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Find the values of k for which the following equations have real and equal roots:
$x^2 + k(2x + k - 1) + 2 = 0$
Given:
Given quadratic equation is $x^2 + k(2x+k-1) + 2 = 0$.
To do:
We have to find the values of k for which the roots are real and equal.
Solution:
$x^2 + k(2x+k-1) + 2 = 0$
$x^2+2kx+k^2-k+2=0$
Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,
$a=1, b=2k$ and $c=k^2-k+2$.
The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.
$D=(2k)^2-4(1)(k^2-k+2)$
$D=4k^2-4k^2+4k-8$
$D=4k-8$
The given quadratic equation has real and equal roots if $D=0$.
Therefore,
$4k-8=0$
$4k=8$
$k=\frac{8}{4}$
$k=2$
The value of $k$ is $2$.
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