according to wolfram alpha this is true: $sqrt5+sqrt24 = sqrt3+sqrt2$

But exactly how do you show this? I recognize of no rules that works with enhancement inside square roots.

You are watching: Sqrt(5)/3

I noticed I can do this:

$sqrt24 = 2sqrt3sqrt2$

But ns still don"t see just how I should display this due to the fact that $sqrt5+2sqrt3sqrt2 = sqrt3+sqrt2$ still consists of that addition


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Hint: because they room both hopeful numbers, they space equal if, and also only if, their squares room equal.


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On can conveniently discover the denesting using my an easy radical denesting algorithm.

$ w = 5+sqrt24,$ has norm $,n = ww" = 5^2-24 = 1.,$ Subtracting out $,sqrtn=1,$ yields $,4+sqrt24.$

This has trace $,t = 8,,$ so separating $,sqrtt = 2sqrt2,$ out of $,4+sqrt24=4+2sqrt6,$ yields

$$ frac4+2sqrt62sqrt2,=, frac2+sqrt6sqrt2 ,=, sqrt2+sqrt3$$


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HINT______________$1$: $$sqrt24=2sqrt6.$$

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


You type of have to assume the the nested radical deserve to be rewritten together the amount of 2 surds (or radicals) in the form $sqrta+bsqrtc=sqrtx+sqrty$.

So in her question, we have actually $sqrt5+sqrt24=sqrtx+sqrty$. Squaring both sides gives you: $$5+sqrt24=x+y+2sqrtxy$$

This have the right to be conveniently solved through finding two numbers ($x$ and also $y$) that amount to $5$, and multiply to $6$. Numbers $3$ and also $2$ work; for this reason therefore, $$sqrt5+sqrt24=sqrt3+sqrt2$$

NOTE: You have the right to generalize this and also develop a formula.


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