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.



2 Answers 2


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.


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

  • $\begingroup$ why? elaborate. thanks $\endgroup$
    – Shobhit
    Jul 20, 2013 at 8:11
  • 1
    $\begingroup$ When a quadratic polynomial is to be a perfect square the roots must be the same and this is possible only when the discriminant is zero(why?). OK? $\endgroup$ Jul 20, 2013 at 8:15
  • $\begingroup$ If the roots of the quadratic are $a$ and $b$ the discriminant is $(a-b)^2$ and is zero precisely when the roots are equal. $\endgroup$ Jul 20, 2013 at 8:16
  • $\begingroup$ @Mahdi Khosravi Oh! thank you. $\endgroup$
    – Shobhit
    Jul 20, 2013 at 8:17


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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .