What is the Galois group of $x^4 +5$ over the rationals?

One way is: If $r=(-5)^{1/4}$, so that the splitting field is $L=\mathbb Q(r,i)$, then $Gal(L/\mathbb Q(i))$ is the cyclic group $C_4$, acting via $r\to i^k r$ ($k\in\mathbb Z$ mod 4). To see it: it is certainly a subgroup of this group; if it's a proper subgroup then $r^2$ is fixed, i.e. $r^2\in\mathbb Q(i)$. But that's not possible: $-5=(a+ib)^2$ has evidently no rational solutions. From this we get that $Gal(L/\mathbb Q)$ is $r\mapsto ri^k$, $i\mapsto\pm i$, which is the dihedral group.

[edited to become a 'pure Galois' proof]

This is a tower of quadratic extensions, and there are general tools to deal with this kind of situation. Some time ago, I explained the general machinery in a more complicated case here.

EDIT : In your example :

The number $2$ is not a square in ${\mathbb Q}[\sqrt{5}]$, because neither $2$ nor $2 \times 5$ are squares in $\mathbb Q$ (by “Extension rule 1” , see the reference).

The number $2$ is a square in ${\mathbb Q}[\sqrt[4]{5}]$ iff the equation $x^4+\frac{\sqrt{5}}{4}=0$ has a solution in ${\mathbb Q}[\sqrt{5}]$ (by Extension rule 2) and this is not the case.

So $\sqrt{2}\not\in {\mathbb Q}[\sqrt[4]{5}]$, and the extension $F={\mathbb Q}[\sqrt[4]{5},\sqrt{2}]$ has degree $8$ over $\mathbb Q$.

Since $F \subseteq {\mathbb R}$, we see that $L=F[i]={\mathbb Q}[\sqrt[4]{5},\sqrt{2},i]$ has degree $16$ over $\mathbb Q$.

The elements of ${\sf Gal}(L/{\mathbb Q})$ are easily described by their action on $\sqrt[4]{5},\sqrt{2},i$. Then, it is not hard to find the subgroup fixing $E$, and to deduce that $[E:\mathbb Q]=8$.

This may not be very useful to you, but here goes anyway.

We see that $-5$ is of order four in the field $K=\mathbb{F}_{13}$. Therefore its fourth roots are necessarily of order sixteen in some extension field of $K$. The smallest extension field of $K$ that contains sixteenth roots of unity is the field $\mathbb{F}_{13^4}$. Therefore $x^4+5$ is irreducible in $K[x]$. By a well known result this implies that the action of the Galois group of $x^4+5$ over $\mathbb{Q}$ as a group of permutations on its roots contains a 4-cycle. This rules out the Klein four group, and settles your question.