# For what value of k,

Question:

For what value of $k,(4-k) x^{2}+(2 k+4) x+(8 k+1)=0$, is a perfect square.

Solution:

The given quadric equation is $(4-k) x^{2}+(2 k+4) x+(8 k+1)=0$, and roots are real and equal

Then find the value of $k$.

Here, $a=(4-k), b=(2 k+4)$ and,$c=(8 k+1)$

As we know that $D=b^{2}-4 a c$

Putting the value of $a=(4-k), b=(2 k+4)$ and, $c=(8 k+1)$

$=(2 k+4)^{2}-4 \times(4-k) \times(8 k+1)$

$=4 k^{2}+16 k+16-4\left(4+31 k-8 k^{2}\right)$

$=4 k^{2}+16 k+16-16-124 k+32 k^{2}$

$=36 k^{2}-108 k+0$

$=36 k^{2}-108 k$

The given equation will have real and equal roots, if $D=0$

Thus,

$36 k^{2}-108 k=0$

$18 k(2 k-6)=0$

$k(2 k-6)=0$

Now factorizing of the above equation

$k(2 k-6)=0$

So, either

$k=0$

or

$(2 k-6)=0$

$2 k=6$

$k=\frac{6}{2}$

$=3$

Therefore, the value of $k=0,3$

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