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 $\frac{a+b}{a-b}$

We have $$\frac{a^2+b^2}{2ab}=\frac31$$
Applying componendo and dividendo,  $$\frac{a^2+b^2+2ab}{a^2+b^2-2ab}=\frac{3+1}{3-1}$$
$$\implies \left(\frac{a+b}{a-b}\right)^2=2$$
