Answer
$x = -30~cm$
Work Step by Step
The coordinate on the x axis must be to the left of particle 1. Then the electric field due to particle 1 is directed to the left while the electric field due to Particle 2 is directed to the right.
Let $d$ be the distance of the x coordinate to the left of Particle 1.
We can find $d$:
$\frac{1}{4~\pi~\epsilon_0}~\frac{\vert q_1 \vert}{d^2} = \frac{1}{4~\pi~\epsilon_0}~\frac{\vert q_2 \vert}{(d+0.50)^2}$
$\frac{1}{4~\pi~\epsilon_0}~\frac{\vert q_1 \vert}{d^2} = \frac{1}{4~\pi~\epsilon_0}~\frac{\vert 4.00 q_1 \vert}{(d+0.50)^2}$
$(d+0.50)^2 = 4.00d^2$
$d^2+d+0.25 = 4.00d^2$
$4d^2+4d+1 = 16d^2$
$12d^2-4d-1 = 0$
We can use the quadratic formula:
$d = \frac{4 \pm \sqrt{(-4)^2-(4)(12)(-1)}}{(2)(12)}$
$d = \frac{4 \pm \sqrt{64}}{24}$
$d = -0.17~m, 0.50~m$
We can choose the positive value as the distance to the left of Particle 1.
We can find the x coordinate:
$x = 20~cm-50~cm$
$x = -30~cm$
