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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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