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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.
Given quadratic equation is
${x^2} + k\left( {2x + k - 1} \right) + 2 = 0 \Rightarrow {x^2} + \left( {2k} \right)x + k\left( {k - 1} \right) + 2 = 0 \Rightarrow {x^2} + \left( {2k} \right)x + {k^2} - k + 2 = 0{\text{ }} \to {\text{(1)}}$
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.
On comparing equations (1) and (2), we get
 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.

Note- In these types of problems, we have to compare the given quadratic equation with the general form of any quadratic equation and then in order to have equal and real roots put the discriminant as zero.

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