Find the values of k for which the roots are real and equal in each of the following equations:

$(k+1)x^2 - 2(3k+1)x + 8k+1 = 0$

Given:

Given quadratic equation is $(k+1)x^2 - 2(3k+1)x + 8k+1 = 0$.

To do:

We have to find the values of k for which the roots are real and equal.

Solution:

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=k+1, b=-2(3k+1)$ and $c=8k+1$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

$D=[-2(3k+1)]^2-4(k+1)(8k+1)$

$D=4(3k+1)^2-(4k+4)(8k+1)$

$D=4(9k^2+6k+1)-32k^2-4k-32k-4$

$D=36k^2+24k+4-32k^2-36k-4$

$D=4k^2-12k$

The given quadratic equation has real and equal roots if $D=0$.

Therefore,

$4k^2-12k=0$

$4(k^2-3k)=0$

$k^2-3k=0$

$k(k-3)=0$

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

$k=0$ or $k=3$

The values of $k$ are $0$ and $3$.