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

$4x^2-2(k+1)x+(k+1)=0$

Given:

Given quadratic equation is $4x^2 – 2(k + 1)x + (k + 1) = 0$.

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,

$a=4, b=-2(k+1)$ and $c=k+1$.

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

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

$D=4(k+1)^2-(16)(k+1)$

$D=4(k^2+2k+1)-16k-16$

$D=4k^2+8k+4-16k-16$

$D=4k^2-8k-12$

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

Therefore,

$4k^2-8k-12=0$

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

$k^2-2k-3=0$

$k^2-3k+k-3=0$

$k(k-3)+1(k-3)=0$

$(k-3)(k+1)=0$

$k-3=0$ or $k+1=0$

$k=3$ or $k=-1$

The values of $k$ are $-1$ and $3$.