Page:Encyclopædia Britannica, Ninth Edition, v. 1.djvu/581

Rh QUADRATIC EQUATIONS.] ALGEBRA 543 x 5 = 0, then x = 5 ; but if x + 7 = 0, then x 7 ; so that the two values of x, or two roots of the equation a? + 2x = 35, are + 5 and 7, as we have already found in a different manner. 75. What has been shown in a particular case is true of any quadratic equation whatever; that is, if x 2 +px = q, or, by bringing all the terms to one side, x 2 +px -(7 = 0, it is always possible to find two factors x-a, and x + b, such, that x 2 +px q = (x - a) (x + b), where a and b are known quantities, which depend only upon p and q, the given numbers in the equation ; and since that to have (x -a)(x + b) = 0, we may either assume x - a = OT x + b = Q, it evidently follows that the condi tions of the equation x 2 +px q = 0, or x 2 +px = q, are alike satisfied by taking x= +a or x -b, From these considerations it follows, that x can have only two values in a quadratic equation; for if it could be supposed to have three or more values, then it would be possible to resolve x^+px q into as many factors, x - c, x d, &c. ; but the product of more than two factors must necessarily contain the third or higher powers of x, and as x&quot; +px q contains no higher power than the second, therefore no such resolution can take place. 70. Solution of Questions which produce Quadratic Equations. Ex. 1. It is required to divide the number 10 into two Buch parts that the sum of their squares may be 58. Let x be the one part; Then, since their sum is 10, the other is 10 - x; .. by the question, x 1 + (10 - x) 2 = 58 ; That is, 2 + 100-20o; + a; 2 = 58, Or 2;r 2 - 20^ = 58-100 = -42; Hence x 2 -Wx= -21. And completing the square, x 2 - lOx + 25 = 25 - 21 = 4 ; Hence, by extracting the root, x - 5 = ^4 = =t 2 , And x = 5 2 , That is, x 7, or x = 3. If we take the greater value of x, viz. 7, the other number 10 - x will be 3 ; and if we take the less value of x, viz. 3, then the other number is 7. Thus it appears, that the greater value of the one number corresponds to the less value of the other ; and indeed this must neces sarily be the case, seeing that both are alike concerned in the question. Hence, the only numbers that will answer the conditions of the question are 7 and 3. Ex. 2. A grazier bought as many sheep as cost him GO, out of which he reserved 15, and sold the remainder for 54, gaining 2s. each upon them. How many sheep did he buy, and what did each cost him ? Suppose that he bought x sheep. Then each would cost him - shillings. x Therefore, since after reserving 15, he sold each of the 1200 remaining x - 15 for + 2 shillings, he would receive for them (x - + 2 j shillings. And, because 54 = 1080 shillings, we have by the question, which, by proper reduction, becomes x* + 45 x = 9000- . 195 45 whence x= -___-_ . And, taking the positive root, shillings, the price of each. 1200 ~^&quot; Ex. 3. It is required to find two numbers, of which the product shall be G, and the sum of their cubes 35. Let x be the one number; then - will be the other. x 216 Therefore, by the question, x 5 + -5- = 35 ; Hence Or a; 6 - 35a; 3 = -216. This equation, by putting X s = y, becomes y2_35y=-216; Hence we find y = 27, ory = S. And since x z = y, .: x = 3, or x = 2 . If x = 3, then the other number is 2, and if x = 2, the other number is 3; so that 2 and 3 are the numbers .re quired. In general, if it be required to find two numbers which are exactly alike concerned in a question that produces a quadratic equation, they will be the roots of that equation. A similar observation applies to any number of quantities which require for their determination the resolution of an equation of any degree whatever. 77. On some Anomalies in the Solution of a Problem ivldcli results in an Equation. From what has preceded, it will be evident that a root of an equation may be a very different thing from the solution of the problem on which the equation is based. It will bo proper to give a few illustrations of this difference before passing on to consider equations in general. (1.) A solution may be inapplicable to the problem as a problem of arithmetic, applying only to the algebraic pro blem. Ex. Find a number such that if it be first increased by 10, and then diminished by 10, the difference of the square roots of the results shall be equal to 10. Let x be the number; then the problem requires that Transposing and squaring, we get Transposing and squaring again, there results x - 10 = 1C, x = 26. Now, it is obvious that 26 does not satisfy the condi tions of the problem, but that it is the solution of another problem, viz., that which substitutes &quot; sum&quot; for &quot; differ ence&quot; in the enunciation. Generally we may remark that an algebraic statement is not definite like an arithmetical one. The algebraic square root of a quantity being + or -, algebra cannot, as arithmetic does, distinguish between the two. The equation &amp;gt;Jx+lO- Jx- 10 = 10 is alge braically the same as /# + 10 + &amp;gt;Jx -10 = 10, &c. (2.) A solution may be inverted, or rather may invert the statement. Ex. Divide 15 into two such parts that the greater shall exceed three times the less by as much as half the less exceeds 3. Let x be the greater, and. . 16 -x the less. The state ment produces the equation, x-3 (15-ff) = - -(15-#)-3, which gives at once ar=ll, so that 11 is the greater, 4 the less part. But, on trying the solution, we find it is not that of the problem given, but of another problem, in Avhich &quot; exceeds&quot; is replaced by &quot; falls short of.&quot; Algebra cannot, in every case, as arithmetic does, distinguish the order of subtraction in stating a difference. Ex. 2. Find a number such that the square root of the difference between its fourth power and its square being
 * 1) = 75, the number of sheep; and consequently