Find the value of k so that one of the solutions of kx^2-3x+10 = 0 is twice the other

Answer 1

x = 3/2k + or - (1/2k)sqr(9-40k)

3/2k + (1/2k)sqr(9-40k) = 3/k - 1/ksqr(9-40k)

3/2k-3/k = (1/ksqr(9-40k))/1/2ksqr(90-40k))

-3/2k = 2sqr(9-40k)sqr(90-4k) = 2

k = 2(-2/3) = -4/3