Finding $\frac{a+b}{a-b}$ such that $a^2+b^2=6ab$ For $a,b > 0$ such that $a^2+b^2=6ab$ .How to find $\frac{a+b}{a-b}$

$$(a-b)^2=4 a b = (a+b)^2-(a-b)^2$$
which means that
$$1 = \left ( \frac{a+b}{a-b}\right)^2 - 1$$
or
$$\frac{a+b}{a-b} =\pm \sqrt{2}$$
depending whether $a > b$ or not.
