Question

The sum of the values of k for which the roots are real and equal of the following equation
$4x^2-2(k+1)x+(k+4)=0$ is

Answer 1

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

$\Rightarrow$ Here, $a=4,b=-2(k+1),c=k+4$

$\Rightarrow$ It is given that roots are real and equl.

$\therefore$ $b^2-4ac=0$

$\Rightarrow$ $[-2(k+1){]}^2-4(4\left)\right(k+4)=0$

$\Rightarrow$ $4(k^2+2k+1)-16k-64=0$

$\Rightarrow$ $4k^2+8k+4-16k-64=0$

$\Rightarrow$ $4k^2-8k-60=0$

$\Rightarrow$ $k^2-2k-15=0$

$\Rightarrow$ $k^2-5k+3k-15=0$

$\Rightarrow$ $k(k-5)+3(k-50=0$

$\Rightarrow$ $(k-5)(k+3)=0$

$\therefore$ $k=5$ and $k=-3$

$\Rightarrow$ Sum of the values of $k=5+(-3)=2$