$u = i\sqrt{5-2\sqrt{5}}$, if $f$ is the minimal polynomial of $u$ on $\mathbb{Q}$ and $\mathbb{E}$ is the splitting field of $f$ on the rationals, then $\mathbb{E} = \mathbb{Q}[u]$.
Now $f=x^4+10x^2+5$ and its zeros are $u, -u, v=i\sqrt{5+2\sqrt{5}}, -v$, so the idea is to show that $v \in \mathbb{E}$. It's easy to prove that $\sqrt{5} \in \mathbb{Q}[u]$ and I want to write $v = a + bu$ with $a,b \in \mathbb{Q}$, but I only get the contraddiction $-5 + 2\sqrt{5} = -b^2(5+2\sqrt{5})$. What am I doing wrong?
It is easy to find out that $$x^4+10x^2+5=\left(x+i\sqrt{5-2\sqrt5}\right)\left(x-i\sqrt{5-2\sqrt5}\right)\left(x+i\sqrt{5+2\sqrt5}\right)\left(x+i\sqrt{5+2\sqrt5}\right)$$ where each of the factors are linear in $\mathbb C[x]$. It follows the only fact you have to show is that $$\mathbb Q\left(i\sqrt{5-2\sqrt5}\right)= \mathbb Q\left(\pm i\sqrt{5-2\sqrt5},\space\pm i\sqrt{5+2\sqrt5}\right)$$ for which it is enough to verify that $$10+\left(i\sqrt{5-2\sqrt5}\right)^2=5+2\sqrt5$$