Find the values of k for which the following equations have real and equal roots:
$k^2x^2 - 2(2k - 1)x + 4 = 0$

Given:

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

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,

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

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

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

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

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

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

$D=-16k+4$

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

Therefore,

$-16k+4=0$

$16k=4$

$k=\frac{4}{16}$

$k=\frac{1}{4}$

The value of $k$ is $\frac{1}{4}$.