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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.
Updated on: 10-Oct-2022
9K+ Views
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