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 $a^2+b^2=6ab$.
To both sides, add $2ab$ to obtain, $(a+b)^2 = 8ab$.
Similarly, subtract $2ab$ to obtain, $(a-b)^2=4ab$.
Thus, $\left(\dfrac{a+b}{a-b}\right)^2 = 2$.
So ultimately $\frac{a+b}{a-b}= \pm\sqrt2$.
