Question

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

Answer 1

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

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

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

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

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

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

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

$\Rightarrow$ $4k^2-12k=0$

$\Rightarrow$ $4k(k-3)=0$

$\Rightarrow$ $4k=0$ and $k-3=0$

$\therefore$ $k=3$