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.

Thanx

2. Hi,
Originally Posted by xwrathbringerx
Eventually, I got 4b^2 - 4ak - 4k^2 - 4ac - 4kc = 0 but I don't know how to apply this to the problem.
This equality can be written $k^2+(a+c)k+ac-b^2=0$... this is a quadratic equation.

3. ah so the roots of this quadratic equation are the two real values that will make ax^2 + 2bx + c + k(x^2 + 1) a perfect square unless a = c and b = 0 where it is only k^2 = 0? (or have I got the wrong idea)

4. Originally Posted by xwrathbringerx
ah so the roots of this quadratic equation are the two real values that will make ax^2 + 2bx + c + k(x^2 + 1) a perfect square unless a = c and b = 0 where it is only k^2 = 0?
Yes.