Page:Elementary algebra (1896).djvu/508

490 Rh Substituting the values of k, a and b, we have the two equations

x^2 -x -1=+ (2x-2);

that is a x^2-3x+1=0, and x^2 + x-3=0;

whence the roots are {3 \pm 5}{2}, {-1 \pm 13}{2}

623. The following solution was given by Descartes in 1637.

Suppose that the biquadratic equation is reduced to the form

x^4 +q x^2 +r x+s=0;

assume x^4 +q x^2 +r x+s= (x^2 +kx+l) (x^2 -k x+m); then by equating coefficients, we have

l+m-k^2=q, k(m -l)=r, lm=s.

From the first two of these equatioas, we obtain

2m=K^2+q+{r}{k}, 2l=k^2+q-{r}{k}

hence substituting in the third equation,

(8 + gk +7)(k + gk -7)=4 sh’, or 4 29k +(¢ -4s)? -7 =0.

This is a cubic in A? which always has one real positive solution [Art. 600]; thus when k? is known the values of / and m are determined, and the solution of the biquadratic is obtained by solving the two quadratics

x^2 +kx+l=0, and x^2 -k x+m=0.

Ex. Solve the equation x^4 -2x^2  +8x -3=0.

Assume x^4 -2x^2  +8x -3=(x^2  +kx+l)(x^2 -k x+m);

then by equating coefficients, we have

l+m-k^2=-2, k(m -l)=8, lm=-8;

whence we obtain (k^2 -2k + 8)(k^3 -2k -8)=- 12k^2,

or k^6-4k^4 +16k^2 -64 =0.