Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
What makes you think you can? –  Michael Albanese Jul 12 at 14:47
    
@MichaelAlbanese It's just that the square root inside another made it look more "simplifiable". –  b16db0 Jul 12 at 14:54

3 Answers 3

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.

share|improve this answer
    
Why did you specifically use the form with $\sqrt6$, not any $\sqrt n$? –  Ruslan Jul 12 at 16:29
1  
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 at 16:39
1  
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 at 16:42
2  
@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 at 17:50
    
Thanks for explanation. –  Ruslan Jul 12 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.)

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.