a2,

ing part of the square of a binomial; namely, x2, which is the square of the first term x, and px, or double the first multiplied by the second, which second is consequently only half of p, or p. To complete the square of the binomial x + p, there must be also the square of the second term, p; but this square may be formed, since p and p are known quantities, and it may be added to the first member, if, to preserve the equality of the two members, it be added at the same time to the second; and this last member will still be a known quantity.

As the square of p is p2, if we add it to the two members of the proposed equation,

we shall have

x2 + px = 4,

x2 + px + { p2 = q + { p2.

The first member of this result is the square of x+p; taking then the root of the two members, we have

x + { p = ± √q+ip2, (106);

by transposition this becomes

I have prefixed the sign + to the second term p, of the root of the first member of the above equation, because the second term of this member is positive; the sign - is to be prefixed in the contrary case, because the х3 square 2 ax+a2 answers to the binomial x a.

Any equation whatever of the second degree may be resolved, by referring it to the general formula,

x2 + px = 9;

or more expeditiously, by performing immediately upon the equation the operations represented under this formula, which, expressed in general terms, are as follows;

To make the first member of the proposed equation a perfect square, by adding to it, and also to the second, the square of half the given quantity, by which the first power of the unknown quantity is multiplied; then to extract the square root of each member, observing, that the root of the first member is composed of the unknown

quantity, and half of the given number, by which the unknown quantity in the second term is multiplied, taken with the sign of this quantity, and that the root of the second member must have the double sign, and be indicated by the sign, if it cannot be obtained directly.

See this illustrated by examples.

110. To find a number such, that if it be multiplied by 7, and this product be added to its square, the sum will be 44.

The number sought being represented by x, the equation will evidently be

x2+7x 44.

In order to resolve this equation, we take 1, half of the coefficient 7, by which x is multiplied; raising it to its square we obtain; this added to each member gives

49

x2 + 7 x + Y = 44 + Y;

reducing the second member to a single term, we have

x2 + 7x + 49 = 225.
7x+49

The root of the first member, according to the rule given above, is, and we find for that of the second ; whence arises the equation

The first value of x solves the question in the sense, in which it was enunciated, since we have by this value

As to the second value of x, since it is affected with the sign the term 7 x, which becomes

must be subtracted from x2, so that the enunciation of the ques

tion resolved by the number 11 is this,

To find a number such, that 7 times this number being subtracted from its square, the remainder will be 44.

The negative value then here modifies the question in a man ner, analogous to what takes place, as we have already seen, in equations of the first degree.

If we put the question, as enunciated above, into an equation, we obtain

ร =

4.

The negative value of x becomes positive, as it satisfies precisely the new enunciation, and the positive value, which does not thus satisfy it, becomes negative.

Hence we see, that in equations of the second degree, algebra unites under the same formula two questions, which have a certain analogy to each other.

111. Sometimes enunciations, which produce equations of the second degree, admit of two solutions. The following is an example;

To find a number such, that if 15 be added to its square, the sum will be equal to 8 times this number.

Let be the number sought; the equation arising from the problem is then

x2 + 15 = 8 x.

This equation reduced to the form prescribed in art. 108, be

There are therefore two different numbers 5 and 3, which fulfil the conditions of the question.

112. Questions sometimes occur, which cannot be resolved precisely in the sense of the enunciation, and which require to be modified. This is the case, when the two roots of the equation are negative, as in the following example,

This equation, which denotes, that the square of the number sought, augmented by 5 times this number, and also by 6, will give a sum equal to 2, evidently cannot be verified by addition, as is implied, since 6 already exceeds 2.

find successively

Indeed, if we resolve it, we

From the sign

x=

with which the numbers 1 and 4 are affected, it may be seen, that the term 5 x must be subtracted from the others, and that the true enunciation for both values is,

To find a number such, that if 5 times this number be subtracted from its square, and 6 be added to the remainder, the result will be 2. This enunciation furnishes the equation,

which gives for x the two positive values 1 and 4.

113. Again, let the following problem be proposed;

To divide a number p into two parts, the product of which shall be equal to q.

If we designate one of these parts by x, the other will be expressed by px, and their product will be p x-x2; we have then the equation

that is, one of the parts will be 7, and the other consequently

If on the contrary, we take 3 for x, the other part will be

10-3 or 7; so that the enunciation, as it stands, admits, strictly speaking, of only one solution, since the second amounts simply to a change in the order of the parts.

If we examine carefully the value of x in the question we have been considering, we shall see that we cannot take any numbers

exceed 2,

indifferently for p and q, for if q exceed

the quantity

q becomes negative, and we are presented

with that species of absurdity mentioned in art. 107.

If we take, for example,

the problem then with these assumptions is impossible.

114. The absurdity of questions, which lead to imaginary roots, is discovered only by the result, and we may wish to determine by characters, which are found nearer to the enunciation, in what consists the absurdity of the problem, which gives rise to that of the solution; this we shall be enabled to do by the fol lowing consideration.

Let d be the difference of the two parts of the proposed num

ber; the greater part will be 2+

d

the less

Р

d

2'

(3); but it

2

therefore, the product of the two parts of the proposed number,

4

whatever they may be, will always be less than 2, or than the square of half their sum, so long as d is any thing but zero; when d is nothing, each of the two parts being equal to their product

ра

will be only. It is then absurd to require it to be greater; and it is just, that algebra should answer in a manner contradictory to established principles, and thereby show, that what is sought does not exist.

What has been proved concerning the equation

x2

furnished by the preceding question, is true of all those of the Alg.

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