Prove that $K$ is normal over $\mathbb{Q}$ and $K(i)=\mathbb{Q}(i, \sqrt[4]{A})$ Let $K=\mathbb{Q}(\sqrt{-13+2\sqrt{13}})$

Prove that $K$ is normal over $\mathbb{Q}$

Need to show that $K$ is a splitting field of some polynomials in $\mathbb{Q}[X]$. Let $X=\sqrt{-13+2\sqrt{13}}$, then:
$$(X^2+13)^2-52=0$$
Do we just need to show this polynomial is irreducible in $\mathbb{Q}[X]$? 

Prove that there is an $A \in \mathbb{Q}$ such that $K(i)=\mathbb{Q}(i, \sqrt[4]{A})$

$K(i)$ is a splitting field of $(X^2-A)(X^2+A)(X-i)$ in $\mathbb{Q}[X]$ so $K(i)/\mathbb{Q}$ is Galois
I am not sure where to go with this one...
 A: An algebraic field extension $\;K/F\;$ is normal iff $\;K\;$ is the splitting field of a set of polynomials in $\;F[x]\;$ . In this case, the polynomial $\;f(x)=(x^2+13)^2-52=x^4+26x^2+117\in\Bbb Q[x]\;$ is a good candidate, and its roots are:
$$(x^2+13)^2=52\implies x^2+13=\pm2\sqrt13\implies x=\pm\sqrt{-13\pm2\sqrt{13}}$$
Now, if we put $\;\omega:=\sqrt{-13+2\sqrt{13}}\;$ , then clearly $\;\pm\omega\in K:=\Bbb Q(\omega)\;$ , but also:
$$\frac1\omega=\frac{\sqrt{-13-2\sqrt{13}}}{\sqrt{169-52}}=\frac{\sqrt{-13-2\sqrt{13}}}{3\sqrt{13}}\;\;\color{red}{(**)}$$
Now, observe that
$$\omega\in K\implies \omega^2=-13+2\sqrt{13}\in K\implies\sqrt{13}\in K$$
since trivially $\;\Bbb Q\subset K\;$, and from here that
$$\color{red}{(**)}\implies\sqrt{-13-2\sqrt{13}}=\frac{3\sqrt{13}}\omega\in K$$
and this proves that all the roots of $\;f(x)\;$ indeed are in K. Now, why is this field the minimal one that contains all these roots? Well, using Eisenstein's Test for $\;f(x)\;$ with the prime $\;13\;$ we get at once that this is an irreducible poliynomial in $\;\Bbb Q[x]\;$ , so $\;[K=\Bbb Q(\omega):\Bbb Q]=\deg f(x)=4\;$ , and this is the minimal possible degree an extension of the rationals containing all the roots of an irreducible quartic can have.
A: To prove that $K$ is normal over $\mathbb Q$, you need to prove that $$P(X) = (X^2+13)^2 -52$$ is irreducible AND that all its roots lie in $K$.
