Question

Find $k$, so that the quadratic equation $(k+1)x^2-2(k+1)x+1=0$ has equal roots.

Answer 1

Since roots are equal.

$\therefore d=0$....(1)

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

$d=b^2-4ac$

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

$d=(-2k+2{)}^2-4k-4$

$d=4k^2+4-8k-4k-4$

$(\because (a+b{)}^2=a^2+b^2+2ab)$

$d=4k^2-12k$

From (1),d=0

$\therefore$ Equation will be

$0=4k^2-12k$

$4k^2=12k$

$k^2=\frac{12}{4}$

$k^2=3k$

$k^2-3k=0$

$k(k-3)=0$

$k=0$ or $k-3=0$

$k=3$

$\therefore$ Values of k are 0,3