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# Find the values of k for which the roots are real and equal in each of the following equations:

$(4-k)x^2 + (2k+4)x + (8k + 1) = 0$

Given:

Given quadratic equation is $(4-k)x^2 + (2k+4)x + (8k + 1) = 0$.

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,

$a=4-k, b=2k+4$ and $c=8k+1$.

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

$D=(2k+4)^2-4(4-k)(8k+1)$

$D=(2k+4)^2-(16-4k)(8k+1)$

$D=(4k^2+16k+16)-128k-16+32k^2+4k$

$D=4k^2+16k+16+32k^2-124k-16$

$D=36k^2-108k$

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

Therefore,

$36k^2-108k=0$

$36(k^2-3)=0$

$k^2-3=0$

$k(k-3)=0$

$k=0$ or $k-3=0$

$k=0$ or $k=3$

The values of $k$ are $0$ and $3$.

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