# Find the value of k for which the equation

Question:

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

Solution:

Let $x^{2}+k(2 x+k-1)+2=0$ be a quadratic equation.

$x^{2}+k(2 x+k-1)+2=0$

$x^{2}+2 x k+k^{2}-k+2=0$

It is given that, it has real and equal roots.

$\Rightarrow$ Discriminant $=0$

$\Rightarrow b^{2}-4 a c=0$

$\Rightarrow(2 k)^{2}-4(1)\left(k^{2}-k+2\right)=0$

$\Rightarrow 4 k^{2}-4 k^{2}+4 k-8=0$

$\Rightarrow 4 k=8$

$\Rightarrow k=2$

Hence, the value of k is 2.