Question

The values of k for which the roots are real and equal of the following equation$(4-k)x^2+(2k+4)x+(8k+1)=0$ are 0 and -----.

Answer 1

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

$\Rightarrow$ Here, $a=4-k,b=2k+4,c=8k+1$

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

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

$\Rightarrow$ $(2k+4{)}^2-4(4-k\left)\right(8k+1)=0$

$\Rightarrow$ $4k^2+16k+16-4(32k+4-8k^2-k)=0$

$\Rightarrow$ $4k^2+16k+16-128k-16+32k^2+4k=0$

$\Rightarrow$ $36k^2-108k=0$

$\Rightarrow$ $36k(k-3)=0$

$\Rightarrow$ $36k=0$ and $k-3=0$

$\therefore$ $k=0$ and $k=3$