Find the values of k for which the following equations have real roots
$x^2 - 4kx + k = 0$

Given:

Given quadratic equation is $x^2-4kx+k=0$.

To do:

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

Solution:

$x^2-4kx+k=0$

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

$a=1, b=-4k$ and $c=k$.

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

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

$D=16k^2-4k$

The given quadratic equation has real roots if $D≥0$.

Therefore,

$16k^2-4k≥0$

$4k(4k-1)≥0$

$4k≥0$ and $4k-1≥0$

$k≥0$ and $4k≥1$

$k≥0$ and $k≥\frac{1}{4}$

This implies,

$k≥\frac{1}{4}$

The value of k is greater than or equal to $\frac{1}{4}$.