Question

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

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Hint- Here, we will be proceeding by using a discriminant method for quadratic equations.
${x^2} + k\left( {2x + k - 1} \right) + 2 = 0 \Rightarrow {x^2} + \left( {2k} \right)x + k\left( {k - 1} \right) + 2 = 0 \Rightarrow {x^2} + \left( {2k} \right)x + {k^2} - k + 2 = 0{\text{ }} \to {\text{(1)}}$
Since we know that for any general quadratic equation $a{x^2} + bx + c = 0{\text{ }} \to {\text{(2)}}$, the roots will be real and equal only if the value of the discriminant is zero.
For the given quadratic equation, $a = 1$, $b = 2k$ and $c = {k^2} - k + 2$
Discriminant of any quadratic equation is given by $d = {b^2} - 4ac$
Using above formula, discriminant of the given quadratic equation is given by $d = {\left( {2k} \right)^2} - 4 \times 1 \times \left( {{k^2} - k + 2} \right) = 4{k^2} - 4{k^2} + 4k - 8 \\ \Rightarrow d = 4k - 8 \\$
Now for the given quadratic equation to have real and equal root, put $d = 0$
$\Rightarrow d = 0 \Rightarrow 4k - 8 = 0 \Rightarrow 4k = 8 \Rightarrow k = 2$
Therefore, the value of $k$ for which the given quadratic equation will have real and equal roots is 2.