Find the values of ' $k$ ', for which the quadratic equation $k x^{2} - (8 k + 4) x + 81 = 0$ has real and equal roots?

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Solution

$k x^{2} - (8 k + 4) x + 81 = 0$
Since the equation has real and equal roots, $Δ = 0$ .
That is, $b^{2} - 4 a c = 0$
Here, $a = k, b = - (8 k + 4), c = 81$
That is,
$[- (8 k + 4)]^{2} - 4 (k) (81) = 0$
$\begin{matrix} 64 k^{2} + 64 k + 16 - 324 k = 0 64 k^{2} - 260 k + 16 = 0 \end{matrix}$
Dividing by 4 we get $16 k^{2} - 65 k + 4 = 0$
$(16 k - 1) (k - 4) = 0 then, k = \frac{1}{16} or k = 4$ .

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