TITLE: Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$
QUESTION [9 upvotes]: I would like to know how to solve part $ii)$ of the following problem:

Let $K /\mathbb{Q}$ be a splitting field for $f(X) =X^4-3X^2+5$.
i) Prove that $f(X)$ is irreducible in $\mathbb{Q}[X]$
ii) Prove that $K$ has degree $8$ over $\mathbb{Q}$.
iii) Determine the Galois group of the extension $K/\mathbb{Q}$ and show how it acts on the roots of $f$.

I've done part i), and have found the roots of $f$ explicitly as:
$$\pm\bigg(\frac{3\pm\sqrt{-11}}{2}\bigg)^{1/2}$$
but am not sure how to show that the extension has degree $8$. If $x_1$ is the root where both of the $\pm$ signs above are $+$ and $x_2$ is the root where only the outer sign is a $+$, then $K = \mathbb{Q}(x_1,x_2)$. By part $i)$, $x_1$ has degree $4$ over $\mathbb{Q}$ and then $x_2$ has degree $1$ or $2$ over $\mathbb{Q}(x_1)$, but I'm not sure how to show that this degree is $2$, or prove the result by other means.
Due to the ordering of the parts, I would expect there to be an answer for ii) that doesn't require computing the entire Galois group of the extension, so would appreciate something along these lines.

REPLY [4 votes]: After some thought, I've found a very short answer that uses a minimal amount of computation:
With notation as in the question, we have: 
$$x_1x_2 = \sqrt{5}, x_1^2 = \frac{3+\sqrt{-11}}{2}.$$ Thus $K$ contains the subfield $F = \mathbb{Q}(\sqrt{5},\sqrt{-11})$ which is Galois and degree $4$ over $\mathbb{Q}$, with Galois group $G'\cong V_4$, generated by $\sigma$ and $\tau$, where $\sigma$ fixes $\sqrt{5}$ and permutes $\pm\sqrt{-11}$ and $\tau$ fixes $\sqrt{-11}$ and permutes $\pm\sqrt{5}$.
If $F=K$, then $x_i \in F$ and then the relations above immediately give that $\sigma\tau(x_1) = \pm x_2$ and $\sigma\tau(x_2)=\mp x_1$. But then $\sigma\tau \in G'$ has order $4$, a contradiction, so we must have that $K$ is strictly larger than $F$, so must be of degree $8$ over $\mathbb{Q}$ as desired.