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For what value of k, $(4 - k)x^2 + (2k + 4)x + (8k + 1) = 0$, is a perfect square.
Given:
Given quadratic equation is $(4 - k)x^2 + (2k + 4)x + (8k + 1) = 0$.
To do:
We have to find the value of k for which the given quadratic equation is a perfect square.
Solution:
$(4 - k)x^2 + (2k + 4)x + (8k + 1) = 0$
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 $D=b^2-4ac$.
$D=(2k+4)^2-4(4-k)(8k+1)$
$D=4k^2+16k+16-(16-4k)(8k+1)$
$D=4k^2+16k+16-128k-16+32k^2+4k$
$D=36k^2-108k$
The given quadratic equation is a perfect square if $D=0$.
Therefore,
$36k^2-108k=0$
$36k(k-3)=0$
$36k=0$ or $k-3=0$
$k=0$ or $k=3$
The values of k are $0$ and $3$.
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