrt{2} = s \); and \( r + \frac{n}{2} \); or \( \frac{5}{2\sqrt{2}} \), in the nearest integers, is \( 2 = t \). And lastly, \( ts = 2 \sqrt{2} \), \( \sqrt{t^2s^2} = n = 1 \), and \( \sqrt{Q} = \sqrt{1} = 1 \). Whence \( 2\sqrt{2} + 1 \) is the root, whose cube, upon trial, I find to be \( \sqrt{968} + 25 \).

**Exampl. II.** To find the cube root of \( 68 - \sqrt{4374} \), we have \( A^2 - B^2 = 250 \), whose divisors are \( 5, 5, 2 \).

Thence \( n = 5 \times 2 = 10 \), and \( Q = 4 \), and \( \sqrt{A + B} \times \sqrt{Q} \), or \( \sqrt{68} + \sqrt{4374} \times 2 \) is nearly \( 7 = r \); again \( A \sqrt{Q} \), or \( 68 \times \sqrt{4} = 136 \times \sqrt{1} \), that is, \( s = 1 \), and \( r + \frac{n}{2} \), or \( \frac{7 + \frac{5}{2}}{2} \), is nearly \( 4 = t \). Therefore \( ts = 4 \), \( \sqrt{t^2s^2} = n = \sqrt{6} \), and \( \sqrt{Q} = \sqrt{4} = \sqrt{2} \), whence the root to be tried is \( \frac{4}{\sqrt{2}} \).

**Chap. XIV. Of the Genesis and Resolution of Equations in general; and the number of Roots an Equation of any Degree may have.**

After the same manner, as the higher powers are produced by the multiplication of the lower powers of the same root, equations of superior orders are generated by the multiplication of equations of inferior orders involving the same unknown quantity. And "an equation of any dimension may be considered as produced by the multiplication of as many simple equations as it has dimensions, or of any other equations whatever, if the sum of their dimensions is equal to the dimension of that equation." Thus, any cubic equation may be conceived as generated by the multiplication of three simple equations, or of one quadratic and one simple equation. A biquadratic is generated by the multiplication of four simple equations, or of two quadratic equations, or, lastly, of one cubic and one simple equation.

If the equations which you suppose multiplied by one another are the same, then the equation generated will be nothing else but some power of those equations, and the operation is merely involution; of which we have treated already: and, when any such equation is given, the simple equation by whose multiplication it is produced is found by evolution, or the extraction of a root.

But when the equations that are supposed to be multiplied tiplied by each other are different, then other equations than powers are generated; which to resolve into the simple equations whence they are generated is a different operation from involution, and is what is called, the resolution of equations.

But as evolution is performed by observing and tracing back the steps of involution; so to discover the rules for the resolution of equations, we must carefully observe their generation.

Suppose the unknown quantity to be $x$, and its values in any simple equations to be $a$, $b$, $c$, $d$, &c., then those simple equations, by bringing all the terms to one side, become $x-a=0$, $x-b=0$, $x-c=0$, &c. And, the product of any two of these, as $x-a \times x-b=0$ will give a quadratic equation, or an equation of two dimensions. The product of any three of them, as $x-a \times x-b \times x-c=0$, will give a cubic equation, or one of three dimensions. The product of any four of them will give a biquadratic equation, or one of four dimensions, as $x-a \times x-b \times x-c \times x-d=0$. And, in general, "in the equation produced, the highest dimension of the unknown quantity will be equal to the number of simple equations that are multiplied by each other."

When any equation, equivalent to this biquadratic $x-a \times x-b \times x-c \times x-d=0$, is proposed to be resolved, the whole difficulty consists in finding the simple equations $x-a=0$, $x-b=0$, $x-c=0$, $x-d=0$, by whose multiplication it is produced; for each of these simple equations gives one of the values of $x$, and one solution of the proposed equation. For, if any of the values of $x$, deduced from those simple equations, be substituted in the proposed equation in place of $x$, then all the terms of that equation will vanish, and the whole be found equal to nothing. Because, when it is supposed that $x=a$, or $x=b$, or $x=c$, or $x=d$, then the product $x-a \times x-b \times x-c \times x-d$ does vanish, because one of the factors is equal to nothing. There are therefore four suppositions that give $x-a \times x-b \times x-c \times x-d=0$ according to the proposed equation; that is, there are four roots of the proposed equation. And after the same manner, "any other equation admits of as many solutions as there are simple equations multiplied by one another that produce it, or, as many as there are units in the highest dimension of the unknown quantity in the proposed equation."

But as there are no other quantities whatsoever besides these four ($a$, $b$, $c$, $d$) that substituted in the product $x-a \times x-b \times x-c \times x-d$, in the place of $x$, will make the product vanish; therefore the equation $x-a \times x-b \times x-c \times x-d=0$, cannot possibly have more than these four roots, and cannot admit of more solutions than four. If you substitute in that product a quantity neither equal to $a$, nor $b$, nor $c$, nor $d$, which suppose $e$, then since neither $e-a$, $e-b$, $e-c$, nor $e-d$ is equal to nothing; their product $e-a \times e-b \times e-c \times e-d$ cannot be equal to nothing, but must be some real product; and therefore there is no supposition beside one of the foreaid four that gives a just value of $x$ according to the proposed equation. So that it can have no more than these four roots. And after the same manner it appears, that "no equation can have more roots than it contains dimensions of the unknown quantity."

To make all this still plainer by an example, in numbers; suppose the equation to be resolved, to be $x^4-10x^3+35x^2-50x+24=0$, and that you discover that this equation is the same with the product of $x-1 \times x-2 \times x-3 \times x-4$, then you certainly infer that the four values of $x$ are $1$, $2$, $3$, $4$; seeing any of these numbers placed for $x$ makes that product, and consequently $x^4-10x^3+35x^2-50x+24$, equal to nothing, according to the proposed equation. And it is certain that there can be no other values of $x$ besides these four: since when you substitute any other number for $x$ in those factors $x-1$, $x-2$, $x-3$, $x-4$, none of the factors vanish; and therefore their product cannot be equal to nothing, according to the equation.

It may be useful sometimes to consider equations as generated from others of an inferior sort besides simple ones. Thus a cubic equation may be conceived as generated from the quadratic $x^2-px+q=0$, and the simple equation $x-a=0$, multiplied by each other; whose product

$$x^3 - px^2 + qx - aq = 0$$

may express any cubic equation whose roots are the quantity ($a$) the value of $x$ in the simple equation, and the two roots of the quadratic equation, viz. $\frac{p+\sqrt{p^2-4q}}{2}$ and $\frac{p-\sqrt{p^2-4q}}{2}$; as appears from Chap. 12. And, according as these roots are real or impossible, two of the roots of the cubic equation are real or impossible.

In the doctrine of involution, we shewed, that "the square of any quantity, positive or negative, is always positive;" and therefore "the square root of a negative is impossible or imaginary." For example, the $\sqrt{-a^2}$ is either $+a$, or $-a$; but $\sqrt{-a^2}$ can neither be $+a$ nor $-a$, but must be imaginary. Hence is understood, that "a quadratic equation may have no impossible expression in its coefficients; and yet, when it is resolved into the simple equations that produce it, they may involve impossible expressions." Thus, the quadratic equation $x^2+a^2=0$ has no impossible coefficient; but the simple equations from which it is produced, viz. $x+\sqrt{-a^2}=0$, and $x-\sqrt{-a^2}=0$, both involve an imaginary quantity; as the square $-a^2$ is a real quantity, but its square root is imaginary. After the same manner, a biquadratic equation, when resolved, may give four simple equations, each of which may give an impossible value for the root; and the same may be said of any equation that can be produced from quadratic equations only, that is, whose dimensions are of the even numbers.

But, "a cubic equation (which cannot be generated from quadratic equations only, but requires one simple equation besides to produce it) if none of its coefficients are impossible, will have, at least, one real root," the same with the root of the simple equation whence it is produced. The square of an impossible quantity may be real, as the square of \( \sqrt{-a^2} \) is \(-a^2\); but "the cube of an impossible quantity is still impossible," as it still involves the square root of a negative; as, \( \sqrt{-a^2} \times \sqrt{-a^2} \times \sqrt{-a^2} = \sqrt{-a^6} = a^3 \sqrt{-1} \), is plainly imaginary. From which it appears, that though two simple equations involving impossible expressions, multiplied by one another, may give a product where no impossible expression may appear; yet "if three such simple equations be multiplied by each other, the impossible expression will not disappear in their product."

And hence it is plain, that though a quadratic equation whose coefficients are all real may have its two roots impossible, yet "a cubic equation whose coefficients are real cannot have all its three roots impossible."

In general, it appears, that the impossible expressions cannot disappear in the equation produced, but when their number is even; that there are never in any equations, whose coefficients are real quantities, single impossible roots, or an odd number of impossible roots, but that the roots become impossible in pairs, and that an equation of an odd number of dimensions has always one real root.

"The roots of equations are either positive or negative, according as the roots of the simple equations whence they are produced are positive or negative." If you suppose \( x = -a, x = -b, x = -c, x = -d, \) &c., then shall \( x + a = 0, x + b = 0, x + c = 0, x + d = 0 \); and the equation \( x + a \times x + b \times x + c \times x + d = 0 \) will have its roots, \(-a, -b, -c, -d, \) &c. negative.

But to know when the roots of equations are positive, and when negative, and how many there are of each kind, shall be explained in the next chapter.

**CHAP. XV. Of the Signs and Coefficients of Equations.**

When any number of simple equations are multiplied by each other, it is obvious that the highest dimension of the unknown quantity in their product is equal to the number of those simple equations; and the term involving the highest dimension is called the first term of the equation generated by this multiplication. The term involving the next dimension of the unknown quantity, less than the greatest by unit, is called the second term of the equation; the term involving the next dimension of the unknown quantity, which is less than the greatest by two, the third term of the equation, &c.; and that term which involves no dimension of the unknown quantity, but is some known quantity, is called the last term of the equation.

"The number of terms is always greater than the highest dimension of the unknown quantity by unit." And when any term is wanting, an asterisk is marked in its place. The signs and coefficients of equations will be understood by considering the following table, where the simple equations \( x = a, x = b, \) &c., are multiplied by one another, and produce successively the higher equations.

From the inspection of these equations it is plain, that the coefficient of the first term is unit.

The coefficient of the second term is the sum of all the roots (a, b, c, d, e,) having their signs changed.

The coefficient of the third term is the sum of all the products that can be made by multiplying any two of the roots (a, b, c, d, e,) by one another.

The coefficient of the fourth term is the sum of all the products that can be made by multiplying into one another any three of the roots, with their signs changed. And after the same manner all the other coefficients are formed.

The last term is always the product of all the roots having their signs changed, multiplied by one another.

Although in the table such simple equations only are multiplied by one another as have positive roots, it is easy to see, that "the coefficients will be formed according to the same rule when any of the simple equations have negative roots." And, in general, if \( x^3 - px^2 + qx - r = 0 \) represent any cubic equation, then shall \( p \) be the sum of the roots; \( q \) the sum of the products made by multiplying any two of them; \( r \) the product of all the three: and, if \(-p, +q, -r, +s, -t, +u, \) &c., be the coefficients of the 2d, 3d, 4th, 5th, 6th, 7th, &c. terms of any equation, then shall \( p \) be the sum of all the roots, \( q \) the sum of the products of any two, \( r \) the sum of the products of any three, \( s \) the sum of the products of any four, \( t \) the sum of the products of any five, \( u \) the sum of the products of any six, &c.

When When therefore any equation is proposed to be resolved, it is easy to find the sum of the roots, (for it is equal to the coefficient of the second term having its sign changed); or to find the sum of the products that can be made by multiplying any determinate number of them.

