Question

If the following quadratic equation has two equal and real roots then find the value of k :
$(k+1)x^2-2(k-1)x+1=0$

Answer 1

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

For equal & real roots, $D=0$

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

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

$\Rightarrow k^2-3k=0$

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

$\Rightarrow k=0,3$.