How do I show that $\sqrt{5+\sqrt{24}} = \sqrt{3}+\sqrt{2}$

Question

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

Answer 1

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

Answer 2

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

Answer 3

Hint: Simply try to square both sides of the equation (since they are both positive numbers).

Answer 4

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

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

Answer 5

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

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

Answer 6

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.