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


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


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

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




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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