Answer
$ x \approx 2.72L $
Work Step by Step
$Q_1 = -5q$ on x-axis
$Q_2 = +2q$ on x-axis
$Q_3 = q $ on unknown position to make sure that $F_{3xnet} = 0$
Now we want to find out what is the $F_{3xnet} = 0$
$F_{3xnet}= F_{31} + F_{32} $
$F_{3x \space net} = \frac{kq_3q_1}{(r_{31})^2}+ \frac{kq_3q_2}{(r_{32})^2}$
$F_{3x \space net} = \frac{kq_3(-5q)}{(x)^2}+ \frac{kq_32q}{(x-L)^2}$
$F_{3x \space net} = kq_3q [\frac{-5}{x^2} + \frac{2}{(x-L)^2}] $
Since $F_{3xnet} = 0$,
$[\frac{-5}{x^2} + \frac{2}{(x-L)^2}] = 0 $
$[\frac{5}{x^2} = \frac{2}{(x-L)^2}] $
$\frac{5}{2} = \frac{x^2}{(x-L)^2}$
$\frac{5}{2} = (\frac{x}{x-L})^2$
$ (\frac{x}{x-L}) = \sqrt{\frac{5}{2}} $
$ (\frac{x}{x-L}) = \sqrt{\frac{5}{2}} $
$ x = \frac{L}{1 - \sqrt{2/5}}$
$ x \approx 2.72L $
