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Simplify the following expression:

2$\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}$

I would be grateful to get a full response.

Hans Lundmark
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Sierra Zulu
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    You might get some good ideas here: https://math.stackexchange.com/questions/196155/strategies-to-denest-nested-radicals – Hans Lundmark Dec 08 '19 at 07:39
  • Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – Martin R Dec 08 '19 at 08:00

2 Answers2

4

Hint

$$13+\sqrt{48}=12+1+2\sqrt{12}=(\sqrt{12}+1)^2$$

Alternatively

We need $$a^2+b^2=13,a^2b^2=12$$ where $a,b>0$

So, $a^2,b^2$ are the roots of $$t^2-13t+12=0$$

lab bhattacharjee
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Since $13+4\sqrt{3}=(1+2\sqrt{3})^2$ and $4-2\sqrt{3}=(\sqrt{3}-1)^2$, your surd is$$2\sqrt{2+\sqrt{3}}=\sqrt{8+2\sqrt{12}}=\sqrt{2}+\sqrt{6}.$$

J.G.
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