If I let $ \alpha = \sqrt{\sqrt5 + 2}$ then the minimal polynomial of $\alpha$ is:
$$f(x) = x^4 - 4x^2 - 1$$
It's also easy to see that this has two non-real roots: $\beta = \sqrt{2-\sqrt5}$ and the negative.
$\Bbb Q(\alpha)/\Bbb Q$ is a degree 4 extension and because of there are imaginary parts getting added $\Bbb Q(\alpha, \beta)/\Bbb Q (\alpha) \geq 2$.
I think it's equal to $2$ because:
$$\alpha^2\beta^2 = (2+\sqrt5)(2-\sqrt5) = -1$$
$$\beta^2 = -\frac1{\alpha^2}$$
So we're actually just adding the square root of an element we already have.
Here is where I get stuck: how do I determine the Galois group of $\Bbb Q(\alpha, \beta)/\Bbb Q$? It has order 8.
I think it's the order 8 dihedral group because it makes intuitive sense, but this might be wrong.
It has a non-normal subfield $\Bbb Q(\alpha)/\Bbb Q$, and subfields correspond to subgroups (fundamental theorem of Galois theory) so maybe it's something to do with the normal subgroups? I don't know what else I can use, there are quite a few groups of order 8.
Also, what actually are the automorphisms in the Galois group? I know complex conjugation always works which gives me one, but what are the others? If it really is just the dihedral group, then I guess I can just put the roots on a square and see what the automorphisms are by rotating and flipping the square, but how can I see those are really field automorphisms?