Solution
= Solution
Put $q_1$ at $x=-L$, $q_2$ at $x=+L$, and $q_3$ at $x=fL$, with $-1<f<1$. For a nonzero test charge, the force vanishes when the electric field vanishes:
$$
\frac{q_1}{(1+f)^2}=\frac{q_2}{(1-f)^2}.
$$
For equal-sign nonzero $q_1,q_2$, take the positive square root $s=\sqrt{q_2/q_1}$. The distances are positive, so $1-f=s(1+f)$ and
$$
\boxed{f=\frac{1-s}{1+s}.}
$$
For $s>1$ the zero moves toward the weaker charge $q_1$; equal charges give $f=0$.