Minimal polynomial for $\alpha=\sqrt{3-2\sqrt{2}}$ over $\mathbb{Q}$ 
Find the minimal polynomial for $\alpha=\sqrt{3-2\sqrt{2}}$ over $\mathbb{Q}$

$\alpha=\sqrt{3-2\sqrt{2}} \implies -\alpha^2-3=2\sqrt{2} \implies (\frac{-\alpha^2-3}{2})^2-2=0$
$\implies (\frac{-\alpha^2-3}{2})^2-2=\frac{\alpha^4+6\alpha^2+9}{4}-2=\alpha^4+6\alpha^2+1=0$
So I believe $P=\alpha^4+6\alpha^2+1$ is a candidate for a minimal polynomial
Let $x=\alpha^2 \implies P=x^2+6x+1=(x+3)^2-8$
Could P be the minimal polynomial? It clearly has $\alpha$ as root, but I am not sure if there is another polynomial of lower degree also having $\alpha$ as a root
Would appreciate your guidance on this

Hint: Note that $(1-\sqrt 2)^2=3-2\sqrt{2}$.
Therefore, $\alpha = \pm(1-\sqrt 2)$. Since $\alpha>0$, we must have $\alpha = -1+\sqrt2$.
Then $\alpha^2 = 3-2\sqrt{2} = -2\alpha +1$ and so the minimal polynomial of $\alpha$ is $x^2+2x-1$.
Clearly, $\mathbb Q(\alpha)=\mathbb Q(\sqrt 2)$, because $\alpha = -1+\sqrt2 \in \mathbb Q(\sqrt 2)$ and $\sqrt 2 = \alpha+1 \in \mathbb Q(\alpha)$.
