Can this result be simplified?

Question

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?

Answer 1

Sometimes, nested radicals can be un-nested. It depends whether $2+\sqrt{6}$ is the square of something of the form $a+b\sqrt{6}$. You can check this by writing:

$(a+b\sqrt{6})^2 = 2 + \sqrt{6}$,

which simplifies to:

$(a^2+6b^2) + 2ab\sqrt{6} = 2 + \sqrt{6}$.

Equating rational and irrational parts, we obtain two equations: $a^2 + 6b^2 = 2$ and $2ab=1$. This system of equations does not have any rational solutions, so the original radical cannot be expressed in terms of un-nested square roots.

Answer 2

Put $t = x^2$. Then we have the quadratic equation $$t^2 - 4t - 2 = 0$$

$$t_1, t_2 = \frac{4 \pm \sqrt{ 16+8}}2 = 2\pm \sqrt 6$$ Since $t$ represents $x^2$, then assuming $x \in \mathbb R$, we must throw out $2 - \sqrt 6 < 0$, because no real squared number can be negative. So we solve for $x$, to obtain: $$x = \sqrt t = \pm \sqrt{2+ \sqrt 6}.$$ (Note that your posted solution is only "half the story", as $-\sqrt{2 + \sqrt 6}$ is also a solution.)

This is as good as you'll get in terms of simplification (i.e., there is no valid way in this case to "un-nest" the nested root.)

Answer 3

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.