Equations - roots, values and biquadratic

EQUATIONS, biquadratic, solution of, by Des Cartes's method. Any biquadratic may be reduced to the form .1-4+q x' r x + s = d, by taking away the second term. Suppose this to be made up of the two quadratics, x= + e x + f = 0, and x' e x = 0, where -Fe and —e are made the coefficients of the second terms, be cause the second term of the biquadratic is wanting, that is, the sum of its roots is 0. By multiplying these quadratics together we have a-4 + g +f— . x' + e g—e f. x +fg = 0, which equation is made to coincide with the former by equating their coefficients, or making g + f — e" = q, e g — of = r,andfg=_-a; hence, g + f = + e", also g —f = and by taking the sum and difference of these equations, 2 g = q + + T, and 2f = q + e4— therefore 4 f g = + 2 q e4 — = 4 8, and multiply ing by e, and arranging the terms accord ing to the dimensions of e, + 2 q e4 + — 4 8 X e= — r' 0 ; or making y = 0, y3 + 2 qy'-f- 9' — 48 y—e=0.

By the solution of this cubic, a value of y, and therefore of .,/ y, is obtained ; also f and g, which are respectively equal q±e' to — and —, are known 2 2 6.

the biquadratic is thus resolved into two quadratics, whose roots may be found.

It may be observed, that whichever value of y is used, the same values of x are obtained.

This solution can only be applied to those cases, in which two roots of the biquadratic are possible, andtwo impossi ble.

Let the roots be e, b, c, — a b + c; then since e, the coefficient of the second term of one of the reducing quadratics, is the sum of two roots, its different values area+ b a+c, b+ c, —a+ b, a + c, — b + c, and the values of e', or y, are c7:71-.6 ", ", all of which being possible, the cubic cannot be solved by any direct method. Suppose the roots of the biquadratic to be a + — 1: b — 1; — 1 ; the values of e are 2 a, b + c .

— 1, b— c . —1,—b— c. V - + c — 1 and — 2 a; and the three values of y are Di ", — b 17.7--?", which are all possible, as in the preceding case. But if the roots of the quadratic be a + b V — 1, a — b V-1, — a c, — a — c, the values of y are c +b — two of which are impossible ; therefore the cubic may be solved by Cardan's rule.