Page:Elementary algebra (1896).djvu/507

489 Rh 622. The solution of a biquadratic equation was first obtained by Ferrari, a pupil of Cardan, as follows:

Denote the equation by

ot + 2pa? + ga*+2ra+s=0;

add to each side (ax +b)^2, the quantities a and b being determined so as to make the left side a perfect square; then

xt 4 2 pa? + (q+ a’)a? +2 (r+ ab)e+s4 0? = (ax +b)’.

Suppose that the left side of the equation is equal to (x^2 + px +k)^2; then by comparing the coefficients, we have

pt+2k=q+e, pk=r+ab, P=s+';

by eliminating a and b from these equations, we obtain (pk —7rP = (2k+p?—g) (FP —s),

or 21° — gk? +2 (pr —s)k+ p's —qs—7? = 0.

From this cubic equation one real value of k can always be found [Art. 600]; thus a and b are known. Also

(x^2 + px +k)^2 = (ax + b)^2; x^2 + px +k= (ax + b);

and the values of x are to be obtained from the two quadratics

x^2+ (p—a)x + (k— b) =0,

and x^2+ (p+ a)x+ (k+b)=0.

Ex. Solve the equation ot — 293 — 5224 10x%—8=0.

Add a^2x^2 + 2 abx + b^2 to each side of the equation, and assume

xt — 243 + (a? — 5) a2 42(ab+5)¢+ 02-38 = (e?-—x+hk)*;

then by equating coefficients, we have

. @=2k +6, ab=—k—5, PHP +3; o. (2k+ 6) (24+ 8) = (k+ 5); ©. 2338452? —-4k—-7T=0.

By trial, we find that k =— 1; hence a^2=4, b^2 =4, ab =— 4.

But from the assumption, it follows that

(a —% +k)? = (ax + b)?.