$x^{2} + x + 1 + 2 k (x^{2} - x - 1) = 0$ is perfect square for how many values of $k$

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Solution

The correct option is A $2$
Now, $x^{2} + x + 1 + 2 k x^{2} - 2 k x - 2 k = 0$
$(1 + 2) x^{2} + (1 - 2 k) x + (1 - 2 k) = 0$ ____ (1)
Now, $(a x + b)^{2} = a^{2} x^{2} + b^{2} + 2 a b x$
Comparing with (1)
$2 \sqrt{(1 - 2 k) (1 + 2 k)} = (1 - 2 k)$
$4 (1 - 4 k^{2}) = 1 + 4 k^{2} - 4 k$
$4 - 16 k^{2} = 1 + 4 k^{2} - 4 k$
$20 k^{2} - 4 k - 3 = 0$
$k = \frac{4 \pm \sqrt{16 + 240}}{40}$
$k = \frac{1}{2}$ or $\frac{- 3}{10}$
$∴$ two values

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