Explain why if $u=\sqrt{i+2}$ is in $\mathbb{Q}(i)$, an extension of the rational numbers, there exists b... Explain why if $u=\sqrt{i+2}$ is in $\mathbb{Q}(i)$, an extension of the rational numbers, there exists $b \in \mathbb{Q}(i)$ which is a root of $a(x)=-1+8x^2+4x^4$.
I have looked at the minimum polynomial for $u$, and I can easily show why $u$ is not actually in $\mathbb{Q}(i)$, just by showing that the degree of $min(i,\mathbb{Q})$ does not equal the degree of $min(u,\mathbb{Q})$. I can't figure out how $a(x)$ is even related.
Keep in mind this is only my third term of abstract, so I only know the basics.

Well, if $u\in\Bbb Q(i),$ then $u=x+iy$ for some $x,y\in\Bbb Q,$ yes? Now, squaring both sides, we have $$i+2=x^2-y^2+2xyi.$$ Hence, $x^2-y^2=2$ and $y=\frac1{2x},$ so....
