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Does anyone have a quick method of denesting square roots?

Problem: To quickly denest $\sqrt{5+2 \sqrt{6}} \qquad \tag{1}$

As an aside, this comes from solving the polynomial

$$x^4-10x^2+1=0$$

Since the elements of the Galois group must send an element of the kernel to another element of the kernel, we can show that four different permutations give us exactly the four-group of Klein.

The roots of the $(1)$ can be shown to be $\{\sqrt{2}+\sqrt{3},\sqrt{2}-\sqrt{3},-\sqrt{2}+\sqrt{3},-\sqrt{2}-\sqrt{3}\}$ all of which arise from the denesting of $\pm \sqrt{5\pm2 \sqrt{6}}$.

Other than brute force, does anyone have a quick way to denest square roots, I find this part of the the solution to Galois problems takes a long time, and detracts from the most fun part, the permutations of the symmetries of the roots.

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You could always use the general formula below to denest, $$\sqrt {a \pm \sqrt{b} } = \sqrt{ {a + \sqrt{a^2 -b}}\over2} \pm \sqrt{ {a - \sqrt{a^2 -b}}\over2} $$

Apply to $\sqrt{5+2 \sqrt{6}}=\sqrt{5+\sqrt{24}}$, with $a=5$ and $b=24$.

Or, a quick way to denest $\sqrt{A+2 \sqrt{B}}$ is to inspect whether you could quickly identify $m$ and $n$, such that,

$$A=m+n, \>\>\> B=mn$$

If so,

$$\sqrt{A+2 \sqrt{B}} = \sqrt{m}+\sqrt{n}$$

Apply to $\sqrt{5+2 \sqrt{6}}$, with $m=3$ and $n=2$.

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  • $\begingroup$ Thank you for picking up the typo - I was ahead of myself in it's simplification! $\endgroup$ – higgs Sep 10 '19 at 14:46
  • $\begingroup$ Ah yes, I like your edit, it's a rearrangement of $(a+b)^2=a^2+b^2+2ab$ - that works well in this case. So does your general formula above work in the case of $\sqrt{a \pm c \sqrt{b}}$ for some $c \geq 2$? $\endgroup$ – higgs Sep 10 '19 at 14:56
  • $\begingroup$ yah, I normally prefer the shortcut myself $\endgroup$ – Quanto Sep 10 '19 at 15:01
  • $\begingroup$ For c, you could simply absorb it into b. So, the new $\sqrt{b}$ would be $\sqrt{bc^2}$ $\endgroup$ – Quanto Sep 10 '19 at 15:02
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$$ \sqrt {5 + 2\sqrt 6 } = \sqrt {3 + 2\sqrt 6 + 2} = \sqrt {\left( {\sqrt 3 + \sqrt 2 } \right)^2 } = \sqrt 3 + \sqrt 2 $$

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  • $\begingroup$ Yes Luca, thank you for this. In the instance one can spot a "cross term" it denests quite easily! $\endgroup$ – higgs Sep 10 '19 at 17:40
  • $\begingroup$ Yes. In many cases it is not to tricky to find the right terms which get a square. As last chance there is the general formula above. $\endgroup$ – Luca Goldoni Ph.D. Sep 10 '19 at 19:14
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Try the ansatz $$\sqrt{5+2\sqrt{3}}=a+b\sqrt{3}$$

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  • $\begingroup$ Thanks you, I take it the lowest 'nested' root is always part of the 'denest' if you follow that garbled question? (i.e., always will be a combination of the $\sqrt{x}$ in $\sqrt{\ldots\sqrt{\ldots + a_{n}\sqrt{x}}}$? $\endgroup$ – higgs Sep 10 '19 at 14:47

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