Find the values of k for which the quadratic equation

Solution:

(i)

The given equation is $(3 k+1) x^{2}+2(k+1) x+1=0$.

This is of the form $a x^{2}+b x+c=0$, where $a=3 k+1, b=2(k+1)$ and $c=1$.

$\therefore D=b^{2}-4 a c$

$=[2(k+1)]^{2}-4 \times(3 k+1) \times 1$

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

$=4 k^{2}+8 k+4-12 k-4$

$=4 k^{2}-4 k$

The given equation will have real and equal roots if D = 0.

$\therefore 4 k^{2}-4 k=0$

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

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

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

Hence, 0 and 1 are the required values of k.

(ii)

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

$x^{2}+2 k x+k^{2}-k+2=0$

For real and equal roots $D=0$

$D=b^{2}-4 a c=0$

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

$D=4 k^{2}-4 k^{2}+4 k-8=0$

$\Rightarrow 4 k-8=0$

$\Rightarrow k=2$