<s>[INST] Can this result be simplified? I was trying to solve the value of $x$ in $x^4-4x^2-2 = 0$ in terms of radical. The answer I got is $x=\sqrt{2+\sqrt{6}}$. How can this value be simplified even more, while still expressing it in terms of radical?
 [/INST] Note: I get four solutions
$$
\begin{align}
x^4 - 4x^2 - 2 &= 0 \iff \\
(x^2 - 2)^2 &= 6 \iff \\
x^2 - 2 &= \pm \sqrt{6} \iff \\
x &= \pm \sqrt{2 \pm \sqrt{6}} \\
  &\in \left\{ \pm \sqrt{\sqrt{6} + 2}, \pm i \sqrt{\sqrt{6} - 2} \right\}
\end{align}
$$
which is possible for a 4-th degree polynomial.
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