# 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

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

• why? elaborate. thanks Jul 20, 2013 at 8:11
• 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? Jul 20, 2013 at 8:15
• 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. Jul 20, 2013 at 8:16
• @Mahdi Khosravi Oh! thank you. Jul 20, 2013 at 8:17

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