Question

For what value of $k,(4-k)x^2+(2k+4)x+(8k+1)=0$ is a perfect square:

Answer 1

$(4-k)x^2+(2k+4)x+8k+1=0$

Here, $a=4+k,b=2k+4,c=8k+1$

The given equation will be a perfect square, i.e., real and equal roots if

$D=0$

$b^2-4ac=0$

${(2k+4)}^2-4(4-k)(8k+1)=0$

$4k^2+16+16k-4(32k-8k^2+4-k)=0$

$k^2+4+4k-32k+8k^2-4+k=0$

$9k^2-27k=0$

$k^2-3k=0$

$k(k-3)=0$

$k=0$ or $k=3$

Hence for $k=0$ or $k=3$, the given equation will be a perfect square.