Show that element is in field extension Call $\alpha = \sqrt{6 + \sqrt{11}}$. 
Show that $\alpha \in \mathcal{Q}(\sqrt{q_1}, \sqrt{q_2})$ for two rational numbers $q_1$ and $q_2$.
I have knowledge of splitting fields and Galois-extensions, but I don't thinkt they're applicable here. Is there a way to use this? Or is this done in a 'random' other way?
 A: $$\sqrt{6+\sqrt{11}}=\frac{1}{\sqrt{2}}\cdot\sqrt{12+2\sqrt{11}}=\frac{\sqrt{2}}{2} \cdot \sqrt{1 + 2\sqrt{11}+11} = \\ = \frac{\sqrt{2}}{2} \cdot \sqrt{(1+\sqrt{11})^2} = \frac{\sqrt{2}}{2} \cdot (1+\sqrt{11})$$
Thus, $\alpha \in \mathbb Q(\sqrt{2}, \sqrt{11})$
A: Here's a possible beginning of a more abstract Galois-theoretic argument:
Suppose you also set $\beta = \sqrt{6 - \sqrt{11}}$; then $\alpha \beta = \sqrt{6^2 - 11} = 5$.  Therefore, $\beta \in \mathbb{Q}(\alpha)$.  Furthermore, both $\pm \alpha$ and $\pm \beta$ are roots of the equation $(x^2 - 6)^2 - 11 = 0$.  Also, the equation $\alpha \beta = 5$ will give restrictions on possible automorphisms of the splitting field of this polynomial; for example, you cannot have an automorphism which sends $\alpha \mapsto \alpha$ and $\beta \mapsto -\beta$.
Continuing to argue along these lines, you should eventually be able to get enough information about the Galois group of the splitting field to conclude that this splitting field is a compositum of two quadratic extensions.