But it is also easy "to find the sum of the squares, or "of any powers, of the roots.

The sum of the squares is always \( p^2 - 2q \). For calling the sum of the squares \( B \), since the sum of the roots is \( p \); and "the square of the sum of any quantities is "always equal to the sum of their squares added to "double the products that can be made by multiplying "any two of them," therefore \( p^2 = B + 2q \), and consequently \( B = p^2 - 2q \). For example, \( a + b + c + d = a^2 + b^2 + c^2 + 2ab + ac + ad + bc + bd + cd \), that is, again, \( p^2 = B + 2q \), or \( B = p^2 - 2q \). And so for any other number of quantities. In general therefore, "B the "sum of the squares of the roots may always be found "by subtracting \( 2q \) from \( p^2 \); the quantities \( p \) and \( q \) being always known, since they are the coefficients in the proposed equation.

"The sum of the cubes of the roots of any equation is equal to \( p^3 - 3pq + 3r \), or to \( Bp - pq + 3r \)." For \( B - q \times p \) gives always the excess of the sum of the cubes of any quantities above the triple sum of the products that can be made by multiplying any three of them.

Thus, \( a^3 + b^3 + c^3 - ab - ac - bc \times a + b + c = (B - q \times p) = a^3 + b^3 + c^3 - 3abc \). Therefore if the sum of the cubes is called \( C \), then shall \( B - q \times p = C - 3r \), and \( C = Bp - pq + 3r \) (because \( B = p^3 - 2q \)).

After the same manner, if \( D \) be the sum of the 4th powers of the roots, you will find that \( D = pC - qB + pr - 4s \), and if \( E \) be the sum of the 5th powers, then shall \( E = pD - qC + rB - ps + 5t \). And after the same manner the sum of any powers of the roots may be found; the progression of these expressions of the sum of the powers being obvious.

As for the signs of the terms of the equation produced, it appears, from inspection, that the signs of all the terms in any equation in the table are alternately \(+ \) and \(-\): these equations are generated by multiplying continually \( x - a, x - b, x - c, x - d \), &c., by one another. The first term is always some pure power of \( x \), and is positive; the second is a power of \( x \) multiplied by the quantities \(-a, -b, -c, &c.\) and since these are all negative, that term must therefore be negative. The third term has the products of any two of these quantities \((-a, -b, -c, &c.)\) for its coefficient; which products are all positive, because \(-x-\) gives \(+\). For the like reason, the next coefficient, consisting of all the products made by multiplying any three of these quantities must be negative, and the next positive. So that the coefficients, in this case, will be positive and negative by turns. But, "in this case the roots are all positive;" since \( x = a, x = b, x = c, x = d, x = e, &c. \) are the assumed simple equations. It is plain then, that "when all the "roots are positive, the signs are alternately \(+ \) and \(-\).

But if the roots are all negative, then \( x + a \times x + b \times x + c \times x + d, &c. = 0 \). will express the equation to be produced; all whose terms will plainly be positive; so that, "when all the roots of an equation are negative, it is "plain there will be no changes in the signs of the terms "of that equation."

In general, "there are as many positive roots in any e- "quation as there are changes in the signs of the terms from \(+ \) to \(-\), or from \(- \) to \(+\); and the remaining "roots are negative." The rule is general, if the impossible roots be allowed to be either positive or negative; and may be extended to all kinds of equations.

In quadratic equations, the two roots are either both positive, as in this

\[ (x - a)(x - b) = x^2 - ax + ab = 0, \]

where there are two changes of the signs: Or they are both negative, as in this

\[ (x + a)(x + b) = x^2 + ax + ab = 0, \]

where there is not any change of the signs: Or there is one positive and one negative, as in

\[ (x - a)(x + b) = x^2 - ax - ab = 0, \]

where there is necessarily one change of the signs; because the first term is positive, and the last negative, and there can be but one change whether the 2d term be \(+\) or \(-\).

Therefore the rule given in the last paragraph extends to all quadratic equations.

In cubic equations, the roots may be,

1°. All positive, as in this, \( x - a \times x - b \times x - c = 0 \), in which the signs are alternately \(+ \) and \(-\), as appears from the table; and there are three changes of the signs.

2°. The roots may be all negative, as in the equation \( x + a \times x + b \times x + c = 0 \), where there can be no change of the signs. Or,

3°. There may be two positive roots and one negative, as in the equation \( x - a \times x - b \times x + c = 0 \); which gives

\[ x^3 - ax^2 + bx + c = 0. \]

Here there must be two changes of the signs; because if \( a + b \) is greater than \( c \), the second term must be negative, its coefficient being \(-a - b + c\).

And if \( a + b \) is less than \( c \), then the third term must be negative, its coefficient \(-ab - ac - bc(ab - cxa + b)\) being in that case negative. And there cannot possibly be three changes of the signs, the first and last terms having the same sign.

4°. There may be one positive root and two negative, as in the equation \( x + a \times x + b \times x - c = 0 \), which gives

\[ x^3 + ax^2 + bx - c = 0. \]

where

* Because the rectangle \( axb \) is less than the square \( a + b \times a + b \), and therefore much less than \( a + b \times c \). where there must be always one change of the signs, since the first term is positive and the last negative. And there can be but one change of the signs, since if the second term is negative, or \(a + b\) less than \(c\), the third must be negative also, so that there will be but one change of the signs. Or, if the second term is affirmative, whatever the third term is, there will be but one change of the signs. It appears therefore, in general, that in cubic equations, there are as many affirmative roots as there are changes of the signs of the terms of the equation.

There are several conjectaries of what has been already demonstrated, that are of use in discovering the roots of equations. But before we proceed to that, it will be convenient to explain some transformations of equations, by which they may often be rendered more simple, and the investigation of their roots more easy.

**CHAP. XVI. Of the Transformation of Equations; and exterminating their intermediate Terms.**

We now proceed to explain the transformation of equations that are most useful: and, first, "The affirmative roots of an equation are changed into negative roots of the same value, and the negative roots into affirmative, by only changing the signs of the terms alternately, beginning with the second." Thus, the roots of the equation \(x^4 - x^3 - 19x^2 + 49x - 30 = 0\) are \(+1, +2, +3, -5\); whereas the roots of the same equation having only the signs of the second and fourth terms changed, viz. \(x^4 + x^3 - 19x^2 + 49x - 30 = 0\) are \(-1, -2, -3, +5\).

To understand the reason of this rule, let us assume an equation, as \(x^4 - ax^3 - bx^2 - cx - dx - e = 0\), whose roots are \(+a, +b, +c, +d, +e, \&c.\) and another, having its roots of the same value, but affected with contrary signs, as \(x^4 + ax^3 + bx^2 + cx + dx + e = 0\). It is plain, that the terms taken alternately, beginning from the first, are the same in both equations, and have the same sign, "being products of an even number of the roots;" the product of any two roots having the same sign as their product when both their signs are changed; as \(+a \times -b = -ax + b\).

But the second terms, and all taken alternately from them, because their coefficients involve always the products of an odd number of the roots, will have contrary signs in the two equations. For example, the product of four, viz. \(abcd\), having the same sign in both, and one equation in the fifth term having \(abcd \times c\), and the other \(abcd \times -e\), it follows, that their product \(abcde\) must have contrary signs in the two equations: these two equations therefore that have the same roots, but with contrary signs, have nothing different but the signs of the alternate terms, beginning with the second. From which it follows, "that if any equation is given, and you change the signs of the alternate terms, beginning with the second, the new equation will have roots of the same value, but with contrary signs."

It is often very useful "to transform an equation into another that shall have its roots greater or less than the roots of the proposed equation by some given difference."

Let the equation proposed be the cubic \(x^3 - px^2 + qx - r = 0\). And let it be required to transform it into another equation whose roots shall be less than the roots of this equation by some given difference \((e)\), that is, suppose \(y = x - e\), and consequently \(x = y + e\); then, instead of \(x\) and its powers, substitute \(y + e\) and its powers, and there will arise this new equation.

\[ (A) \begin{cases} y^3 + 3ey^2 + 3e^2y + e^3 \\ - py^2 - 2pey - pe^2 \\ + qy + qe \\ - r \end{cases} = 0, \]

whose roots are less than the roots of the preceding equation by the difference \((e)\).

If it had been required to find an equation whose roots should be greater than those of the proposed equation by the quantity \((e)\), then we must have supposed \(y = x + e\), and consequently \(x = y - e\), and then the other equation would have had this form.

\[ (B) \begin{cases} y^3 - 3ey^2 + 3e^2y - e^3 \\ - py^2 + 2pey - pe^2 \\ + qy - qe \\ - r \end{cases} = 0. \]

If the proposed equation be in this form \(x^3 + px^2 + qx + r = 0\), then, by supposing \(x = y - z\), there will arise an equation agreeing in all respects with the equation \((A)\), but that the second and fourth terms will have contrary signs.

And by supposing \(x = y + z\), there will arise an equation agreeing with \((B)\) in all respects, but that the second and fourth terms will have contrary signs to what they have in \((B)\).

The first of these suppositions gives this equation,

\[ (C) \begin{cases} y^3 - 3ey^2 + 3e^2y - e^3 \\ + py^2 - 2pey + pe^2 \\ + qy - qe \\ + r \end{cases} = 0. \]

The second supposition gives the equation,

\[ (D) \begin{cases} y^3 + 3ey^2 + 3e^2y + e^3 \\ + py^2 + 2pey + pe^2 \\ + qy + qe \\ + r \end{cases} = 0. \]

The first use of this transformation of equations is to shew "how the second (or other intermediate) term may be taken away out of an equation."

It is plain, that in the equation \((A)\) whose second term is \(3e - pxy\), if you suppose \(e = \frac{p}{3}\), and consequently \(3e - p = 0\), then the second term will vanish.

In the equation \((C)\) whose second term is \(-3e + pxy^2\), supposing \(e = \frac{p}{3}\), the second term also vanishes.

Now the equation \((A)\) was deduced from \(x^3 - px^2 + qx - r = 0\), by supposing \(y = x - e\); and the equation \((C)\) was deduced from \(x^3 + px^2 + qx + r = 0\); by supposing \(y = x + e\). From which this rule may easily be deduced for exterminating the second term out of any cubic equation.

**Rule.** Rule. Add to the unknown quantity of the given equation the third part of the coefficient of the second term with its proper sign, viz., \( \frac{1}{2}p \), and suppose this aggregate equal to a new unknown quantity \( y \). From this value of \( y \) find a value of \( x \) by transposition, and substitute this value of \( x \) and its powers in the given equation, and there will arise a new equation that shall want the second term.

Example. Let it be required to exterminate the second term out of this equation, \( x^3 - ox^2 + 26x - 34 = 0 \), suppose \( x - 3 = y \), or \( y + 3 = x \); and substituting according to the rule, you will find

\[ \begin{align*} y^3 + 2y^2 + 27y + 27 \\ -9y^2 - 54y - 81 \\ +26y + 78 \\ -34 \\ y^3 * -y - 10 = 0. \end{align*} \]

In which there is no term where \( y \) is of two dimensions, and an asterisk is placed in the room of the 2d term, to show it is wanting.

Let the equation proposed be of any number of dimensions represented by \( n \); and let the coefficient of the second term with its sign prefixed be \( -p \), then supposing \( x - \frac{p}{n} = y \), and consequently \( x = y + \frac{p}{n} \), and substituting this value for \( x \) in the given equation, there will arise a new equation that shall want the second term.

