Degree of the splitting field of $X^4-3X^2+5$ over $\mathbb{Q}$
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.
Over $\mathbb{Q}(\sqrt{-11})$ you have two polynomials $$x^2-3-\sqrt{-11}=0$$ and $$y^2-3+\sqrt{-11}=0$$
Now verify that $$(xy)^2=20$$ so that if $y\in\mathbb{Q}(\sqrt{-11},x)$ then $\sqrt{5} \in \mathbb{Q}(\sqrt{-11},x)$. So it only remains to verify that this cannot happen.
For this one approach is to assume $$\sqrt{5} =a+b\sqrt{-11}+(c+d\sqrt{-11})x$$ and solving this for $x$ we are reduced to $x\in \mathbb{Q}(\sqrt{-11},\sqrt{5})$.
Which would mean $$\sqrt{3+\sqrt{-11}}=a+b\sqrt{-11}+c\sqrt{5}+d\sqrt{-11}\sqrt{5}$$
lets write this as $$\sqrt{3+\sqrt{-11}}=p+q\sqrt{5}$$ where $p=a+b\sqrt{-11}$ and $q=c+d\sqrt{-11}$ elements of $\mathbb{Q}(\sqrt{-11})$.
Squaring we have $$3+\sqrt{-11}=p^2+5q^2+2qp\sqrt{5}$$ if $pq\neq 0$ then $\sqrt{5}\in \mathbb{Q}(\sqrt{-11})$. So $pq=0$. If $q=0$ then $\sqrt{3+\sqrt{-11}}\in \mathbb{Q}(\sqrt{-11})$. And if $p=0$ then $\sqrt{3+\sqrt{-11}}=(c+d\sqrt{-11})\sqrt{5}$ which is also impossible.
Rightly $$ x_1=\sqrt{\frac{3+\sqrt{-11}}{2}}= \frac{1}{2}(\sqrt{\sqrt{20}+3}+i\sqrt{\sqrt{20}-3}) $$ and $$ x_2=\sqrt{\frac{3-\sqrt{-11}}{2}}= \frac{1}{2}(\sqrt{\sqrt{20}+3}-i\sqrt{\sqrt{20}-3}) $$ (any determination thereof) suffice to generate the splitting field.
If $x_2\in\mathbb{Q}(x_1)$, also $x_1+x_2=\sqrt{\sqrt{20}+3}$ belongs to $\mathbb{Q}(x_1)$; however, $\alpha=\sqrt{\sqrt{20}+3}$ has degree $4$ over $\mathbb{Q}$ and therefore $\mathbb{Q}(x_1)=\mathbb{Q}(\alpha)$ would be a subset of the reals, which it is not.
Why does $\alpha$ have degree $4$? We clearly have $\alpha^2\in\mathbb{Q}(\sqrt{5})$, so we just need to show that $\alpha$ cannot be written as $$ \alpha=a+b\sqrt{5} $$ for rational $a$ and $b$. This means $$ 2\sqrt{5}+3=a^2+5b^2+2ab\sqrt{5} $$ and so $b=a^{-1}$, hence $$ a^4-3a^2+5=0 $$ which is exactly the equation we started with and that has no rational root.