<s>[INST] Degree of the extension $\mathbb{Q}(\sqrt{3 + 2\sqrt{2}})$. The following question is from Abstract Algebra by Dummit and Foote. 

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

This question has been answered here and the correct answer is $2$. But I got the answer as $4$ and couldn't find my mistake. 
My attempt : I started out by constructing a polynomial  $f (x)\in \Bbb{Q}[x]$ such that $\sqrt{3 +
2\sqrt{2}}$ is a root of $f(x)$.
$$x=\sqrt{3 +2\sqrt{2}}\;\implies x^2=3 +2\sqrt{2}.$$
Taking $3$ to the LHS and squaring gives$$x^4-6x^2+9=8.$$
Hence $\sqrt{3 +2\sqrt{2}}$ is a root of  $$f (x)=x^4-6x^2+1.$$
By Rational Root Theorem, the only possible rational roots are $\pm1$ and neither of these satisfy  $f(x)=0$. Hence $f(x)$ is a monic irreducible polynomial which has $\sqrt{3 +2\sqrt{2}}$ as a root. By the uniqueness property of minimal polynomial, $f(x)$ is the minimal polynomial for $\sqrt{3 +2\sqrt{2}}$. Since the degree of the extension is same as the degree of minimal polynomial, we have $[\mathbb{Q}(\sqrt{3 +2\sqrt{2}}):\Bbb Q]=4.$
So,where did I go wrong? Please help me find out my mistake. 
Thank you. 
 [/INST] Because $$3+2\sqrt2=(1+\sqrt2)^2$$ and
$$x^4-6x^2+1=x^4-2x^2+1-4x^2=(x^2-2x-1)(x^2+2x-1).$$
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