Question

If the roots of $(k+1)x^2-2(k-1)x+1=0$ are real and equal ; then find the value of $k$ .

Answer 1

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

This equation has real and equal roots

$⟹$Discriminant $=0$

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

$⟹(k-1{)}^2=(k+1)$

$⟹k^2-2k+1=k+1$

$⟹k^2=3k$

$⟹k=0$ or $3$