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

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Solution

The correct options are
A $0$
D $3$
$(4 - k) x^{2} + (2 k + 4) x + (8 k + 1) = 0$ is a quadratic equation
Any quadratic equation is a perfect square if, discriminant $D = b^{2} - 4 a c = 0$
$b^{2} - 4 a c = 0$
$(2 k + 4)^{2} - 4 (4 - k) (8 k - 1) = 0$
$4 k^{2} + 16 k + 16 - 4 [32 k + 4 - 8 k^{2} - k] = 0$
$4 k^{2} + 16 k + 16 - 128 k - 16 + 32 k^{2} + 4 k = 0$
$\Rightarrow 36 k^{2} - 108 k = 0$
$\Rightarrow 36 k [k - 3] = 0$
$\Rightarrow k = 0$ or $k = 3$

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