How do I solve this square root problem? I need to solve the following problem:
$$\frac{\sqrt{7+\sqrt{5}}}{\sqrt{7-\sqrt{5}}}=\,?$$
 A: Squaring the fraction gives
$$\frac{7+\sqrt5}{7-\sqrt5}=\frac{1}{44}(7+\sqrt 5)^2$$
so by taking the square root we find
$$\frac{7+\sqrt 5}{2\sqrt{11}}$$
A: $\frac{7+\sqrt{5}}{7-\sqrt{5}}=\frac{7+\sqrt{5}}{7-\sqrt{5}}\cdot\frac{7+\sqrt{5}}{7+\sqrt{5}}=\frac{(7+\sqrt{5})^{2}}{49-5}=\frac{(7+\sqrt{5})^{2}}{44}$
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.
A: \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}
