Question

find the values of k for which the quadratic equation (3k + 1)$x^2$ + 2(k + 1) x + 1 = 0 has equal roots. Also, find the roots.

Answer 1

Given equation is $y=(3k+1)x^2+2(k+1)x+1=0$

Also, it is given that the equation has equal roots.

Then $D=0\Rightarrow b^2-4ac=0$

$a=3k+1$ $b=2(k+1)$ $c=1$

$b^2-4ac=\left[2\right(k+1){]}^2-4(3k+1)=0$

$\Rightarrow 4k^2+4+8k-12k-4=0$

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

$\Rightarrow 4k(k-1)=0$

$\Rightarrow k=0$ $k=1$

When $k=0$

equation $y=x^2+2x+1=0$

Roots are $x=-1,-1$

When $k=1$

equation $4x^2+4x+1=0$

Roots are $x=\frac{-1}{2},\frac{-1}{2}$