It is plain from what was demonstrated in chap. 15, that the sum of the roots of the proposed equation is \( +p \); and since we suppose \( y = x - \frac{p}{n} \), it follows, that, in the new equation, each value of \( y \) will be less than the respective value of \( x \) by \( \frac{p}{n} \); and, since the number of the roots is \( n \), it follows, that the sum of the values of \( y \) will be less than \( +p \), the sum of the values of \( x \), by \( n \times \frac{p}{n} \), the difference of any two roots, that is, by \( +p \); therefore the sum of the values of \( y \) will be \( +p - p = 0 \).

But the coefficient of the second term of the equation of \( y \) is the sum of the values of \( y \), viz., \( +p - p \), and therefore that coefficient is equal to nothing; and consequently, in the equation of \( y \), the second term vanishes. It follows then, that the second term may be exterminated out of any given equation by the following

Rule. Divide the coefficient of the second term of the proposed equation by the number of dimensions of the equation; and assuming a new unknown quantity \( y \), add to it the quotient having its sign changed. Then suppose this aggregate equal to \( x \) the unknown quantity in the proposed equation; and for \( x \) and its powers, substitute the aggregate and its powers, so that the new equation that arises want its second term.

If the proposed equation is a quadratic, as \( x^2 - px + q \)

\[ \begin{align*} y^2 + py + \frac{1}{2}p^2 \\ - \frac{1}{2}p^2 \\ + q \\ y^2 * - \frac{1}{2}p^2 + q = 0. \end{align*} \]

And from this example the use of exterminating the 2d term appears: for commonly the solution of the equation that wants the 2d term is more easy. And, if you can find the value of \( y \) from this new equation, it is easy to find the value of \( x \), by means of the equation \( y + \frac{1}{2}p = x \).

For example,

Since \( y^2 + q - \frac{1}{4}p^2 = 0 \), it follows, that

\[ \begin{align*} y^2 &= \frac{1}{4}p^2 - q, \\ y &= \pm \sqrt{\frac{1}{4}p^2 - q}, \\ x &= y + \frac{1}{2}p = \pm \sqrt{\frac{1}{4}p^2 - q}; \end{align*} \]

which agrees with what we demonstrated, chap. 12.

If the proposed equation is a biquadratic, as \( x^4 - px^3 + qx^2 - rx + s = 0 \), then, by supposing \( x - \frac{p}{2} = y \) or \( x = y + \frac{p}{2} \), an equation shall arise having no second term. And if the proposed equation is of 5 dimensions, then you must suppose \( x = \frac{p}{5} \). And so on.

When the second term in any equation is wanting, it follows, that "the equation has both affirmative and negative roots," and that "the sum of the affirmative roots is equal to the sum of the negative roots:" by which means the coefficient of the 2d term, which is the sum of all the roots of both sorts, vanishes, and makes the second term vanish.

In general, "The coefficient of the 2d term is the difference between the sum of the affirmative roots and the sum of the negative roots:" and the operations we have given serve only to diminish all the roots when the sum of the affirmative is greatest, or increase the roots when the sum of the negative is greatest, so as to balance them, and reduce them to an equality.

It is obvious, that in a quadratic equation that wants the second term, there must be one root affirmative and one negative; and these must be equal to one another.

In a cubic equation that wants the second term, there must be either two affirmative roots equal, taken together to a third root that must be negative; or, two negative equal to a third that must be positive.

"Let an equation \( x^3 - px^2 + qx - r = 0 \) be proposed, and let it be now required to exterminate the third term."

By supposing \( y = x - e \), the coefficient of the 3d term in the equation of \( y \) is found (see equation A) to be \( 3e^2 - 2pe + q \). Suppose that coefficient equal to nothing, and by resolving the quadratic equation \( 3e^2 - 2pe + q = 0 \), you will find the value of \( e \), which substituted for it in the equation \( y = x - e \), will show how to transform the proposed equation into one that shall want the third term.

The quadratic \( 3e^2 - 2pe + q = 0 \), gives \( e = \frac{p \pm \sqrt{p^2 - 3q}}{3} \).

So that the proposed cubic will be transformed into an equation wanting the third term by supposing \( y = x - \frac{p \pm \sqrt{p^2 - 3q}}{3} \), or \( y = x - \frac{p \pm \sqrt{p^2 - 3q}}{3} \). If the proposed equation is of \( n \) dimensions, the value of \( c \), by which the 3d term may be taken away, is had by resolving the quadratic equation \( x^2 + \frac{2p}{n}x + \frac{n-1}{n} = 0 \), supposing \( -p \) and \( +q \) to be the coefficients of the 2d and 3d terms of the proposed equation.

The 4th term of any equation may be taken away by solving a cubic equation, which is the coefficient of the 4th term in the equation, when transformed, as in the former part of this chapter. The fifth term may be taken away by solving a biquadratic; and after the same manner, the other terms can be exterminated if there are any.

There are other transmutations of equations that, on some occasions, are useful.

An equation, as \( x^3 - px^2 + qx - r = 0 \), "may be transformed into another that shall have its roots equal to the roots of this equation multiplied by a given quantity," as \( f \), by supposing \( y = fx \), and consequently \( x = \frac{y}{f} \), and substituting this value for \( x \) in the proposed equation, there will arise \( y^3 - py^2 + qy - r = 0 \), and multiplying all by \( f^3 \), \( y^3 - fpy^2 + f^2qy - f^3r = 0 \), where the coefficient of the 2d term of the proposed equation multiplied into \( f \), makes the coefficient of the 2d term of the transformed equation; and the following coefficients are produced by the following coefficients of the proposed equation, (as \( q, r, \) &c.) multiplied into the powers of \( f \) (\( f^2, f^3, \) &c.).

Therefore "to transform any equation into another whose roots shall be equal to the roots of the proposed equation multiplied by a given quantity (\( f \))," you need only multiply the terms of the proposed equation, beginning at the 2d term, by \( f, f^2, f^3, f^4, \) &c. and putting \( y \) instead of \( x \), there will arise an equation having its roots equal to the roots of the proposed equation multiplied by (\( f \)) as required.

The transformation mentioned above is of use when the highest term of the equation has a coefficient different from unity; for, by it, the equation may be transformed into one that shall have the coefficient of the highest term unit.

If the equation proposed is \( ax^3 - px^2 + qx - r = 0 \), then transform the equation into one whose roots are equal to the roots of the proposed equation multiplied by (\( a \)).

That is, suppose \( y = ax \) or \( x = \frac{y}{a} \) and there will arise \( ay^3 - py^2 + qy - ra^2 = 0 \); so that

\[ y^3 - py^2 + qy - ra^2 = 0. \]

From which we easily draw this

**Rule.** Change the unknown quantity \( x \) into another \( y \), prefix no coefficient to the highest term, pass the 2d, multiply the following terms, beginning with the 3d, by \( a, a^2, a^3, a^4, \) &c. the powers of the coefficient of the highest term of the proposed equation, respectively.

Thus the equation \( 3x^3 - 13x^2 + 14x + 16 = 0 \), is transformed into the equation \( y^3 - 13y^2 + 14y + 16 = 0 \).

Then finding the roots of this equation, it will easily be discovered what are the roots of the proposed equation, since \( 3x = y \), or \( x = \frac{y}{3} \). And therefore, since one of the values of \( y \) is \(-2\), it follows, that one of the values of \( x \) is \(-\frac{2}{3}\).

By the last rule, "an equation is easily cleared of fractions." Suppose the equation proposed is \( x^3 - \frac{p}{m}x^2 + \frac{q}{n}x - \frac{r}{e} = 0 \). Multiply all the terms by the product of the denominators, you find

\[ mnex^3 - nepx^2 + mneqx - mnr = 0. \]

Then (as above) transforming the equation into one that shall have unit for the coefficient of the highest term, you find

\[ y^3 - nepy^2 + mne^2qy - mnr^2 = 0. \]

Or, neglecting the denominator of the last term \( \frac{r}{e} \), you need only multiply all the equation by \( mn \), which will give

\[ mnex^3 - nepx^2 + mneqx - mnr = 0. \]

And then

\[ y^3 - nepy^2 + mne^2qy - mnr^2 = 0. \]

Now after the values of \( y \) are found, it will be easy to discover the values of \( x \); since, in the first case, \( x = \frac{y}{mn} \); in the second, \( x = \frac{y}{mn^2} \).

For example, the equation \( x^3 - \frac{3}{4}x^2 - \frac{1}{4}x - \frac{3}{8} = 0 \), is first reduced to this form \( x^3 - 4x^2 - 9x - 12 = 0 \), and then transformed into \( y^3 - 12y^2 - 146 = 0 \).

Sometimes, by these transformations, "surds are taken away." As for example,

The equation \( x^3 - p\sqrt{a}x^2 + qx - r/\sqrt{a} = 0 \), by putting \( y = \sqrt{a}x \), or \( x = \frac{y}{\sqrt{a}} \), is transformed into this equation,

\[ y^3 - p\sqrt{a}y^2 + qy - r/\sqrt{a} = 0. \]

Which, by multiplying all the terms by \( \sqrt{a} \), becomes \( y^3 - pay^2 + qay - ra = 0 \), an equation free of surds. But in order to make this succeed, the surd (\( \sqrt{a} \)) must enter the alternate terms, beginning with the second.

"An equation, as \( x^3 - px^2 + qx - r = 0 \), may be transformed into one whose roots shall be the quantities reciprocal of \( x \); by supposing \( y = \frac{1}{x} \), and \( y = \frac{z}{r} \), or (by one supposition), \( x = \frac{r}{z} \), becomes \( z^3 - qz^2 + prz - r^2 = 0 \).

In the equation of \( y \), it is manifest, that the order of the coefficients is inverted; so that, if the second term had been wanting in the proposed equation, the last but one should have been wanting in the equations of \( y \) and \( z \). If the 3d had been wanting in the equation proposed, the last but two had been wanting in the equations of \( y \) and \( z \).

Another use of this transformation is, that the greatest left root in the one is transformed into the least root in the other. For since \( x = \frac{1}{y} \), and \( y = \frac{1}{x} \), it is plain, that when the value of \( x \) is greatest, the value of \( y \) is least, and conversely.

How an equation is transformed so as to have all its roots affirmative, shall be explained in the following chapter.

**Chap. XVII. Of finding the Roots of Equations when two or more of the Roots are equal to each other.**

§1. Before we proceed to explain how to resolve equations of all sorts, we shall first demonstrate how an equation that has two or more roots equal, is depressed to a lower dimension; and its resolution made, consequently, more easy. And shall endeavour to explain the grounds of this and many other rules we shall give in the remaining part of this treatise, in a more simple and concise manner than has hitherto been done.

In order to this, we must look back to the last chapter, where we find, that if any equation, as \( x^3 - px^2 + qx - r = 0 \), is proposed, and you are to transform it into another that shall have its roots less than the values of \( x \) by any given difference, as \( e \), you are to assume \( y = x - e \), and substituting for \( x \) its value \( y + e \), you find the transformed equation,

\[ \begin{align*} y^3 + 3ey^2 + 3e^2y + e^3 \\ - py^2 - 2epy - pe^2 \\ + qy + qe \\ \end{align*} \]

Where we are to observe,

1°. That the last term \( (e^3 - pe^2 + qe - r) \) is the very equation that was proposed, having \( e \) in place of \( x \).

2°. The coefficient of the last term but one is \( 3e^2 - 2pe + q \), which is the quantity that arises by multiplying every term of the last coefficient \( e^3 - pe^2 + qe - r \) by the index of \( e \) in each term, and dividing the product \( 3e^2 - 2pe + qe \) by the quantity \( e \) that is common to all the terms.

