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

Given:  The equation $x^{2}+k( 2x+k-1)+2=0$ has real and equal roots.

To do: To find the value of $k$.

Solution: 

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

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

$\Rightarrow a=1,\ b=2k\ and\ c=k^{2}-k+2$

For equal roots discriminant, $D=0$

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

$\Rightarrow ( 2k)^{2}-4\times1\times( k^{2}-k+2)$

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

$\Rightarrow 4k-8=0$

$\Rightarrow 4k=8$

$\Rightarrow k=\frac{8}{4}$

$\Rightarrow k=2$

Thus, for $k=2$, the given equation has real and equal roots.
 

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Updated on: 10-Oct-2022

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