When $\frac{1}{p}-\frac{1}{q}-\frac{1}{p+q}=0$, find the value of $\frac{q}{p}+\frac{p}{q}$.


Solution 1 From the given, we have
$$
(p+q) q-(p+q) p-p q=0 \text {. }
$$

which simplifies to $q^{2}-p q-p^{2}=0$.
Dividing by $p^{2}$, we get $\left(\frac{q}{p}\right)^{2}-\left(\frac{q}{p}\right)-1=0$.
$$
\therefore \quad \frac{q}{p}=\frac{1 \pm \sqrt{5}}{2} \text {. }
$$

When $\frac{q}{p}=\frac{1+\sqrt{5}}{2}$,
$$
\frac{q}{p}+\frac{p}{q}=\frac{1+\sqrt{5}}{2}+\frac{2}{1+\sqrt{5}}=\sqrt{5},
$$

When $\frac{q}{p}=\frac{1-\sqrt{5}}{2}$,
$$
\frac{q}{p}+\frac{p}{q}=\frac{1-\sqrt{5}}{2}+\frac{2}{1-\sqrt{5}}=-\sqrt{5} .
$$

Solution 2 From the above, we get $q^{2}-p q-p^{2}=0$,
then $q=\frac{1 \pm \sqrt{5}}{2} p$,
$$
\therefore \quad \frac{q}{p}+\frac{p}{q}= \pm \sqrt{5} .
$$