2. Let $a$ and $b$ be real numbers such that $a>b>0$ and $a^{2}+b^{2}=6ab$. Determine the value of the expression $\frac{a+b}{a-b}$.
Solution. The equation $a^{2}+b^{2}=6 a b$ is equivalent to the equations

$$
(a+b)^{2}=8 a b \text { and }(a-b)^{2}=4 a b
$$

Now, since $a>b>0$ we get $\frac{a+b}{a-b}>0$, so

$$
\frac{a+b}{a-b}=\sqrt{\frac{(a+b)^{2}}{(a-b)^{2}}}=\sqrt{\frac{8 a b}{4 a b}}=\sqrt{2}
$$