Question

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

• A. $1,-2$
• B. $0,4$
• C. $-1,3$
• D. $0,3$

Answer 1

The correct option is D $0,3$

$(k+1)x^2-2(k-1)x+1=0$
$D=b^2-4ac$
$=4{(k-1)}^2-4(k+1)$
$=4(k^2-3k)$
For equal roots, $D=0$
$4(k^2-3k)=0$
$\Rightarrow$ $k(k-3)=0$
$\Rightarrow$ $k=0$ or $3$