Question

Find the values of k for which the given equation has real and equal roots: [3 MARKS]

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

Answer 1

Concept : 1 Mark
Application : 2 Marks

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

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

Then the discriminant D is,

$D=b^2-4ac$

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

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

The given equation will have real and equal roots, if

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

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

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