Prove that there are always two real values of k for which ax^2 + 2bx + c + k(x^2 + 1) is a perfect square, unless a =c, and b=0.
What I did was:
For the equation to be a perfect square, its roots must be equal.
For equal roots, the discriminant must be equal to 0.
Thus, discriminant = (2b)^2 - 4(a+k)(k+c) = 0
Eventually, I got 4b^2 - 4ak - 4k^2 - 4ac - 4kc = 0 but I don't know how to apply this to the problem.
Please, any help?