<s>[INST] Minimal polynomial of $\sqrt{3+2\sqrt{3}}$ Let $a=\sqrt{3+2\sqrt{3}}$. Then
\begin{align*}
&a=\sqrt{3+2\sqrt{3}}\\
&\implies a^2=3+2\sqrt{3}\\
&\implies a^2-3 = 2\sqrt{3}\\
&\implies (a^2-3)^2 = 4\cdot 3=12\\
&\implies (a^2-3)^2-12=0.
\end{align*}
So, $a$ a root of
$$(x^2-3)^2-12=x^4-6x^2-3.$$
By Eisenstein's criterion, $x^4-6x^2-3$ is irreducible over $\mathbb Q$. So, $a$ is algebraic of degree $4$.

However, I've seen that the degree of $\mathbb Q(\sqrt{3+2\sqrt{3}})$ over $\mathbb Q$ is $2$.
What is wrong with my attempt?
 [/INST] I think you are right.
Another way:
$$\sqrt{3+2\sqrt3}=\sqrt[4]3\sqrt{2+\sqrt3}=\frac{\sqrt[4]3}{\sqrt2}\sqrt{4+2\sqrt3}=$$ $$=\frac{\sqrt[4]3}{\sqrt2}\sqrt{(1+\sqrt3)^2}=\frac{\sqrt[4]3}{\sqrt2}(1+\sqrt3),$$ which gives the same answer.
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