Answer
At $~~x = -\frac{d}{2},~~$ then $V_{net} = 0$
Work Step by Step
We can write a general equation for the net electric potential:
$V_{net} = \frac{1}{4\pi~\epsilon_0}\sum \frac{q_i}{r_i}$
It is given that $V_{net} = 0$ at the point $x = \frac{d}{4}$
We can write an expression for the net electric potential at a point due to these two charges:
$V_{net} = \frac{1}{4\pi~\epsilon_0}~(\frac{q_1}{r_1}+\frac{q_2}{r_2})$
$V_{net} = \frac{1}{4\pi~\epsilon_0}~(\frac{q_1}{d/4}+\frac{q_2}{3d/4})$
$V_{net} = \frac{1}{4\pi~\epsilon_0}~(\frac{4q_1}{d}+\frac{4q_2}{3d})$
If $V_{net} = 0$ at $x = \frac{d}{4},$ then $(\frac{4q_1}{d}+\frac{4q_2}{3d}) = 0$
Then: $q_2 = -3q_1$
If $(\frac{q_1}{r_1}-\frac{3q_1}{r_2}) = 0$, then $V_{net} = 0$
If $~~r_2 = 3r_1~~$ then $V_{net} = 0$
Suppose $x$ is a negative value of $x$ where $V_{net} = 0$
Then $~~r_1 = -x~~$ and $~~r_2 = d-x$
We can find $x$:
$r_2 = 3r_1$
$d-x = -3x$
$-2x = d$
$x = -\frac{d}{2}$
At $~~x = -\frac{d}{2},~~$ then $V_{net} = 0$