3°. The coefficient of the last term but two is \( 3e - p \), which is the quantity that arises by multiplying every term of the coefficient last found \( (3e^2 - 2pe + q) \) by the index of \( e \) in each term, and dividing the whole by \( 2e \).

§2. These same observations extend to equations of all dimensions. If it is the biquadratic \( x^4 - px^3 + qx^2 - rx + s = 0 \) that is proposed, then by supposing \( y = x - e \), it will be transformed into this other,

\[ \begin{align*} y^4 + 4ey^3 + 6e^2y^2 + 4e^3y + e^4 \\ - py^3 - 3epy^2 - 3pe^2y - pe^3 \\ + gy^2 + 2qey + qe^2 \\ - ry - re \\ \end{align*} \]

Where again it is obvious, That the last term is the equation that was proposed, having \( e \) in place of \( x \): That the last term but one has for its coefficient the quantity

that arises by multiplying the terms of the last quantity by the indices of \( e \) in each term, and dividing the product by \( e \): That the coefficient of the last term but two, (viz. \( 6e^2 - 3pe^2 + qe - r \)) is deduced in the same manner from the term immediately following, that is, by multiplying every term of \( 4e^3 - 3pe^2 + 2qe - r \) by the index of \( e \) in that term, and dividing the whole by \( e \) multiplied into the index of \( y \) in the term sought, that is, by \( e \times 2 \): And the next term is \( 4e - p = \frac{6e^2 \times 2 - 3pe \times 1}{3e} \).

The demonstration of this may easily be made general by the theorem for finding the powers of a binomial, since the transformed equation consists of the powers of the binomial \( y + e \) that are marked by the indices of \( e \) in the last term, multiplied each by their coefficients \( 1, -p, +q, -r, +s, \) &c. respectively.

§3. From the last two articles we can easily find the terms of the transformed equation without any involution. The last term is had by substituting \( e \) instead of \( x \) in the proposed equation; the next term, by multiplying every part of that last term by the index of \( e \) in each part, and dividing the whole by \( e \); and the following terms in the manner described in the foregoing article; the respective divisors being the quantity \( e \) multiplied by the index of \( y \) in each term.

The demonstration for finding when two or more roots are equal will be easy, if we add to this, that "when the unknown quantity enters all the terms of any equation, then one of its values is equal to nothing." As in the equation \( x^3 - px^2 + qx = 0 \), where \( x = 0 \) being one of the simple equations that produce \( x^3 - px^2 + qx = 0 \), it follows that one of the values of \( x \) is 0. In like manner, two of the values of \( x \) are equal to nothing in this equation \( x^3 - px^2 = 0 \); and three of them vanish in the equation \( x^4 - px^3 = 0 \).

It is also obvious (conversely) that "if \( x \) does not enter all the terms of the equation, i.e. if the last term be not wanting, then none of the values of \( x \) can be equal to nothing," for if every term be not multiplied by \( x \), then \( x = 0 \) cannot be a divisor of the whole equation, and consequently 0 cannot be one of the values of \( x \). If \( x^2 \) does not enter into all the terms of the equation, then two of the values of \( x \) cannot be equal to nothing. If \( x^3 \) does not enter into all the terms of the equation, then three of the values of \( x \) cannot be equal to nothing, &c.

§4. Suppose now that two values of \( x \) are equal to one another, and to \( e \); then it is plain that two values of \( y \) in the transformed equation will be equal to nothing: since \( y = x - e \). And consequently, by the last article, the two last terms of the transformed equation must vanish.

Suppose it is the cubic equation of §1, that is proposed, viz. \( x^3 - px^2 + qx - r = 0 \); and because we suppose \( x = e \), therefore the last term of the transformed equation, viz. \( e^3 - pe^2 + qe - r \) will vanish. And since two values of \( y \) vanish, the last term but one, viz. \( 3e^2 - 2pe + qe \) will vanish at the same time. So that \( 3e^2 - 2pe + qe = 0 \). But, by supposition, \( e = x \); therefore, when two values of \( x \), in the equation, \( x^3 - px^2 + qx - r = 0 \), are equal, it follows, that \( 3x^2 - 2px + q = 0 \). And thus, thus, "the proposed cubic is depressed to a quadratic that has one of its roots equal to one of the roots of that cubic."

If it is the biquadratic that is proposed, viz., \(x^4 - px^3 + qx^2 - rx + s = 0\), and two of its roots be equal; then supposing \(e = x\), two of the values of \(y\) must vanish, and the equation of § 2, will be reduced to this form.

\[ \begin{align*} y^4 + 4ey^3 + 6e^2y^2 \\ - py^3 - 3py^2 \\ + qy^2 \\ - ry = 0. \end{align*} \]

So that

\[ 4e^3 - 3pe^2 + 2qe - r = 0; \quad \text{or, because } e = x, \]

\[ 4x^3 - 3px^2 + 2qx - r = 0. \]

In general, when two values of \(x\) are equal to each other, and to \(e\), the two last terms of the transformed equation vanish; and consequently, "if you multiply the terms of the proposed equation by the indices of \(x\) in each term, the quantity that will arise will be \(= 0\),

and will give an equation of a lower dimension than the proposed, that shall have one of its roots equal to one of the roots of the proposed equation."

That the last two terms of the equation vanish when the values of \(x\) are supposed equal to each other, and to \(e\), will also appear by considering, that since two values of \(y\) then become equal to nothing, the product of the values of \(y\) must vanish, which is equal to the last term of the equation; and because two of the four values of \(y\) are equal to nothing, it follows also that one of any three that can be taken out of these four must be \(= 0\); and therefore, the products made by multiplying any three must vanish; and consequently the coefficient of the last term but one, which is equal to the sum of these products, must vanish.

§ 5. After the same manner, if there are three equal roots in the biquadratic \(x^4 - px^3 + qx^2 - rx + s = 0\), and if \(e\) be equal to one of them, three values of \(y\) \((x-e)\) will vanish, and consequently \(y^3\) will enter all the terms of the transformed equation; which will have this form,

\[ \begin{align*} y^4 + 4ey^3 \\ - py^3 \\ + qy^2 \\ - ry = 0. \end{align*} \]

So that here

\[ 6e^2 - 3pe + q = 0; \quad \text{or, since } e = x, \text{ therefore,} \]

\[ 6x^2 - 3px + q = 0; \quad \text{and one of the roots of this quadratic will be equal to one of the roots of the proposed biquadratic.} \]

In this case, two of the roots of the cubic equation \(4x^3 - 3px^2 + 2qx - r = 0\) are roots of the proposed biquadratic, because the quantity \(6x^2 - 3px + q\) is deduced from \(4x^3 - 3px^2 + 2qx - r\), by multiplying the terms by the indexes of \(x\) in each term.

In general, "whatever is the number of equal roots in the proposed equation, they will all remain but one in the equation that is deduced from it, by multiplying all the terms by the indexes of \(x\) in them; and they will all remain but two in the equation deduced in the same manner from that;" and so of the rest.

§ 6. What we observed of the coefficients of equations transformed by supposing \(y = x - e\), leads to this easy demonstration of this rule; and will be applied in the next chapter to demonstrate the rules for finding the limits of equations.

It is obvious, however, that though we make use of equations whose signs change alternately, the same reasoning extends to all other equations.

It is a consequence also of what has been demonstrated, that "if two roots of any equation, as \(x^3 - px^2 + qx - r = 0\), are equal, then multiplying the terms by any arithmetical series, as \(a + 3b, a + 2b, a + b, a\), the product will be \(= 0\)."

For, since

\[ ax^3 - apx^2 + aqx - ar = 0; \quad \text{and} \]

\[ 3x^2 - 2px + qx = 0, \quad \text{it follows that} \]

\[ ax^3 + 3bx^2 - apx^2 + aqx + bxq - ar = 0. \]

Which is the product that arises by multiplying the terms of the proposed equation by the terms of the series, \(a + 3b, a + 2b, a + b, a\); which may represent any arithmetical progression.

**Chap. XVIII. Of the Limits of Equations.**

We now proceed to show how to discover the limits of the roots of equations, by which their solution is much facilitated.

Let any equation, as \(x^3 - px^2 + qx - r = 0\) be proposed; and transform it, as above, into the equation

\[ \begin{align*} y^3 + 3ey^2 + 3e^2y + e^3 \\ - py^2 - 2epy - pe^2 \\ + qy + qe \\ - r = 0. \end{align*} \]

Where the values of \(y\) are less than the respective values of \(x\) by the difference \(e\). If you suppose \(e\) to be taken such as to make all the coefficients of the equation of \(y\) positive, viz., \(e^3 - pe^2 + qe - r, 3e^2 - 2ep + q, 3e - p\); then there being no variation of the signs in the equation, all the values of \(y\) must be negative; and consequently, the quantity \(e\), by which the values of \(x\) are diminished, must be greater than the greatest positive value of \(x\); and consequently must be the limit of the roots of the equation \(x^3 - px^2 + qx - r = 0\).

It is sufficient therefore, in order to find the limit, to inquire what quantity substituted for \(x\) in each of these expressions \(x^3 - px^2 + qx - r, 3x^2 - 2px + q, 3x - p\), will give them all positive;" for that quantity will be the limit required.

How these expressions are formed from one another, was explained in the beginning of the last chapter.

**Examp.** If the equation \(x^5 - 2x^4 - 10x^3 + 30x^2 + 63x + 120 = 0\) is proposed; and it is required to determine the limit that is greater than any of the roots; you are to inquire what integer number substituted for \(x\) in the proposed equation, and following equations deduced from it by § 3, chap. 17, will give, in each, a positive quantity,

\[ \begin{align*} 5x^4 - 8x^3 - 30x^2 + 60x + 63 \\ 5x^3 - 6x^2 - 15x + 15 \\ 5x^2 - 4x - 5 \\ 5x - 2. \end{align*} \]

The least integer number which gives each of these positive, is 2; which therefore is the limit of the roots of of the proposed equation; or a number that exceeds the greatest positive root.

If the limit of the negative roots is required, you may (by chap. 16.) change the negative into positive roots; and then proceed as before to find their limits. Thus, in the example, you will find, that $-3$ is the limit of the negative roots. So that the five roots of the proposed equation are between $-3$ and $+2$.

Having found the limit that surpasses the greatest positive root, call it $m$. And if you assume $y = m - x$, and for $x$ substitute $m - y$, the equation that will arise will have all its roots positive; because $m$ is supposed to surpass all the values of $x$, and consequently $m - x$ ($= y$) must always be affirmative. And, by this means, any equation may be changed into one that shall have all its roots affirmative.

Or, if $-n$ represent the limit of the negative roots, then by assuming $y = x + n$, the proposed equation shall be transformed into one that shall have all its roots affirmative; for $+n$ being greater than any negative value of $x$, it follows, that $y = x + n$ must be always positive.

The greatest negative coefficient of any equation increased by unit, always exceeds the greatest root of the equation.

To demonstrate this, let the cubic $x^3 - px^2 - qx - r = 0$ be proposed; where all the terms are negative except the first. Assuming $y = x - e$, it will be transformed into the following equation:

$$ (A) \quad y^3 + 3ey^2 + 3e^2y + e^3 \\ - py^2 - 2py - pe \\ - qy - qe \\ - ry - re = 0. $$

1°. Let us suppose that the coefficients $p$, $q$, $r$, are equal to each other; and if you also suppose $e = p + r$, then the last equation becomes

$$ (B) \quad y^3 + 2py^2 + p^2y + 1 \\ + 3y^2 + 3py \\ + 3y = 0; $$

where all the terms being positive, it follows that the values of $y$ are all negative, and that consequently $e$, or $p + r$, is greater than the greatest value of $x$ in the proposed equation.

