How do I solve this square root problem?

Question

I need to solve the following problem:

$$\frac{\sqrt{7+\sqrt{5}}}{\sqrt{7-\sqrt{5}}}=\,?$$

Answer 1

$\frac{7+\sqrt{5}}{7-\sqrt{5}}=\frac{7+\sqrt{5}}{7-\sqrt{5}}\cdot\frac{7+\sqrt{5}}{7+\sqrt{5}}=\frac{(7+\sqrt{5})^{2}}{49-5}=\frac{(7+\sqrt{5})^{2}}{44}$

Then you are taking the root of the quotient (note for $a,b >0$ $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$) and you get

$\frac{7+\sqrt{5}}{\sqrt{44}}=\frac{7+\sqrt{5}}{2\sqrt{11}}=\frac{\sqrt{11}(7+\sqrt{5})}{22}$

None of the answers proposed is correct: we can use the squared value we have calculated

$\frac{(7+\sqrt{5})^{2}}{44}=\frac{3}{11}+\frac{\sqrt{35}}{22}$

As you can see it is not rational, so you exclude $1$ and $2$

Then $(6 \pm \sqrt{35})^2= 36+35 \pm 12 \sqrt{35}$ and you can see that both of them are incorrect.

Answer 2

Squaring the fraction gives $$\frac{7+\sqrt5}{7-\sqrt5}=\frac{1}{44}(7+\sqrt 5)^2$$ so by taking the square root we find $$\frac{7+\sqrt 5}{2\sqrt{11}}$$

Answer 3

\begin{align} \frac{\sqrt{7+\sqrt{5}}}{\sqrt{7-\sqrt{5}}}&=\frac{\sqrt{7+\sqrt{5}}}{\sqrt{7-\sqrt{5}}}\cdot \frac{\sqrt{7+\sqrt{5}}}{\sqrt{7+\sqrt{5}}}\\ &=\frac{(\sqrt{7+\sqrt{5}})^2}{\sqrt{(7-\sqrt{5})(7+\sqrt{5})}}\\ &=\frac{7+\sqrt{5}}{\sqrt{7^2-(\sqrt{5})^2}}\\ &=\frac{7+\sqrt{5}}{\sqrt{49-5}}\\ &=\frac{7+\sqrt{5}}{\sqrt{44}}\\ &=\frac{7+\sqrt{5}}{2\sqrt{11}}\cdot\frac{\sqrt{11}}{\sqrt{11}}\\ &=\frac{7\sqrt{11}+\sqrt{5\cdot11}}{2(\sqrt{11})^2}\\ &=\frac{7\sqrt{11}+\sqrt{55}}{22} \end{align}