Computing $\sqrt{2}+\sqrt[4]{2}$ over $\mathbb{F}_3$ I'm computing different minimal polynomials in $\mathbb{F}_3$. I found that I can look at elements of the form $\sqrt[4]{2}$ in $\mathbb{F}_3[\sqrt{2}]$ and  in particular I found the four fourth roots of two in that field are $1+2\sqrt{2},1-2\sqrt{2},2+\sqrt{2},2-\sqrt{2}$ while the square roots of 2 are $\sqrt{2},-\sqrt{2}$. 
Now I'm trying to calculate the minimal polinomial of $\alpha = \sqrt{2}+\sqrt[4]{2}$ over $\mathbb{F}_3$ and I obtain that any $\alpha$  must be a root of $$(X-1)^2(X^2+2X+2)$$ (see the edit below to understand how I obtain it) This gives me at most 3 possible $\alpha$. However, if I sum the elements above I get six different values.
What am I doing wrong?
Edit:
\begin{align*}
\alpha &= \sqrt{2}+\sqrt[4]{2}\\
(\alpha - \sqrt{2})^2 &= \sqrt{2}\\(\alpha^2+2)^2 &= ((2 \alpha + 1)\sqrt{2})^2\\\alpha^4 - 4 \alpha^2 - 8 \alpha + 2 &=0\\\alpha^4 - \alpha^2 + \alpha+2 &=0\\(\alpha-1)^2(\alpha^2+2\alpha^2+2)&=0
\end{align*}
 A: Let me suggest an approach that might be precisely orthogonal to your strategy. We’re tourists in the land of characteristic three, in which $2=-1$, so your first adjunction is of $i=\sqrt{-1}=\sqrt2$. So you’ve adjoined a primitive fourth root of unity, which lies in the field $\Bbb F_9$, the (unique) quadratic extension of $\Bbb F_3$.
But the multiplicative group of $\Bbb F_9$ is of order eight, so that you have the eighth roots of unity free of charge by adjoining $i$. The nonzero elements of $\Bbb F_9$ that aren’t in the set $\{\pm1,\pm i\}$ are $\pm1\pm i$, four in number, and there is your total list of the four possibilities for what you called $\sqrt[4]2$. Just take one of them, $1+i$, whose $\Bbb F_3$-conjugate is $1-i$. Their common minimal polynomial is clearly $X^2-X-1$. The other two are $-1\pm i$, and their common polynomial is $X^2+X-1$.
I think that with this explicit description of the elements of $\Bbb F_9$, you can answer your question. Note that one of the one of the elements “$\sqrt2+\sqrt[4]2$” is in fact equal to $1$.
