Question

Find the value of $k$ for which the quadratic equation $(3k+1)x^2-2(k+1)x+1=0$ has equal and real roots.

Answer 1

From question,
$(3k+1)x^2-2(k+1)x+1=0$
On comparing the given equation with general form of quadratic equation $a{x}^2+bx+c=0$, we get
$a=(3k+1),b=-2(k+1)$ and $c=1$

We know that,
when $D=0$, then the equation real equal roots.

$D=b^2-4ac=0$

$\Rightarrow (-2(k+1){)}^2-4(3k+1\left)\right(1)=0$

$\Rightarrow 4(k+1{)}^2-4(3k+1)=0$

$\Rightarrow k^2+1+2k-3k-1=0[\because (a-b{)}^2=a^2+b^2-2ab]$

$\Rightarrow k^2-k=0$

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

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

Therefore, the value of $k$ is either $0$ or $-1$.