Find the Galois group of $x^4-5x^2+6$ over $\mathbb{Q}$ 
Find the Galois group of $x^4-5x^2+6$ over $\mathbb{Q}$

$$x^4-5x^2+6 = a^2-5a+6 = (a-3)(a-2) = (x^2-3)(x^2-2)$$
The roots are $\pm \sqrt{3}, \pm \sqrt{2}$, and the Galois group is made of the automorphisms that mixes the roots of $x^2-3$ and $x^2-2$. The possibilities of automorphisms:
$$e: \sqrt{2}\to\sqrt{2}, \sqrt{3}\to\sqrt{3}\\a: \sqrt{2}\to-\sqrt{2}, \sqrt{3}\to\sqrt{3}\\b: \sqrt{2}\to\sqrt{2}, \sqrt{3}\to-\sqrt{3}\\c: \sqrt{2}\to-\sqrt{2}, \sqrt{3}\to-\sqrt{3}$$
Is this the Galois group? 
 A: Good work so far.
When asked to find the Galois group, you generally want to find the descriptive "common name" for the group $\text{Gal}(f)$; this is useful to determine things like whether the group is solvable, etc.  For example, $\text{Gal}(f) \cong S_n$ or $\text{Gal}(f) \cong \mathbb{Z}_n$.  You have the elements of the Galois group correct, but you'll want to determine which group of order $4$ it actually is.
If you haven't already, think about why, for example, $\sqrt{2} \mapsto \sqrt{3}$ is invalid.  Basically, you want to justify why the maps in your list actually are automorphisms and why the list is exhaustive.  
One approach would be to recognize that $f(x) = (x^2 - 2)(x^2 - 3)$ splits inside $\mathbb{Q}(\sqrt{2}, \sqrt{3})$; try to justify that this is the smallest field inside which $f$ splits$^\dagger$.  From here, once you've shown $[\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}] = 4$, then the fact that $|\text{Gal}(f)| = [\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}]$ shows that the list is exhaustive.
In greater generality, there is a "formulaic" approach to classifying the Galois groups of products of irreducible polynomials.  See discussion here.

$^\dagger$ In order to apply the fact that $|\text{Gal}(K/\mathbb{Q})| = [K:\mathbb{Q}]$, we need to know that $K/\mathbb{Q}$ is a Galois extension.  By definition, an extension is Galois $\iff$ it is both separable and normal.  As every extension of $\mathbb{Q}$ is separable, it suffices to show that $K$ is normal; that is, $K$ is the splitting field for a polynomial in $\mathbb{Q}(x)$.  In this case, the polynomial is, of course, $(x^2-2)(x^2-3)$.
A: The roots of $x^4-5x^2+6$ are $\pm\sqrt2$ and $\pm\sqrt3$ as you have pointed out.
Let $r$ be an automorphism of $\Bbb Q(\sqrt2,\sqrt3)$ that fixes $\Bbb Q$.
Then, $r(\sqrt2)^2 = r(2)=2$, so $r(\sqrt2)=\pm\sqrt2$.
Similarly, $r(\sqrt3)=\pm\sqrt3$.
Since an automorphism of a field is completely determiend by its generators, there are only $4$ possible automorphisms. There are three involutions, i.e. three elements with order $2$, so the group is $V_4$.