2°. If $q$ and $r$ be not $= p$, but less than it, and for $e$ you still substitute $p + r$ (since the negative part $(-qy - qe)$ becomes less, the positive remaining undiminished,) a fortiori, all the coefficients of the equation $(A)$ become positive. And the same is obvious if $q$ and $r$ have positive signs, and not negative signs, as we supposed. It appears therefore, "that if, in any cubic equation, $p$ be the greatest negative coefficient, then $p + r$ must surpass the greatest value of $x$."

3°. By the same reasoning it appears, that if $q$ be the greatest negative coefficient of the equation, and $e = q + r$, then there will be no variation of the signs in the equation of $y$: for it appears from the last article, that if all the three ($p$, $q$, $r$) were equal to one another, and $e$ equal to any one of them increased by unit, as to $q + r$, then all the terms of the equation $(A)$ would be positive. Now if $e$ be supposed still equal to $q + r$, and $p$ and $r$ to be less than $q$, then, a fortiori, all these terms will be positive, the negative part, which involves $p$ and $r$ being diminished, while the positive part and the negative involving $q$ remain as before.

4°. After the same manner it is demonstrated, that if $r$ is the greatest negative coefficient in the equation, and $e$ is supposed $= r + 1$, then all the terms of the equation $(A)$ of $y$ will be positive; and consequently $r + 1$ will be greater than any of the values of $x$.

What we have said of the cubic equation $x^3 - px^2 - qx - r = 0$, is easily applicable to others.

In general, we conclude, that "the greatest negative coefficient in any equation increased by unit, is always a limit that exceeds all the roots of that equation."

But it is to be observed at the same time, that the greatest negative coefficient increased by unit, is very seldom the nearest limit: that is best discovered by the rule in the beginning of this chapter.

Having shown how to change any proposed equation into one that shall have all its roots affirmative; we shall only treat of such as have all their roots positive, in what remains relating to the limits of equations.

Any such equation may be represented by $x - a \times x - b \times x - c \times x - d$, &c., $= 0$, whose roots are $a$, $b$, $c$, $d$, &c.

And of all such equations two limits are easily discovered from what precedes, viz. $\sigma$, which is less than the least, and $e$, found as directed in the beginning of this chapter, which surpasses the greatest root of the equation.

But, besides these, we shall now show how to find other limits betwixt the roots themselves. And, for this purpose, will suppose $a$ to be the least root, $b$ the second root, $c$ the third, and so on; it being arbitrary.

If you substitute $o$ in place of the unknown quantity, putting $x = o$, the quantity that will arise from that supposition is the last term of the equation, all the others that involve $x$ vanishing.

If you substitute for $x$ a quantity less than the least root $a$, the quantity resulting will have the same sign as the last term; that is, will be positive or negative according as the equation is of an even or odd number of dimensions. For all the factors $x - a$, $x - b$, $x - c$, &c., will be negative, and their product will be positive or negative according as their number is even or odd.

If you substitute for $x$ a quantity greater than the least root $a$, but less than all the other roots, then the sign of the quantity resulting will be contrary to what it was before; because one factor ($x - a$) becomes now positive, all the others remaining negative as before.

If you substitute for $x$ a quantity greater than the two least roots, but less than all the rest, both the factors $x - a$, $x - b$, become positive, and the rest remain as they were. So that the whole product will have the same sign as the last term of the equation. Thus successively placing instead of $x$ quantities that are limits betwixt the roots of the equation, the quantities that result will have alternately the signs $+$ and $-$. And, conversely, "if you find quantities which, substituted in place of $x$ in the proposed equation, do give alternately positive and negative results, those quantities are the limits of that equation." It is useful to observe, that, in general, "when, by substituting any two numbers for \( x \) in any equation, the results have contrary signs, one or more of the roots of the equation must be betwixt those numbers." Thus, in the equation, \( x^3 - 2x^2 - 5 = 0 \), if you substitute 2 and 3 for \( x \), the results are -5, +4; whence it follows, that the roots are betwixt 2 and 3: for when these results have different signs, one or other of the factors which produce the equations must have changed its sign; suppose it is \( x = e \), then it is plain that \( e \) must be betwixt the numbers supposed equal to \( x \).

Let the cubic equation \( x^3 - px^2 + qx - r = 0 \) be proposed, and let it be transformed, by assuming \( y = x - e \), into the equation

\[ \begin{align*} y^3 + 3ey^2 + 3e^2y + e^3 \\ - py^2 - 2pey - pe^2 \\ + qy + qe \\ - r \end{align*} = 0. \]

Let us suppose \( e \) equal successively to the three values of \( x \), beginning with the least value; and because the last term \( e^3 - pe^2 + qe - r \) will vanish in all these suppositions, the equation will have this form,

\[ \begin{align*} y^3 + 3ey^2 + 3e^2y \\ - py^2 - 2pey \\ + qy \\ - r \end{align*} = 0; \]

where the last term \( 3e^2 - 2pe + q \) is, from the nature of equations, produced of the remaining values of \( y \), or of the excesses of two other values of \( x \) above what is supposed equal to \( e \); since always \( y = x - e \). Now,

1°. If \( e \) be equal to the least value of \( x \), then those two excesses being both positive, they will give a positive product, and consequently \( 3e^2 - 2pe + q \) will be, in this case, positive.

2°. If \( e \) be equal to the second value of \( x \), then, of those two excesses, one being negative and one positive, their product \( 3e^2 - 2pe + q \), will be negative.

3°. If \( e \) be equal to the third and greatest value of \( x \), then the two excesses being both negative, their product \( 3e^2 - 2pe + q \) is positive. Whence,

If in the equation \( 3e^2 - 2pe + q = 0 \), you substitute successively in the place of \( e \), the three roots of the equation \( e^3 - pe^2 + qe - r = 0 \), the quantities resulting will successively have the signs +, -, +; and consequently the three roots of the cubic equation are the limits of the roots of the equation \( e^3 - pe^2 + qe - r = 0 \). That is, the least of the roots of the cubic is less than the least of the roots of the other; the second root of the cubic is a limit between the two roots of the other; and the greatest root of the cubic is the limit that exceeds both the roots of the other.

We have demonstrated, that the roots of the cubic equation \( e^3 - pe^2 + qe - r = 0 \) are limits of the quadratic \( 3e^2 - 2pe + q \); whence it follows (conversely) that the roots of the quadratic \( 3e^2 - 2pe + q = 0 \) are the limits between the first and second, and between the second and third roots of the cubic \( e^3 - pe^2 + qe - r = 0 \). So that if you find the limit that exceeds the greatest root of the cubic, by the beginning of this chapter you will have (with \( o \), which is the limit less than any of the roots) four limits for the three roots of the proposed cubic.

It was demonstrated in chap 17, §3, how the quadratic \( 3e^2 - 2pe + q \) is deduced from the proposed cubic \( e^3 - pe^2 + qe - r = 0 \), viz., by multiplying each term by the index of \( e \) in it, and then dividing the whole by \( e \); and what we have demonstrated of cubic equations is easily extended to all others; so that we conclude "that the last term but one of the transformed equation is the equation for determining the limits of the proposed equation." Or, that the equation arising by multiplying each term by the index of the unknown quantity in it, is the equation whose roots give the limits of the proposed equation; if you add to them the two mentioned in p. 109, col. 2, par. 4.

For the same reason, it is plain that the root of the simple equation \( 3e - p = 0 \), (i.e., \( \frac{p}{3} \)) is the limit between the two roots of the quadratic \( 3e^2 - 2pe + q = 0 \). And, as \( 4e^3 - 3pe^2 + 2qe - r = 0 \) gives three limits of the equation \( e^3 - pe^2 + qe - r = 0 \), so the quadratic \( 6e^2 - 3pe + q = 0 \) gives two limits that are betwixt the roots of the cubic \( 4e^3 - 3e^2 + 2qe - r = 0 \); and \( 4e - p = 0 \) gives one limit that is betwixt the two roots of the quadratic \( 6e^2 - 3pe + q = 0 \). So that we have a complete series of these equations arising from a simple equation to the proposed, each of which determines the limits of the following equation.

If two roots in the proposed equation are equal, then "the limit that ought to be betwixt them must, in this case, become equal to one of the equal roots themselves." Which perfectly agrees with what was demonstrated in the last chapter, concerning the rule for finding the equal roots of equations.

And, the same equation that gives the limits, giving also one of the equal roots, when two or more are equal, it appears, that "if you substitute a limit in place of the unknown quantity in an equation," and, instead of a positive or negative result, it be found = 0, then you may conclude, that "not only the limit itself is a root of the equation, but that there are two roots in that equation equal to it and to one another."

It having been demonstrated, that the roots of the equation \( x^3 - px^2 + qx - r = 0 \) are the limits of the roots of the equation \( 3x^2 - 2px + q = 0 \), the three roots of the cubic equation, which suppose to be \( a, b, c \), substituted for \( x \) in the quadratic \( 2x^2 - 2px + q \), must give the results positive and negative alternately. Suppose these three results to be \( +N, -M, +L \); that is, \( 3a^2 - 2pa + q = N \), \( 3b^2 - 2pb + q = -M \), \( 3c^2 - 2pc + q = L \); and since \( a^3 - pa^2 + qa - r = 0 \), and \( 3a^2 - 2pa + q = N \times a \), subtracting the former multiplied into 3 from the latter, the remainder is \( pa^2 - 2qa + 3r = Nx \times a \). In the same manner \( pb^2 - 2qb + 3r = -Mx \times b \), and \( pc^2 - 2qc + 3r = Lx \times c \). Therefore \( px^2 - 2qx + 3r \) is such a quantity, that if, for \( x \), you substitute in it successively \( a, b, c \), the results will be \( +Nx, -Mx, +Lx \). Whence \( a, b, c \), are limits of the equation \( px^2 - 2qx + 3r = 0 \), by p. 109, col. 2, par. 8. and, conversely, the roots of the equation \( px^2 - 2qx + 3r = 0 \) are limits between the first and second, and between the second and third roots of the cubic \( x^3 - px^2 + qx - r = 0 \). Now the equation \( px^2 - 2qx + 3r = 0 \) arises rises from the proposed cubic by multiplying the terms of this latter by the arithmetical progression \(0, -1, -2, -3\). And, in the same manner, it may be shewn that the roots of the equation \(px^3 - 2qx^2 + 3rx - 4s = 0\) are limits of the equation \(x^4 - px^3 + qx^2 - rx + s = 0\).

Or, multiply the terms of the equation

\[ \frac{x^3 - px^2 + qx - r}{a + 3b, a + 2b, a + b, a} \]

by \(a + 3b, a + 2b, a + b, a\)

\[ \frac{ax^3 - apx^2 + aqx - ar}{a^3 + 3a^2b + 3ab^2 + b^3} \quad \text{or} \quad (a = 0) \]

\[ + 3bx^3 - 2bp^2x + bq^2x = (3x^3 - 2px + qx) \times bx. \]

Any arithmetical series where \(a\) is the least term, and \(b\) the common difference, and the products (if you substitute for \(x\), successively, \(a, b, c\), the three roots of the proposed cubic) shall be \(N \times bx, M \times bx + L \times bx\). For the first part of the product \(ax^3 - px^2 + qx - r = 0\); and \(a, b, c\), being limits in the equation \(3x^3 - 2px + qx = 0\), their substitution must give results \(N, M, L\), alternately positive and negative.

