RECIPROCAL EQUATIONS.] A L G E 13 K A 551 SECT. XIIT. SOLUTION OF EQUATIONS IN WHICH CERTAIN RELATIONS ARE KNOWN TO EXIST AMONGST THE ROOTS. 102. When the coefficients of the terms of an equation form the same numerical series, whether taken in a direct or an inverted order, as in this example,
- 4 + px 3 + qx 2 +px +1=0,
it may always be transformed into another of a degree de noted by half the exponent of the highest power of the unknown quantity, if that exponent be an even number; or half the exponent diminished by unity, if it be an odd number. The same observation will also apply to any equation of this form, x 4 +pax 3 + qa 2 x 2 +pa 3 x + a 4 = . 103. That we may effect the proposed transformation upon the equation X 1 +px 3 + qx 2 +px +1=0, let every two terms which are equally distant from the extremes be collected into one, and the whole be divided by x 2 , then x" + -5 +p(x + - ) + q = . X* X/ Let us assume , 1 then x- + 2 + = 2 , and x- + -^ = z 2 - 2 x 2 x 2 Thus the equation x 2 + +p(x + -) + q = Q , " 1 I becomes and since x + - = z, therefore x 2 zx + 1 = . Hence, to determine the roots of the biquadratic equation -qx" 2 we form the quadratic 2 2 +jt>* + 2-2 = 0, and find its roots, which, let us suppose denoted by z and / ; then the four roots of the proposed equation will be found by resolving two quadratic equations, viz. x 2 - z x+l =0, x 1 - z"x +1=0. 104. It may be observed, respecting these two quadratic equations, that since the last term of each is unity, if we put a, a to denote the roots of the one, and b, b those of the other, we have from the theory of equations, aa = 1, and therefore a = - : also bb = 1, and b = 7 : now a, a, b, a b b are also the roots of the equation x* +JJX 3 + qx 2 +px +1=0. Hence it appears that the proposed equation has this property, that one-half of its roots are the reciprocals of the other half ; and to that circumstance we are indebted for the simplicity of its resolution. 105. If the greatest exponent of the unknown quantity in a reciprocal equation is an odd number, as in this example, x 5 +px* + qx 3 + qx 2 +px + 1 = 0, the equation will always be satisfied by substituting - 1 for x; hence, - 1 must be a root of the equation, and therefore the equation must be divisible by#+l. Ac cordingly, if the division be actually performed, we shall have in the present case another reciprocal equation, in which the greatest exponent of x is an even number, and therefore resolvable in the manner we have already explained. 106. As an application of the theory of reciprocal equa tions, ]et it be proposed to find x from this equation, where a denotes a given number. Every expression of the form x n + 1 is divisible by x + 1 when n is an odd number. In the present case, the numerator and denominator being divided by x+ 1, the equation becomes X 1 X 3 + X and this again, by proper reduction, becomes (a - I)x* + (4a + I)*- 3 + (6 - 1> 2 + (4a + 1)* + a - 1 = ; 4a+l 6a-l and, putting p = ,q = -! -!
- 4 +px 3 + qx 2 +px +1=0;
a reciprocal equation, resolvable into two quadratics. EQUATIONS WHICH HAVE EQUAL BOOTS. 107. When an equation has two or more equal roots, these may always be discovered, and the equation reduced to another of an inferior degree, by a method of resolution which is peculiar to this class of equations. Although the method of resolution we are to employ will apply alike to equations of every degree, having equal roots, yet, for the sake of brevity, we shall take a biqua dratic equation, x* +px 3 + qx 2 + rx + s = , the roots of which may be generally denoted by a, b, c, and d. Thus we have, from the theory of equations, (x - a)(x - b)(x - c}(x -d} = X* +px 3 + qx 2 + rx + s . Let us put A = (x - a)(x - b)(x - c), A" = (x- a)(x - c)(x- d) , A! = (x - ci)(x - b)(x - d), A " = (x - b)(x - c)(x - d) : then, by actual multiplication, we have A = x 3 - a } - 6 x 2 +ac > x-abc , -o + bc } A! = x 3 a ) + ab ] -b} x 2 +ad > x-abd , -d} + bd } A." y? a + ac } c > x 2 +ad / x acd t -d) + cd } -3c -3d x 2 + bd > x - bed : +cd) and taking the sum of these four equations ; +2ab ~] 2 +2ac +2ad ( +2bc f x + 2bd + 2cd j But since a, b, c, d are the roots of the equation x* + px 3 + qx 2 + rx + s = , we have -3(a + b + c + d) = 3p, 2(ab + ac + ad + bc + bd+ cd) - 2q , (abc + abd + acd + bed) = r ; -abc -abd -acd -bed. This result expressed in its most general, form is as follows : Let A represent the product of all the differences x-a, &c., except one, 2A, the sum of all such products; then 2A = iix n ~ l + (n - l)px n ~ 2 + (n - 2)qx*~ 3 - &c. 108. Let us now suppose that the proposed biquadratic equation has two equal roots, or a = b; then x a = x - b, and since one or other of these equal factors enters each of the four products A, A , A", A ", it is evident that A + A + A" + A ", or ^x 3 + 3px 2 + 2qx + r must be divisible by x - a, or x - b. Thus it appears that if the proposed
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