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According to wolfram alpha this is true: $\sqrt{5+\sqrt{24}} = \sqrt{3}+\sqrt{2}$

But how do you show this? I know of no rules that works with addition inside square roots.

I noticed I could do this:

$\sqrt{24} = 2\sqrt{3}\sqrt{2}$

But I still don't see how I should show this since $\sqrt{5+2\sqrt{3}\sqrt{2}} = \sqrt{3}+\sqrt{2}$ still contains that addition

Alice Ryhl
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    Welcome to the wonderful world of denesting radicals! There do exist methods for doing this (ex: Landau' Algorithm) but none are particularly fast or trivial. Your best bet in general is just to assume the answer has a particular form (ex: if i assume $\sqrt{5 + \sqrt{24}} = \sqrt{a} + \sqrt{b}$) then you can find it – Sidharth Ghoshal Apr 17 '14 at 15:46
  • possible duplicate of How to simplify a square root – MJD Apr 17 '14 at 15:47
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    This is a particular case of the algebraic identity ($x,y>0$)

    $$\sqrt{x+\sqrt{y}}=\sqrt{\frac{x+\sqrt{x^2-y}}{2}}+\sqrt{\frac{x-\sqrt{x^2-y}}{2}}$$

    $$\sqrt{5+\sqrt{24}}=\sqrt{\frac{5+\sqrt{5^2-24}}{2}}+\sqrt{\frac{5-\sqrt{5^2-24}}{2}}=\sqrt{3}+\sqrt{2}$$

     – Américo Tavares Apr 17 '14 at 16:02

6 Answers6

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Hint: Since they are both positive numbers, they are equal if, and only if, their squares are equal.

Git Gud
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$$5+\sqrt{24}=(\sqrt3)^2+(\sqrt2)^2+2\cdot\sqrt2\cdot\sqrt3=(\sqrt3+\sqrt2)^2$$

lab bhattacharjee
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On can easily discover the denesting using my simple radical denesting algorithm.

$\ w = 5+\sqrt{24}\,$ has norm $\,n = ww' = 5^2-24 = 1.\,$ Subtracting out $\,\sqrt{n}=1\,$ yields $\,4+\sqrt{24}.$

This has trace $\,t = 8,\,$ so dividing $\,\sqrt{t} = 2\sqrt{2}\,$ out of $\,4+\sqrt{24}=4+2\sqrt{6}\,$ yields

$$ \frac{4+2\sqrt{6}}{2\sqrt{2}}\,=\, \frac{2+\sqrt{6}}{\sqrt{2}\ } \,=\, \sqrt{2}+\sqrt{3}$$

Community
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Bill Dubuque
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  • So, you're saying that $\sqrt{a+b\sqrt{n}} = \frac{a+b\sqrt{n} - \sqrt{a^2-nb^2}}{2a}$ or am I confused? – Alice Ryhl Apr 17 '14 at 17:49
  • @Darksonn See the linked answer. The trace is of the intermediate result $,w-n.$ – Bill Dubuque Apr 17 '14 at 17:53
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    Can you correct it in your other answers too? I was a bit confused when I first stumbled upon one of them, since you put it in a way which suggested that one needs to divide out by the trace of the original expression under the square root, instead of the intermediate result. And it didn't work. – SasQ Dec 15 '14 at 15:01
  • Also, what if the norm turns out to be a square root too? Continuing the calculations lead to the answer 1/1 then, which is incorrect. (I see that it works that way when the expression under the square root is not a perfect square. But this has to be mentioned upfront: in which cases your algorithm does apply.) – SasQ Dec 15 '14 at 15:03
  • @Sasq I don't see anything that needs correction. If you obtain another square root then it cannot be denested (by any method). It is a simple exercise to verify that it works. – Bill Dubuque Dec 17 '14 at 17:07
  • My second comment was about a separate problem. The first one was about the problem that when one calculates the trace from the original expression under the square root (as your other comments suggest), the result will be incorrect, because the trace has to be calculated from the intermediate result after the first step of your algorithm (as you correctly stated here in this answer, but incorrectly in others). That's what I meant that needs correction. – SasQ Dec 18 '14 at 00:11
  • @SasQ The English summary $\rm \ , \color{blue}{subtract\ out}\ \sqrt{norm}:,\ \ {\it then}\ \ \color{#0a0}{divide\ out}\ \sqrt{trace}\ \ $ is meant to be memorable. I do realize that, at first sight, this summary it is a bit ambiguous, but this is clarified once one sees the examples. I see nothing incorrect anywhere. – Bill Dubuque Dec 18 '14 at 00:18
  • Sure, now I understand it. But let me show you what exactly was confusing for me at first: http://math.stackexchange.com/a/196181/16191 There you use the letter $w$ in calculating both the norm and the trace, so it seems as if they had to be calculated from the same number (the one under the original square root). Are you sure it doesn't need any correction? Then it will remain confusing. – SasQ Dec 18 '14 at 00:28
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    @SasQ Thanks for elaborating - I didn't realize that was the source of confusion. There $,w,$ is used as a common name for the (complex) argument of the norm and trace functions, just like $,x,y,$ are used as common names of arguments of bivariate functions (e.g. addition and multiplication). – Bill Dubuque Dec 18 '14 at 01:09
  • Yeah, about that... I've seen several of your answers, but I've nowhere seen the theory behind these two functions (norm and trace) explained. Is it elaborated somewhere a bit more? I can see how the norm formula is similar to how the norm of a complex number is calculated, but there are no imaginaries in nested radicals, so I don't quite understand WHY your magic works so well. And why is the norm being subtracted instead of divided out, and the trace quite contrary. Am I missing some obvious part of complex analysis or something? I'd like to know some background of how did you discover that. – SasQ Dec 18 '14 at 01:53
  • @SasQ Norm and trace are met when one studies algebraic extension fields, e.g. in algebraic number theory. If you go on to study Galois theory then that will help you understand better the innate algebraic structure, and why the additive and multiplicative objects are intertwined in the rule. – Bill Dubuque Dec 18 '14 at 02:32
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Hint: Simply try to square both sides of the equation (since they are both positive numbers).

sirfoga
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You kind of have to assume that the nested radical can be rewritten as the sum of two surds (or radicals) in the form $\sqrt{a+b\sqrt{c}}=\sqrt{x}+\sqrt{y}$.

So in your question, we have $\sqrt{5+\sqrt{24}}=\sqrt{x}+\sqrt{y}$. Squaring both sides gives you: $$5+\sqrt{24}=x+y+2\sqrt{xy}$$

This can be easily solved by finding two numbers ($x$ and $y$) that sum to $5$, and multiply to $6$. Numbers $3$ and $2$ work; so therefore, $$\sqrt{5+\sqrt{24}}=\sqrt{3}+\sqrt{2}$$

NOTE: You can generalize this and develop a formula.

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HINT______________$1$: $$\sqrt{24}=2\sqrt{6}.$$

HINT______________$2$: $$a^2=b^2\Leftrightarrow a=b\,\vee a=-b.$$

Jika
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