Find the value of $k$ for which the equation $x^{2} + k (2 x + k - 1) + 2 = 0$ has real and equal roots.

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Solution

The given equation is $x^{2} + k (2 x + k - 1) + 2 = 0$

$\Rightarrow x^{2} + 2 k x + (k^{2} - k + 2) = 0$

So, $a = 1, b = 2 k, c = k^{2} - k + 2$

We know $D = b^{2} - 4 a c$

$= {(2 k)}^{2} - 4 \times 1 \times (k^{2} - k + 2)$

$= 4 k^{2} - 4 (k^{2} - k + 2)$

$= 4 k^{2} - 4 k^{2} + 4 k - 8$

$= 4 k - 8 = 4 (k - 2)$

For equal roots, $D = 0$

Thus, $4 (k - 2) = 0$

So, $k = 2$

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