Question

Find the values of $k$ for which the given equation has real and equal roots
$(k+1)x^2-2(k-1)x+1=0$

Answer 1

The given equation is $(k+1)x^2-2(k-1)x+1=0$

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

Let $D$ be the discriminant of the given equation. Then,

$\therefore D=b^2-4ac={4(k-1)}^2-4(k+1)=4(k^2-3k)$

The given equation will have real and equal roots if

$D=0\Rightarrow 4(k^2-3k)=0\Rightarrow k(k-3)=0\Rightarrow k=0,3$