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# Find the values of k for which the following equations have real and equal roots:

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

Given:

Given quadratic equation is $x^2 – 2(k + 1)x + k^2 = 0$.

To do:

We have to find the values of k for which the roots are real and equal.

Solution:

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=1, b=-2(k+1)$ and $c=k^2$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

$D=[-2(k+1)]^2-4(1)(k^2)$

$D=4(k+1)^2-(4)(k^2)$

$D=4(k^2+2k+1)-4k^2$

$D=4k^2+8k+4-4k^2$

$D=8k+4$

The given quadratic equation has real and equal roots if $D=0$.

Therefore,

$8k+4=0$

$8k=-4$

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

$k=\frac{-1}{2}$

The value of $k$ is $\frac{-1}{2}$.

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