Question: Find the minimal polynomial of $alpha=sqrt{3+2sqrt{2}}$ over $mathbb{Q}$

Thoughts: the “standard” method of starting by squaring (twice) to get rid of the square roots, because I don’t have a nice way of showing the resulting polynomial is irreducible. So..

Attempt: It would be great if I could get our $alpha$ in the form $(a+b)^2=a^2+2ab+b^2=sqrt{3+2sqrt{2}}$. So, $3-2sqrt{2}=(sqrt{2})^2+2sqrt{2}+1=(sqrt{2}+1)^2$. So, $alpha=sqrt{3+2sqrt{2}}=sqrt{2}+1$.

So, $(alpha-1)^2=2 implies alpha^2-2alpha+1=2impliesalpha^2-2alpha-1=0$. So let $f(x)=x^2-2x-1$. Since $f(x)$ is irreducible over $mathbb{Q}$ by the Rational Roots Test (since it has degree $2$), $f(x)$ is monic, and $f(alpha)=0$, we conclude that $f(x)$ is the minimal polynomial of $alpha$ over $mathbb{Q}$.

Does this look okay?