Determine the degree of the extension $\mathbb{Q}(\sqrt{3 + 2\sqrt{2}})/\mathbb Q$. The following is from a set of exercises and solutions.

Determine the degree of the extension $\mathbb{Q}(\sqrt{3 + 2\sqrt{2}})$ over $\mathbb Q$.

The solution says that the degree is $2$ since
$\mathbb{Q}(\sqrt{2}) = \mathbb{Q}(\sqrt{3 + 2\sqrt{2}})$.
I understand that the LHS is a subset of the RHS since
$$
 \sqrt{2} = \frac{(\sqrt{3 + 2\sqrt{2}})^2 - 3}{2}.
$$
How can the RHS be a subset of the LHS?  In other words, how can $\sqrt{3 + 2\sqrt{2}}$ be in $\mathbb{Q}(\sqrt{2})$? 
The author of the solution mentioned that $(1 + \sqrt{2})^2 = 3 + 2\sqrt{2}$ but I do not see how this helps.  
 A: This is Exercise 10 of $\S$13.2 of Dummit and Foote.  The exercise immediately preceding it is the following:

Let $F$ be a field of characteristic $\neq 2$.  Let $a,b$ be elements of the field $F$ with $b$ not a square in $F$.  Prove that a necessary and sufficient condition for $\sqrt{a + \sqrt{b}} = \sqrt{m} + \sqrt{n}$ for some $m$ and $n$ in $F$ is that $a^2 - b$ is a square in $F$.  Use this to determine when the field $\mathbb{Q}(\sqrt{a + \sqrt{b}})$ ($a,b, \in \mathbb{Q}$) is biquadratic over $\mathbb{Q}$.

Taking $a = 3$ and $b=8$, we find that $a^2 - b = 9 - 8 = 1$ is indeed a square, so the extension is biquadratic.  Furthermore, one can actually determine $m$ and $n$ from $a$ and $b$ (I can say more about this if you like):
$$
m = \frac{a + \sqrt{a^2 - b}}{2} \qquad n = \frac{a - \sqrt{a^2 - b}}{2} \, .
$$
Thus for $a = 3$ and $b = 8$, we have $m = 2$ and $n = 1$, as claimed in the solution.
A: Since $\left(1+\sqrt2\right)^2=3+2\sqrt2$, we have $\sqrt{3+2\sqrt2}=1+\sqrt2$.
Therefore, $\mathbb{Q}\!\left(\sqrt{3+2\sqrt2}\right)=\mathbb{Q}\!\left(1+\sqrt2\right)=\mathbb{Q}\!\left(\sqrt2\right)$
A: One may also note that $\mathbb{Q}(\sqrt{2})\subseteq \mathbb{Q}(\sqrt{3+2\sqrt{2}})$ and $x^2-3-2\sqrt{2}\in \mathbb{Q}(\sqrt{2})[x]$ kills$\sqrt{3+2\sqrt{2}}$. So determining the degree of $\mathbb{Q}(\sqrt{3+2\sqrt{2}})$ over $\mathbb{Q}(\sqrt{2})$ (sufficient to answer the original question by multiplicativity of degree) amounts to figuring out if 
$$
x^2-3-2\sqrt{2}
$$
is irreducible in $\mathbb{Q}(\sqrt{2})$, which is easy since this is asking if it has a root in $\mathbb{Q}(\sqrt{2})$. Indeed, a solution to the system of equations derived from
$$
(a+b\sqrt{2})^2-3-2\sqrt{2}
$$
yields $a=b=1$ as a solution. Thus $\mathbb{Q}(\sqrt{2})=\mathbb{Q}(\sqrt{3+2\sqrt{2}})$ and the field extension is quadratic. 
A: $\alpha=\sqrt{3+2\sqrt{2}}\implies \alpha^4-6\alpha^2+1=0$.
So,$x^4-6x^2+1$ annihilates $\alpha$.Now if we could show that this polynomial is irreducible over $\mathbb Q$ then we are done.Now we use mod $3$ irreducibility test as in $\mathbb Z_3$ this polynomial reduces to $x^4+1=0$ which is irreducible over $\mathbb Z_3$ and hence irreducible over $\mathbb Z$ also.Hence it is irreducible over $\mathbb Q$ also by Gauss's lemma.
