Find the value of $^{'} k^{'}$ if the expression $(4 - k) x^{2} + 2 (k + 2) x + 8 k + 1$ is a perfect square

$0$

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$1$

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$2$

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$3$

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Solution

The correct option is D
$3$

$(4 - k) x^{2} + 2 (k + 2) x + 8 k + 1$
For the given expression to be a perfect square, discriminant needs to be $= 0$
$\Rightarrow (2 (k + 2))^{2} - 4 (8 k + 1) (4 - k) = 0$
$\Rightarrow 4 [(k^{2} + 4 k + 4) - (32 k + 4 - 8 k^{2} - k)] = 0$
$\Rightarrow k^{2} + 4 k + 4 - 32 k - 4 + 8 k^{2} + k)] = 0$
$\Rightarrow 9 k^{2} - 27 k = 0$
$\Rightarrow 9 k (k - 3) = 0$
$\Rightarrow k = 0$ or $k = 3$

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