Question

For what value of $k$, $(4-k)x^2+(2k+4)x+(8k+1)$ is a perfect square

Answer 1

The given equation is a perfect square if its discriminant is zero.

$D=b^2-4ac=0$
$\therefore (2k+4{)}^2-4(4-k\left)\right(8k+1)=0$
$\Rightarrow 4(k+2{)}^2-4(4-k\left)\right(8k+1)=0$
$\Rightarrow 4\left[\right(k+2{)}^2-(4-k)(8k+1)]=0$
$\Rightarrow \left[\right(k^2+4k+4)-(-8k^2+31k+4\left)\right]=0\Rightarrow 9k^2-27k=0$
$\Rightarrow 9k(k-3)=0\Rightarrow k=0$ or $k=3$
Hence, the given equation is a perfect square if $k=0$ or $k=3$