The no. of values of k for which $(16x^2+12x+39) + k(9x^2 -2x +11)$ is perfect square is: I wanted to know, how can i determine the no. of values of k for which $(16x^2+12x+39) + k(9x^2 -2x +11)$ is a perfect square.($x \in R$)
I have tried, since $x$ is real the discriminant must be $\geq 0$.
$D = 4(6-k)^2 -4(16+9k)(11k+39)  \geq 0.$
which gives $k \in [-4,-1.5]$. But how can i determine the values where the question will be a perfect square.
Any help appreciated.
Thanks.

we can factor 
$$
ax^2+bx+c
$$
to this:
$$
a\left(x-\frac{-b + \sqrt{b^2-4ac}}{2a}\right)\left(x-\frac{-b - \sqrt{b^2-4ac}}{2a}\right)
$$
Obviously this is a perfect square only when the discriminant is zero.
So 

$$D = 4(6-k)^2 -4(16+9k)(11k+39) = 0.$$

