An algebra problem by Paola Ramírez

$(a+b)^2=a^2+b^2+2ab=4ab+2ab=6ab$

$(a-b)^2=a^2+b^2-2ab=4ab-2ab=2ab$

Then, $\frac{(a+b)^2}{(a-b)^2}=\frac{6ab}{2ab} \Rightarrow \frac{a+b}{a-b}=\sqrt{3}$

Taking $a^2 - 4ab + b^2 = 0$ , we obtain (via the Quadratic Formula):

$a = \frac{4b \pm \sqrt{16b^2 - 4b^2}}{2} \Rightarrow a = (2 \pm \sqrt{3})b$

Since $0 < b < a$ , only $a = (2 + \sqrt{3})b$ is admissible. Substituting this value into $\frac{a+b}{a-b}$ yields:

$\frac{(2+\sqrt{3}) + 1}{(2+\sqrt{3})-1} = \frac{3+\sqrt{3}}{1+\sqrt{3}} \cdot \frac{1-\sqrt{3}}{1-\sqrt{3}} = \frac{-2\sqrt{3}}{-2} = \boxed{\sqrt{3}}.$