For what values of $k, (4 - k) x^{2} + (2 k + 4) x + (8 k + 1) = 0$ , is a perfect square.

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Solution

$(4 - k) x^{2} + (2 k + 4) x + (8 k + 1) = 0$
The equation to be perfect square $\sqrt{b^{2} - 4 a c} = 0$ for equation $a x^{2} + b x + c$
So, $(2 k + 4)^{2} - 4 (4 - k) (8 k + 1) = 0$
$⟹ 36 k^{2} - 108 k = 0$
$⟹ k (36 k - 108) = 0$
$⟹ k = 0$ or $k = 3$

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