In general, the roots of the equation \(x^n - px^{n-1} + qx^{n-2} - rx^{n-3} + \ldots + qx - r = 0\) are limits of the roots of the equation \(x^n - px^{n-1} + qx^{n-2} - rx^{n-3} + \ldots + qx - r = 0\); or of any equation that is deduced from it by multiplying its terms by any arithmetical progression, \(a + b, a + 2b, a + 3b, a + 4b, \ldots\). And, conversely, the roots of this new equation will be limits of the proposed equation

\[ x^n - px^{n-1} + qx^{n-2} + \ldots + qx - r = 0. \]

"If any roots of the equation of the limits are impossible, then must there be some roots of the proposed 'equation impossible.' For as (in p. 110, col. 1, par. 2,) the quantity \(3x^3 - 2px + qx = 0\) was demonstrated to be equal to the product of the excesses of two values of \(x\) above the third supposed equal to \(e\); if any impossible expression be found in those excesses, then there will of consequence be found impossible expressions in these two values of \(x\).

And "from this observation rules may be deduced for discovering when there are impossible roots in equations." Of which we shall treat afterwards.

Besides the method already explained, there are others by which limits may be determined which the root of an equation cannot exceed.

Since the squares of all real quantities are affirmative, it follows, that "the sum of the squares of the roots of any equation must be greater than the square of the greatest root." And the square root of that sum will therefore be a limit that must exceed the greatest root of the equation.

If the equation proposed is \(x^n - px^{n-1} + qx^{n-2} - rx^{n-3} + \ldots + qx - r = 0\), then the sum of the squares of the roots (p. 103, col. 1, par. 1.) will be \(p^2 - 2q\). So that \(\sqrt{p^2 - 2q}\) will exceed the greatest root of that equation.

Or if you find, by p. 103, col. 1, par. 4., the sum of the 4th powers of the roots of the equation, and extract the biquadratic root of that sum, it will also exceed the greatest root of the equation.

If you find a mean proportional between the sum of the squares of any two roots, \(a, b\), and the sum of their biquadrates \((a^4 + b^4)\), this mean proportional will be \(\sqrt[4]{a^6 + a^4b^2 + a^2b^4 + b^6}\). And the sum of the cubes is \(a^3 + b^3\). Now, since \(a^2 - ab - b^2\) is the square of \(a - b\), it must be always positive; and if you multiply it by \(a^2b^2\), the product \(a^4b^2 - 2a^2b^4 + a^2b^4\) will also be positive; and consequently \(a^4b^2 + a^2b^4\) will be always greater than \(2a^2b^2\). Add \(a^6 + b^6\) and we have \(a^6 + a^4b^2 + a^2b^4 + b^6\) greater than \(a^6 + 2a^2b^2 + b^6\); and extracting the root \(\sqrt[4]{a^6 + a^4b^2 + a^2b^4 + b^6}\) greater than \(a^3 + b^3\). And the same may be demonstrated of any number of roots whatever.

Now, if you add the sum of all the cubes taken affirmatively to their sum with their proper signs, they will give double the sum of the cubes of the affirmative roots. And if you subtract the second sum from the first, there will remain double the sum of the cubes of the negative roots. Whence it follows, that "half the sum of the mean proportional betwixt the sum of the squares and the sum of the biquadrates, and of the sum of the cubes of the roots with their proper signs, exceeds the sum of the cubes of the affirmative roots;" and "half their difference exceeds the sum of the cubes of the negative roots." And, by extracting the cube root of that sum and difference, you will obtain limits that shall exceed the sums of the affirmative and of the negative roots. And since it is easy, from what has been already explained, to diminish the roots of an equation so that they all may become negative but one, it appears how, by this means, you may approximate very near to that root. But this does not serve when there are impossible roots.

Several other rules like these might be given for limiting the roots of equations. We shall give one not mentioned by other authors.

In a cubic \(x^3 - px^2 + qx - r = 0\), find \(q^2 - 2pr\), and call it \(e^4\); then shall the greatest root of the equation always be greater than \(\frac{e^4}{3}\), or \(\sqrt[4]{\frac{e^4}{3}}\). And,

In any equation \(x^n - px^{n-1} + qx^{n-2} - rx^{n-3} + \ldots + qx - r = 0\), find \(q^2 - 2pr + \frac{2}{n}\), and extracting the root of the 4th power out of that quantity, it shall always be less than the greatest root of the equation.

**CHAP. XIX. Of the Resolution of Equations, all whose Root are commensurate.**

It was demonstrated in chap. 15, that the last term of any equation is the product of its roots: from which it follows, that the roots of an equation, when commensurable quantities, will be found among the divisors of the last term. And hence we have, for the resolution of equations, this

**Rule.** Bring all the terms to one side of the equation, find all the divisors of the last term, and substitute them successively for the unknown quantity in the equation. So shall that divisor which, substituted in this manner, manner; gives the result = 0, be the root of the pro- posed equation.

For example, suppose this equation is to be resolved,

\[ x^3 - 3ax^2 + 2a^2x - 2a^2b \]

where the last term is \( 2a^2b \), whose simple literal divisors are \( a, b, 2a, 2b \), each of which may be taken either positively or negatively; but as here we find there are variations of signs in the equation, we need only take them positively. Suppose \( x = a \) the first of the divisors, and substituting \( a \) for \( x \), the equation becomes

\[ a^3 - 3a^2 + 2a^2 - 2a^2b \]

or \( 3a^3 - 3a^2 + 3a^2b - 3a^2b = 0 \)

So that, the whole vanishing, it follows, that \( a \) is one of the roots of the equation.

After the same manner, if you substitute \( b \) in place of \( x \), the equation is

\[ b^3 - 3ab^2 + 2a^2b - 2a^2b \]

which vanishing, shews \( b \) to be another root of the equa- tion.

Again, if you substitute \( 2a \) for \( x \), you will find all the terms destroy one another so as to make the sum = 0.

For it will then be

\[ 8a^3 - 12a^3 + 4a^2 - 2a^2b \]

Whence we find, that \( 2a \) is the third root of the equa- tion. Which, after the first two (\( +a, +b \)) had been found, might have been collected from this, that the last term being the product of the three roots, \( +a, +b \), being known, the third must necessarily be equal to the last term divided by the product \( ab \), that is, \( \frac{2a^2b}{ab} = 2a \).

Let the roots of the cubic equation

\[ x^3 - 2x^2 - 33x + 90 = 0 \]

be required.

And first the divisors of 90 are found to be 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. If you substitute 1 for \( x \), you will find \( x^3 - 2x^2 - 33x + 90 = 56 \); so that 1 is not a root of the equation. If you substitute 2 for \( x \), the result will be 24; but, putting \( x = 3 \), you have

\[ x^3 - 2x^2 - 33x + 90 = 27 - 18 - 99 + 90 = 117 - 117 = 0. \]

So that three is one of the roots of the proposed equa- tion. The other affirmative root is \( +5 \); and after you find it, as it is manifest from the equation, that the other root is negative, you are not to try any more divisors ta- ken positively, but to substitute them, negatively taken, for \( x \); and thus you find, that \( -6 \) is the third root.

For putting \( x = -6 \), you have

\[ x^3 - 2x^2 - 33x + 90 = -216 - 72 + 198 + 90 = 0. \]

This last root might have been found by dividing the last term 90, having its sign changed by 15, the product of the two roots already found.

When one of the roots of an equation is found, in order to find the rest with less trouble, divide the propo- sed equation by the simple equation which you are to de- duce from the root already found, and the quotient shall give an equation of a degree lower than the proposed; whose roots will give the remaining roots required.

As for example, the root \( +3 \), first found, gave \( x = 3 \), or \( x - 3 = 0 \), whence dividing thus,

\[ x - 3 \quad x^3 - 2x^2 - 33x + 90 \] \[ x^3 - 3x^2 \] \[ x^2 - 33x + 90 \] \[ x^2 - 3x \] \[ -30x + 90 \] \[ -30x + 90 \]

The quotient shall give a quadratic equation \( x^2 + x - 30 = 0 \), which must be the product of the other two simple equations from which the cubic is generated, and whose roots therefore must be two of the roots of that cubic.

Now the roots of that quadratic equation are easily found, by chap. 12, to be \( +5 \) and \( -6 \). For,

\[ x^2 + x - 30 \] add \( \frac{1}{4} \) \[ x^2 + x + \frac{1}{4} = 30 + \frac{1}{4} = \frac{121}{4} \] \[ \sqrt{\frac{121}{4}} = \frac{11}{2} \] and \( x = \frac{-1}{2} + \frac{11}{2} = +5 \) or \( -6 \).

After the same manner, if the biquadratic \( x^4 - 2x^3 - 25x^2 + 26x + 120 = 0 \) is to be resolved; by substituting the divisors of 120 for \( x \), you will find, that \( +3 \), one of those divisors, is one of the roots; the substitution of 3 for \( x \) giving \( 81 - 54 - 225 + 78 + 120 = 279 - 279 = 0 \).

And therefore, dividing the proposed equation by \( x - 3 \), you must inquire for the roots of the cubic \( x^3 + x^2 - 22x - 40 = 0 \), and finding that \( +5 \), one of the divisors of 40, is one of the roots, you divide that cubic by \( x - 5 \), and the quotient gives the quadratic \( x^2 + 6x + 8 = 0 \), whose two roots are \( -2, -4 \). So that the four roots of the biquadratic are \( +3, +5, -2, -4 \).

This rule supposes that you can find all the divisors of the last term; which you may always do thus.

"If it is a simple quantity, divide it by its least divi- for that exceeds unit, and the quotient again by its least divisor, proceeding thus till you have a quotient that is not divisible by any number greater than unit. This quotient, with these divisors, are the first or simple divisors of the quantity. And the products of the multiplication of any 2, 3, 4, &c. of them are the compound divisor."

As to find the divisors of 60; first I divide by 2, and the quotient 30 again by 2, then the next quo- 15 by 3, and the quotient of this division 5 is not farther divisible by any integer above units; so that the simple divisors are,

\[ 2, 2, 3, 5; \] The products of two, 4, 6, 10, 15. The products of three, 12, 20, 30. The product of all four, . . . . . . . . . 60. The divisors of 90 are found after the same manner;

Simple Simple divisors, 2, 3, 5.

The products of two, 6, 9, 10, 15.

The products of three, 18, 30, 45.

The product of all four, 90.

The divisors of 21ab.

The simple divisors, 3, 7, a, b, b.

The products of two, 21, 3a, 3b, 7a, 7b, ab, bb.

The products of three, 21a, 21b, 3ab, 3bb, 7ab, 7bb, abb.

The products of four, 21ab, 21bb, 3abb, 7abb.

The products of the five, 21abb.

But as the last term may have very many divisors, and the labour may be very great to substitute them all for the unknown quantity, we shall now show how it may be abridged, by limiting to a small number the divisors you are to try. And, first, it is plain, from p. 109, col. 1, par. 4, that "any divisor that exceeds the greatest negative coefficient by unity is to be neglected." Thus, in resolving the equation \(x^4 - 2x^3 - 25x^2 + 26x + 120 = 0\), as 25 is the greatest negative coefficient, we conclude, that the divisors of 120 that exceed 26 may be neglected.

But the labour may be still abridged, if we make use of the rule in the beginning of ch. 18.; that is, if we find the number which substituted in these following expressions,

\[ \begin{align*} & x^4 - 2x^3 - 25x^2 + 26x + 120, \\ & 2x^3 - 3x^2 - 25x + 13, \\ & 6x^2 - 6x - 25, \\ & 2x - 1, \end{align*} \]

will give in them all a positive result: for that number will be greater than the greatest root, and all the divisors of 120 that exceed it may be neglected.

That this investigation may be easier, we ought to begin always with that expression where the negative roots seem to prevail most; as here in the quadratic expression \(6x^2 - 6x - 25\); where finding that 6 substituted for \(x\) gives that expression positive, and gives all the other expressions at the same time positive, I conclude, that 6 is greater than any of the roots, and that all the divisors of 120 that exceed 6 may be neglected.

