$4 \cdot 7$ If the equations $x^{2}+a x+b=0$ and $x^{2}+p x+q=0$ have a non-zero common root, find the quadratic equation whose roots are the distinct roots of these equations.
[Solution] Let the common root be $\alpha$. By Vieta's formulas, the distinct roots are $\frac{b}{\alpha}$ and $\frac{q}{\alpha}$.
Since $\alpha$ is the common root, we have $\left\{\begin{array}{l}\alpha^{2}+a \alpha+b=0, \\ \alpha^{2}+p \alpha+q=0 .\end{array}\right.$
Subtracting the two equations, we get $\quad \alpha=\frac{q-b}{\alpha-p}$.
Thus, the two distinct roots are $\frac{b(a-p)}{q-b}$ and $\frac{q(a-p)}{q-b}$.
Therefore, the required equation is
$$
x^{2}-\frac{(b+q)(a-p)}{q-b} x+\frac{b q(a-p)^{2}}{(q-b)^{2}}=0 .
$$