Is $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ a normal extension? I want to reason as to whether $K=\mathbb{Q}(\sqrt{2}, \sqrt{3})$ is a normal extension over $\mathbb{Q}$. There are competing definitions and so my question is re-phrased as:

Is it true that for every irreducible polynomial $f \in \mathbb{Q}[X]$ which has (at least) one root in $K$, $f$ splits completely in $K$?

Any help that can be offered is appreciated.
Yes, it is. A basic theorem in Galois theory says that a finite extension $E \supseteq F$ is normal and separable if and only if $E$ is the splitting field of a separable polynomial $f(X) \in F[X]$.

It suffices to consider $f(X) = (X^2-2)(X^2-3)$. Its splitting field must contain $\sqrt{2}$ and $\sqrt{3}$, and since $K$ is the smallest field containing both of them, $K$ must in fact be the splitting field of $f(X)$. Therefore $K$ is normal over $\mathbb{Q}$.
If we consider $F=\mathbb{Q}[{2^{\frac{1}{3}}}]$, then $f=x^3-2$ has a root in $F$. However, the other two roots are complex. (Why?) But $F\subseteq \mathbb{R}$.
Hence $f$ cannot split over $F
$.