If the equation \(x^3 + 11x^2 + 10x - 72 = 0\) is proposed, the rule of p. 109, col. 1, par. 4, does not help to abridge the operation; the last term itself being the greatest negative term. But, by chap. 18. we inquire what number substituted for \(x\) will give all these expressions positive.

\[ \begin{align*} & x^3 + 11x^2 + 10x - 72 \\ & 3x^2 + 22x + 10 \\ & 3x + 11. \end{align*} \]

Where the suppositions of \(x = 1\), \(x = 0\), \(x = -1\), give the quantity \(x^3 - x^2 - 10x + 6\) equal to \(-4\), 6, 14; among whose divisors we find only one arithmetical progression

| Suppos. | Result | Divisors | Arith.Prog.decr. | |---------|--------|----------|-----------------| | \(x = 1\) | \(-4\) | 1, 2, 4 | 4 | | \(x = 0\) | 6 | 1, 2, 3 | 6 | | \(x = -1\) | 14 | 1, 2, 7, 14 | |

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; which

Where the labour is very short, since we need only attend to the first expression; and we see immediately that 4 substituted for \(x\) gives a positive result, whence all the divisors of 72 that exceed 4 are to be rejected; and thus by a few trials, we find, that +2 is the positive root of the equation. Then dividing the equation by \(x - 2\), and resolving the quadratic equation that is the quotient of the division, you find the other two roots to be \(-9\), and \(-4\).

But there is another method that reduces the divisors of the last term, that can be useful, still more narrow limits.

Suppose the cubic equation \(x^3 - px^2 + qx - r = 0\) is proposed to be resolved. Transform it to an equation whose roots shall be less than the values of \(x\) by unity, assuming \(y = x - 1\). And the last term of the transformed equation will be \(1 - p + q - r\); which is found by substituting unit, the difference of \(x\) and \(y\), for \(x\), in the proposed equation; as will easily appear from p. 106, col. 1, par. 4, where, when \(y = x - e\), the last term of the transformed equation was \(e^3 - pe^2 + qe - r\).

Transform again the equation \(x^3 - px^2 + qx - r = 0\), by assuming \(y = x + 1\), into an equation whose roots shall exceed the values of \(x\) by unit, and the last term of the transformed equation will be \(-1 - p - q - r\), the same that arises by substituting \(-1\), the difference between \(x\) and \(y\), for \(x\), in the proposed equation.

Now the values of \(x\) are some of the divisors of \(r\), which is the term left when you suppose \(x = 0\); and the values of the \(y\)'s are some of the divisors of \(-1 - p + q - r\), and of \(-1 - p - q - r\), respectively. And these values are in arithmetical progression increasing by the common difference unit; because \(x - 1\), \(x + 1\), are in that progression. And it is obvious the same reasoning may be extended to any equation of whatever degree. So that this gives a general method for the resolution of equations whose roots are commensurable.

**Rule.** Substitute, in place of the unknown quantity, successively the 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 you can find among these divisors, whose common difference is unit; and the values of \(x\) will be among the divisors arising from the substitutions of \(x = 0\) that belong to these progressions. The values of \(x\) will be affirmative when the arithmetical progression increases, but negative when it decreases.

**Examp.** Let it be required to find one of the roots of the equation \(x^3 - x^2 - 10x + 6 = 0\). The operation is thus;

Vol. I. No. 5. succeeding one of its roots must be $-3$. Then dividing the equation by $x+3$, we find the roots of the (quadratic) quotient $x^2 - 4x + 2 = 0$ are $2 \pm \sqrt{2}$.

If it is required to find the roots of the equation $x^3 - 3x^2 - 46x - 72 = 0$, the operation will be thus:

| Suppos. | Results | Divisors | Progressions | |---------|----------|----------|--------------| | $x = 1$ | $1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 20, 24, 30, 40, 60, 120$ | $8, 3, 4, 5$ | | $x = 0$ | $72, 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$ | $9, 2, 3, 4$ | | $x = -1$ | $30, 1, 2, 3, 5, 6, 10, 15, 30$ | $10, 1, 2, 3$ |

Of these four arithmetical progressions having their common difference equal to unit, the first gives $x = 9$, the others give $x = -2, x = -3, x = -4$; all which succeed except $x = -3$; so that the three values of $x$ are $+3, -2, -4$.

**Chap. XX. Of the Resolution of Equations by finding the Equations of a lower Degree that are their Divisors.**

To find the roots of an equation is the same thing as to find the simple equations, by the multiplication of which into one another it is produced, or, to find the simple equations that divide it without a remainder.

If such simple equations cannot be found, yet if we can find the quadratic equations from which the proposed equation is produced, we may discover its roots afterwards by the resolution of these quadratic equations. Or, if neither these simple equations, nor these quadratic equations can be found, yet, by finding a cubic or biquadratic that is a divisor of the proposed equation, we may depress it lower, and make the solution more easy.

Now, in order to find the rules by which these divisors may be discovered, we shall suppose, that

\[ \begin{align*} mx - n \\ mx^2 - nx + r \\ mx^3 - nx^2 + rx - s \end{align*} \]

are the divisors of the proposed equation; and if $E$ represent the quotient arising by dividing the proposed equation by that divisor, then

\[ E \times mx - n \]

Or, $E \times mx^3 - nx^2 + rx - s$, will represent the proposed equation itself. Where it is plain, that "since $m$ is the coefficient of the highest term of the divisors, it must be a divisor of the coefficient of the highest term of the proposed equation."

Next we are to observe, that, supposing the equation has a simple divisor $mx - n$, if we substitute in the equation $E \times mx - n$, in place of $x$, any quantity, as $a$, then the quantity that will result from this substitution will necessarily have $ma - n$ for one of its divisors: since, in this substitution, $mx - n$ becomes $ma - n$.

If we substitute successively for $x$, any arithmetical progression, $a, a-e, a-2e, \ldots$, the quantities that will result from these substitutions will have among their divisors

\[ ma - n, \quad ma - me - n, \quad ma - 2me - n, \]

which are also in arithmetical progression, having their common difference equal to $me$.

If, for example, we substitute for $x$ the terms of this progression, $1, 0, -1$, the quantities that result have among their divisors the arithmetical progression $m - n, -n, -m - n$; or, changing the signs, $n - m, n, n + m$. Where the difference of the terms is $m$, and the term belonging to the supposition of $x = 0$ is $n$.

It is manifest therefore, that when an equation has any simple divisor, if you substitute for $x$ the progression $1, 0, -1$, there will be found amongst the divisors of the sums that result from these substitutions, one arithmetical progression at least; whose common difference will be unit or a divisor $m$ of the coefficient of the highest term, and which will be the coefficient of $x$ in the simple divisor required: and whose term, arising from the supposition of $x = 0$, will be $n$, the other member of the simple divisor $mx - n$.

From which this rule is deduced for discovering such a simple divisor, when there is any.

**Rule.** Substitute for $x$ in the proposed equation successively the numbers $1, 0, -1$. Find all the divisors of the sums that result from this substitution, and take out all the arithmetical progressions you can find amongst them, whose difference is unit, or some divisor of the coefficient of the highest term of the equation. Then suppose $n$ equal to that term of any one progression that arises from the supposition of $x = 0$, and $m$ the foregoing divisor of the coefficient of the highest term of the equation, which $m$ is also the difference of the terms of this progression; so shall you have $mx - n$ for the divisor required.

You may find arithmetical progressions giving divisors that will not succeed; but if there is any divisor, it will be found thus by means of these arithmetical progressions.

If the equation proposed has the coefficient of its highest term $\neq 1$, then it will be $m = 1$, and the divisor will be $x - n$, and the rule will coincide with that given in the end of the last chapter, which we demonstrated after a different manner; for the divisor being $x - n$, the value of $x$ will be $+n$, the term of the progression that is a divisor of the sum that arises from supposing $x = 0$. Of this case we gave examples in the last chapter; and though it is easy to reduce an equation whose highest term has a coefficient different from unit, to one where that coefficient shall be unit, by p. 106, col. 1, par. 6.; yet, without that reduction, the equation may be resolved by this rule, as in the following

**Examp.** Examp. Suppose $8x^3 - 26x^2 + 11x + 10 = 0$, and that it is required to find the values of $x$; the operation is thus;

| Suppos. | Results | Divisors | Progr. | |---------|---------|----------|--------| | $x = 1$ | $+3$ | $3$ | $3$ | | $x = 0$ | $-26$ | $-25$ | $-24$ | | $x = -1$| $-35$ | $-34$ | $-33$ |

The difference of the terms of the last arithmetical progression is $2$, a divisor of $8$, the coefficient of the highest term $x^3$ of the equation, therefore supposing $m = 2$, $n = 5$, we try the divisor $2x - 5$; which succeeding, it follows, that $2x - 5 = 0$, or $x = \frac{5}{2}$.

The quotient is the quadratic $4x^2 - 3x - 2 = 0$, whose roots are $\frac{3 + \sqrt{41}}{8}$, and $\frac{3 - \sqrt{41}}{8}$, so that the three roots of the proposed equation are $\frac{3 + \sqrt{41}}{8}$, $\frac{3 - \sqrt{41}}{8}$, and $2$.

The other arithmetical progression gives $x + 2$ for a divisor; but it does not succeed.

If the proposed equation has no simple divisor, then we are to inquire if it has not some quadratic divisor (if itself is an equation of more than three dimensions.)

An equation having the divisor $mx^2 - nx + r$ may be expressed, as in the first article of this chapter, by $E \times mx^2 - nx + r$; and if we substitute for $x$ any known quantity $a$, the sum that will result will have $ma^2 - na + r$ for one of its divisors; and, if we substitute successively for $x$ the progression $a, a - e, a - 2e, a - 3e$, &c., the sums that arise from this substitution will have

$$ma^2 - na + r$$

$$m \times a^2 - n \times a - e + r$$

$$m \times a^2 - 2e^2 - n \times a - 2e + r$$

$$m \times a^2 - 3e^2 - n \times a - 3e + r,$$

&c.

among their divisors respectively.

These terms are not now, as in the last case, in arithmetical progression; but if you subtract them from the squares of the terms $a, a - e, a - 2e, a - 3e$, &c., multiplied by $m$ a divisor of the highest term of the proposed equation, that is from

$$ma^2$$

$$m \times a^2 - e^2$$

$$m \times a^2 - 2e^2$$

$$m \times a^2 - 3e^2,$$

&c. the remainders,

$$n^2 - r$$

$$n \times a - c - r$$

$$n \times a - 2e - r$$

$$n \times a - 3e - r,$$

&c. shall be in arithmetical progression, having their common difference equal to $nx$.

If, for example, we suppose the assumed progression $a, a - e, a - 2e, a - 3e$, &c. to be $2, 1, 0, -1$, the divisors will be

$$4m - 2n + r$$

$$m - n + r$$

$$+ r$$

$$m + n + r,$$

which subtracted from $4m$, $m$, $0$, $m$, leave $2n - r$

$$n - r$$

$$- r$$

$$- n - r,$$

an arithmetical progression whose difference is $+n$; and whose term, arising from the substitution of $0$ for $x$, is $-r$.

From which it follows, that by this operation, if the proposed equation has a quadratic divisor, you will find an arithmetical progression that will determine to you $n$ and $r$, the coefficient $m$ being supposed known; since it is unit, or a divisor of the coefficient of the highest term of the equation. Only you are to observe, that if the first term $mx^2$ of the quadratic divisor is negative, then in order to obtain an arithmetical progression, you are not to subtract, but add the divisors $-4m - 2n + r$, $-m - n + r$, $+r$, $-m + n + r$, to the terms $4m$, $m$, $0$, $m$.

