<s>[INST] 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.
 [/INST] HINT:
Let  $(16x^2+12x+39) + k(9x^2 -2x +11)=(ax+b)^2$
$\implies (16+9k)x^2+(12-2k)x+3+11k=a^2x^2+2abx+b^2$
Comparing the coefficients of $x^2,x,x^0$ 
we have $a^2=16+9k,2ab=12-2k\implies ab=6-k,  b^2=3+11k$
$\implies (6-k)^2=(ab)^2=(16+9k)(3+11k)$ which is a Quadratic Equation in $k$ on re-arrangement 
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