Question

The product of values of k for which the roots are real and equal of the following equation
(k + 1)$x^2$ - 2(3k + 1)x + 8k + 1 = 0 is

Answer 1

$(k+1)x^2-2(3k+1)x+8k+1=0$

$\Rightarrow$ Here, $a=(k+1),b=-2(3k+1),c=8k+1$

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

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

$\Rightarrow$ $[-2(3k+1){]}^2-4(k+1\left)\right(8k+1)=0$

$\Rightarrow$ $4(9k^2+6k+1)-4(8k^2+k+8k+1)=0$

$\Rightarrow$ $36k^2+24k+4-32k^2-4k-4=0$

$\Rightarrow$ $4k^2+20k=0$

$\Rightarrow$ $4k(k+5)=0$

$\Rightarrow$ $4k=0$ and $k+5=0$

$\therefore$ $k=0$ and $k=-5$

$\Rightarrow$ Required product $=0\times (-5)=0$