The general rule therefore, deduced from what we had said, is,

"Substitute in the proposed equation for $x$ the terms $2, 1, 0, -1, \&c.$ successively. Find all the divisors of the sums that result, adding and subtracting them from the squares of these numbers $2, 1, 0, -1, \&c.$ multiplied by a numerical divisor of the highest term of the proposed equation, and take out all the arithmetical progressions that can be found amongst these sums and differences. Let $r$ be that term in any progression that arises from the substitution of $x = 0$, and let $-n$ be the difference arising from subtracting that term from the preceding term in the progression; lastly, let $m$ be the foreaid divisor of the highest term; then shall $mx^2 - nx + r$ be the divisor that ought to be tried." And one or one of the divisors found in this manner will succeed, if the proposed equation has a quadratic divisor.

Chap. XXI. Of the Methods by which you may approximate to the Roots of Numerical Equations by their Limits.

When any equation is proposed to be resolved, first find the limits of the roots (by chap. 17.) as for example, if the roots of the equation $x^2 - 16x + 55 = 0$ are required, you find the limits are $8, 9$, and $17$, by p. 110, col. 2, par. 2.; that is, the least root is between $0$ and $8$, and the greatest between $8$ and $17$.

In order to find the first of the roots, I consider, that if I substitute $0$ for $x$ in $x^2 - 16x + 55$, the result is positive, viz. $+55$, and consequently any number between $0$ and $8$ that gives a positive result, must be less than the least root, and any number that gives a negative result must be greater. Since $0$ and $8$ are the limits, I try $4$, that is, the mean between them, and supposing $x = 4$, $x^2 - 16x + 55 = 16 - 64 + 55 = 7$, from which I conclude that the root is greater than $4$. So that now we have the root limited between $4$ and $8$. Therefore I next try 6, and substituting it for \( x \) we find \( x^3 - 16x + 55 = 36 - 96 + 55 = -5 \); which result being negative, I conclude that 6 is greater than the root required, which therefore is limited now between 4 and 6. And substituting 5, the mean between them, in place of \( x \), I find \( x^3 - 16x + 55 = 25 - 80 + 55 = 0 \); and consequently 5 is the least root of the equation. After the same manner you will discover 11 to be the greatest root of that equation.

Thus by diminishing the greater, or increasing the lesser limit, you may discover the true root when it is a commensurable quantity. But, by proceeding after this manner, when you have two limits, the one greater than the root, the other lesser, that differ from one another but by unit, then you may conclude the root is incommensurable.

We may however, by continuing the operation in fractions, approximate to it. As if the equation proposed is \( x^3 - 6x + 7 = 0 \), if we suppose \( x = 2 \), the result is \( 4 + 12 + 7 = 23 \), which being negative, and the supposition \( x = 0 \) giving a positive result, it follows that the root is between 0 and 2. Next we suppose \( x = 1 \); whence \( x^3 - 6x + 7 = 1 - 6 + 7 = 2 \), which being positive, we infer the root is betwixt 1 and 2, and consequently incommensurable. In order to approximate to it, we suppose \( x = 1 \frac{1}{2} \), and find \( x^3 - 6x + 7 = 2 \frac{1}{2} - 9 + 7 = \frac{1}{2} \); and this result being positive, we infer the root must be betwixt 1 and 1 \(\frac{1}{2}\). And therefore we try \( x = 1 \frac{1}{2} \), and find \( x^3 - 6x + 7 = 2 \frac{1}{2} - 9 + 7 = \frac{1}{2} \), which is negative; so that we conclude the root to be betwixt 1 \(\frac{1}{2}\) and 1 \(\frac{3}{4}\). And therefore we try next 1 \(\frac{3}{4}\), which giving also a negative result, we conclude the root is betwixt 1 \(\frac{1}{2}\) (or 1 \(\frac{3}{4}\)) and 1 \(\frac{3}{4}\). We try therefore 1 \(\frac{3}{4}\), and the result being positive, we conclude that the root must be betwixt 1 \(\frac{3}{4}\) and 1 \(\frac{3}{4}\), and therefore is nearly 1 \(\frac{3}{4}\).

Or you may approximate more easily by transforming the equation proposed into another whose roots shall be equal to 10, 100, or 1000 times the roots of the former, by p. 106, col. 1, par. 4, and taking the limits greater in the same proportion. This transformation is easy; for you are only to multiply the 2d term by 10, 100, or 1000, the 3d term by their squares, the 4th by their cubes, &c. The equation of the last example is thus transformed into \( x^3 - 600x + 70000 = 0 \), whose roots are 100 times the roots of the proposed equation, and whose limits are 100 and 200. Proceeding as before, we try 150, and find \( x^3 - 600x + 70000 = 22500 - 90000 + 70000 = 2500 \), so that 150 is less than the root. You next try 175, which giving a negative

\[ f^3 = 0.000220226 + 0.0023578 + 0.08409g^4 + g^3 \\ - 12f^2 = -0.00942816 - 0.67272g - 12g^2 \\ + 36f = 1.00908 + 36g \\ - 1 = -1. \]

Of which the first two terms, neglecting the rest, give

\[ 35.329637 \times g = 0.0003261374, \quad \text{and} \quad g = \frac{0.0003261374}{35.329637} = 0.0000923127. \]

So that \( f = 0.02803923127 \); and

\[ x = 1 + f = 1.02803923127; \quad \text{which is very near the true root of the equation that was proposed.} \]

If still a greater degree of exactness is required, suppose \( b \) equal to the difference betwixt the true value of \( g \) and and that we have already found, and proceeding as above you may correct the value of \( g \).

It is not only one root of an equation that can be obtained by this method, but, by making use of the other limits, you may discover the other roots in the same manner. The equation of p. 116, col. 2, par. 1, \( x^3 - 15x^2 + 63x - 5 = 0 \), has for its limits 0, 3, 7, 50. We have already found the least root to be nearly 1.028039.

If it is required to find the middle root, you proceed in the same manner to determine its nearest limits to be 6 and 7; for 6 substituted for \( x \) gives a positive, and 7 a negative result. Therefore you may suppose \( x = 6 + f \), and by substituting this value for \( x \) in that equation, you find \( f^3 + 3f^2 - 9f + 4 = 0 \), so that \( f = \frac{4}{9} \) nearly. Or,

\[ f = \frac{4}{9 - 3f^2} \]

it is (by substituting \( \frac{4}{9} \) for \( f \))

\[ f = \frac{4}{9 - 3 \left( \frac{4}{9} \right)^2} = \frac{34}{81}, \text{ whence } x = 6 + \frac{34}{81} \text{ nearly. Which} \]

value may still be corrected as in the preceding articles.

After the same manner you may approximate to the value of the highest root of the equation.

"In all these operations, you will approximate sooner to the value of the root, if you take the three last terms of the equation, and extract the root of the quadratic equation consisting of these three terms."

Then, in p. 116, col. 2, par. 2, instead of the two last terms of the equation \( f^3 - 12f^2 + 26f - 1 = 0 \), if you take the three last, and extract the root of the quadratic \( 12f^2 - 36f + 1 = 0 \), you will find \( f = 0.28031 \), which is much nearer the true value than what you discover by supposing \( 36f - 1 = 0 \).

It is obvious that this method extends to all equations.

"By assuming equations affected with general coefficients, you may, by this method, deduce general rules or theorems for approximating to the roots of proposed equations of whatever degree."

**Chap. XXII. Of the Rules for finding the Number of impossible Roots in an Equation.**

The number of impossible roots in an equation may, for most part, be found by this

**Rule.** Write down a series of fractions whose denominators are the numbers in this progression, 1, 2, 3, 4, 5, &c., continued to the number which expresses the dimension of the equation. Divide every fraction in the series by that which precedes it, and place the quotients in order over the middle terms of the equation. And, if the square of any term multiplied into the fraction that stands over it gives a product greater than the rectangle of the two adjacent terms, write under the term the sign \( + \), but if that product is not greater than the rectangle, write \( - \); and the signs under the extreme terms being \( + \), there will be as many imaginary roots as there are changes of the signs from \( + \) to \( - \), and from \( - \) to \( + \).

Thus, the given equation being \( x^3 + px^2 + 3px - q = 0 \), I divide the second fraction of the series \( \frac{1}{2}, \frac{2}{3}, \frac{3}{4} \), by the first, and the third by the second, and place the quotients \( \frac{1}{2} \) and \( \frac{1}{3} \), over the middle terms in this manner,

\[ \begin{array}{cccc} x^3 + px^2 + 3px - q = 0 \\ + & + & + & + \end{array} \]

Then because the square of the second term multiplied into the fraction that stands over it, that is, \( \frac{1}{2} \times p^2 \times x^2 \), is less than \( 3p^2 \times x^2 \) the rectangle under the first and third terms, I place under the second term the sign \( - \); but as \( \frac{1}{2} \times 9p^2 \times x^2 = (3p^2 \times x^2) \) the square of the third term multiplied into its fraction is greater than nothing, and consequently much greater than \( -pqx^2 \), the negative product of the adjoining terms, I write under the third term the sign \( + \). I write \( + \) likewise under \( x^3 \) and \( -q \) the first and last terms; and finding in the signs, thus marked, two changes, one from \( + \) to \( - \), and another from \( - \) to \( + \), I conclude the equation has two impossible roots.

When two or more terms are wanting in the equation, under the first of such terms place the sign \( - \), under the second \( + \), under the third \( - \), and so on alternately; only when the two terms to the right and left of the deficient terms have contrary signs, you are always to write the sign \( + \) under the last deficient term.

As in the equations

\[ \begin{array}{cccc} x^5 + ax^4 & * & * & * & +a^5 = 0 \\ + & + & - & + & + \end{array} \]

and

\[ \begin{array}{cccc} x^5 + ax^4 & * & * & * & -a^5 = 0 \\ + & + & - & + & + \end{array} \]

the first of which has four impossible roots, and the other two.

Hence too we may discover if the imaginary roots lie hid among the affirmative, or among the negative roots. For the signs of the terms which stand over the signs below that change from \( + \) to \( - \), and \( - \) to \( + \), show, by the number of their variations, how many of the impossible roots are to be reckoned affirmative; and that there are as many negative imaginary roots as there are repetitions of the same sign.

As in the equation

\[ \begin{array}{cccc} x^5 - 4x^4 + 4x^3 - 2x^2 - 5x - 4 = 0 \\ + & + & - & + & + \end{array} \]

the signs \( (- + - ) \) of the terms \( -4x^4 + 4x^3 - 2x^2 \) which stand over the signs \( + - + \) pointing out two affirmative roots, we infer that two impossible roots lie among the affirmative; and the three changes of the signs in the equation \( (+ - + - - ) \) giving three affirmative roots and two negative, the five roots will be one real affirmative, two negative, and two imaginary affirmatives. If the equation had been

\[ \begin{array}{cccc} x^5 - 4x^4 - 4x^3 - 2x^2 - 5x - 4 = 0 \\ + & + & - & + & + \end{array} \]

the terms \( -4x^4 - 4x^3 \) that stand over the first variation \( + - \), show by the repetition of the sign \( - \), that one imaginary root is to be reckoned negative, and the... terms $-2x^2 - 5x$ that stand over the last variation $- +$, give, for the same reason, another negative impossible root; so that the signs of the equation $(+ - - -)$ giving one affirmative root, we conclude that of the four negative roots two are imaginary.

This always holds good, unless, which sometimes may happen, "there are more impossible roots in the equation than are discoverable by the rule."