Question

The quadratic equation x2 −2kx+(k−1) = 0 has roots a and b such that a2 +b2 = 4. Without solving the equation, find the possible values of the real number k

Answer 1

If the roots are a and b, then the equation is
(x-a)(x-b) = 0
x^2 -ax - bx + ab = 0
x^2 - (a+b)x + ab = 0, but we also know that
x^2 - 2kx + k+1 = 0

by comparison:
ab = k-1
a+b = 2k

We know that (a+b)^2 = a^2 + b^2 + 2ab
(a+b)^2 = 4 + 2ab
subbing in our given,
4k^2 = 4 + 2(k-1)
4k^2 = 4 + 2k - 2
4k^2 - 2k -2 = 0
2k^2 - k - 1 = 0
(2k + 1)(k - 1) = 0
k = -1/2 or k = 1