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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Updated on: 10-Oct-2022

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