Find the minimal polynomial of $\alpha=\sqrt{3+2\sqrt{2}}$ over $\mathbb{Q}$ Question: Find the minimal polynomial of $\alpha=\sqrt{3+2\sqrt{2}}$ 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 $$(a+b)^2=a^2+2ab+b^2=\sqrt{3+2\sqrt{2}}.$$  So, $$3+2\sqrt{2}=(\sqrt{2})^2+2\sqrt{2}+1=(\sqrt{2}+1)^2.$$So,
$$\alpha=\sqrt{3+2\sqrt{2}}=\sqrt{2}+1.$$So,
$$(\alpha-1)^2=2\\
 \alpha^2-2\alpha+1=2\\
\alpha^2-2\alpha-1=0.$$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?

Since the comment has answered the OP's stated question, it is open season.
Alternative approach
If $a$ is rational and $b$ is irrational, then $(a+b)$ is irrational.
Further, if $a$ is irrational, then $\sqrt{a}$ is also irrational.
The proof to the 2nd assertion above is that if $\sqrt{a}$ can be rationally expressed as $\frac{P}{Q}$, this implies that $a$ has the rational expression $\frac{P^2}{Q^2}$, which contradicts the premise that $a$ is irrational.
Using the two assertions, you have that since $\sqrt{2}$ is known to be irrational, so is $\left[3 + 2\sqrt{2}\right]$, and therefore, so is $\sqrt{3 + 2\sqrt{2}}.$
This implies that there can not be any polynomial equation of degree 1, with form $x + b = 0,$ with $(b)$ rational, whose root is $\sqrt{3 + 2\sqrt{2}}.$
