5. If $a, b$ are nonzero real numbers such that $a^{2}+b^{2}=8 a b$, find the value of $\left|\frac{a+b}{a-b}\right|$.
Answer: $\frac{\sqrt{15}}{3}$
Solution: Note that
$$
\left|\frac{a+b}{a-b}\right|=\sqrt{\frac{(a+b)^{2}}{(a-b)^{2}}}=\sqrt{\frac{a^{2}+b^{2}+2 a b}{a^{2}+b^{2}-2 a b}}=\sqrt{\frac{10 a b}{6 a b}}=\frac{\sqrt{15}}{3} .
$$