For what value of k,


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


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 $D=0$


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

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

$k(2 k-6)=0$

Now factorizing of the above equation

$k(2 k-6)=0$

So, either



$(2 k-6)=0$

$2 k=6$



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


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now