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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?

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What makes you think you can? – Michael Albanese Jul 12 '14 at 14:47
@MichaelAlbanese It's just that the square root inside another made it look more "simplifiable". – b16db0 Jul 12 '14 at 14:54
up vote 4 down vote accepted

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.

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Why did you specifically use the form with $\sqrt6$, not any $\sqrt n$? – Ruslan Jul 12 '14 at 16:29
Well, the number $2+\sqrt{6}$ is in the quadratic field $\mathbb{Q}(\sqrt{6})$. Its square root, if it's not in the same field, will be in a quadratic extension of that field, which will be a quartic extension of $\mathbb{Q}$. Therefore, it won't be of the form $a+b\sqrt{n}$, but rather of the form $a+b\theta+c\theta^2+d\theta^3$, where $\theta$ is some quartic irrational with $\sqrt{6}\in\mathbb{Q}(\theta)$. – G Tony Jacobs Jul 12 '14 at 16:39
It's possible that $\theta$ can be expressed in terms of un-nested radicals, but they would be fourth roots, not square roots. Determining whether $\sqrt{\sqrt{6}+2}$ can be written in terms of $\sqrt[4]{6}$ (which seems to be our best fourth-root candidate, because its square is $\sqrt{6}$) is a little trickier, as it involves solving a system of four polynomials in four variables. – G Tony Jacobs Jul 12 '14 at 16:42
@Ruslan, a simpler answer to your question is this: If you square something of the form $a+b\sqrt{n}$, you get something with the same $n$ under the radical. There's no way for $a+b\sqrt{3}$, for example, to have a square of the form $c+d\sqrt{6}$. Just square it out and see. – G Tony Jacobs Jul 12 '14 at 17:50
Thanks for explanation. – Ruslan Jul 12 '14 at 17:54

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.)

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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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