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

Given, $x^{2} + k (2 x + k - 1) + 2 = 0$
Simplify above equation:
$x^{2} + 2 k x + (k^{2} - k + 2) = 0$
Compare given equation with the general form of quadratic equation, which $a x^{2} + b x + c = 0$
here, $a = 1, b = 2 k, c = (k^{2} - k + 2)$
Find discriminant:
$D = b^{2} - 4 a c$
$= (2 k)^{2} - 4 \times 1 \times (k^{2} - k + 2)$
$= 4 k^{2} - 4 k^{2} + 4 k - 8$
$= 4 k - 8$
Since roots are real and equal (given)
Put $D = 0$
$4 k - 8 = 0$
$k = 2$
Hence, the value of $k$ is $2$ .

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