Let $q_1$ be at the origin $(x=0)$ and $q_2$ be at $x=d=0.05$ m. The total potential $U$ at a point $x$ is zero when $U_1 + U_2 = 0$.

$$k\frac{q_1}{|x|} + k\frac{q_2}{|x-d|} = 0 \implies \frac{q_1}{|x|} = -\frac{q_2}{|x-d|}$$ $$\frac{3 \times 10^{-7}}{|x|} = -\frac{-2 \times 10^{-7}}{|x-d|} \implies \frac{3}{|x|} = \frac{2}{|x-d|}$$

Case 1: Between the charges ($0 < x < d$).

$$|x| = x, \quad |x-d| = d-x$$ $$\frac{3}{x} = \frac{2}{d-x} \implies 3(d-x) = 2x \implies 3d = 5x \implies x = \frac{3}{5}d$$

Case 2: On the extension, on the side of the smaller magnitude charge $q_2$ ($x > d$).

$$|x| = x, \quad |x-d| = x-d$$ $$\frac{3}{x} = \frac{2}{x-d} \implies 3(x-d) = 2x \implies 3x - 3d = 2x \implies x = 3d$$

Substituting $d=5$ cm:

$$x_1 = \frac{3}{5}(5 \text{ cm}) = 3 \text{ cm}$$ $$x_2 = 3(5 \text{ cm}) = 15 \text{ cm}$$