# Questions tagged [nested-radicals]

In algebra, a nested radical is a radical expression (one containing a square root sign, cube root sign, etc.) that contains (nests) another radical expression.

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### Confusing regarding a nested radical equation

For all $a\in\Bbb R$ solve the equation $$\sqrt{x^2+4a^2\sqrt{x+a}}=x+2a$$ It is immediate to see that we got the restriction $x\geqslant-a$ (even though not given I assume that this equation is ...
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### Simplify $\sqrt {9 + 2(1 + \sqrt {3})(1 + \sqrt {7})}$

Simplify $\sqrt {9 + 2(1 + \sqrt {3})(1 + \sqrt {7})}$ I know this requires denesting but I don't know how to begin here.
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### Problem with $\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}$

How to simplify $$\frac{\sqrt{6+4\sqrt{2}}}{4+2\sqrt{2}}?$$ Rationalise the denominator $$\frac{\sqrt{6+4\sqrt{2}}}{4}(2-\sqrt{2})$$ This is still not simplify.
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### Four Nested Radicals

For how many positive integers $n$ is $$\sqrt {n+\sqrt {n+\sqrt {n+\sqrt {n}}}}$$ an integer? I still have yet to find a single integer that satisfies this (there's some that get extremely close, ...
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### $\lim\limits_{n \to \infty}\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\cdots+\sqrt{\frac{1}{n^2}}}}}.$

Problem Evaluate $\lim\limits_{n \to \infty}T_n$ where $$T_n=\sqrt{1+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{3^2}+\cdots+\sqrt{\frac{1}{n^2}}}}}.$$ Analysis It's obvious that $T_n$ is increasing with a ...
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### How to use the formula in negative roots?

This formula for nested-radicals: $\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}$, works fine with $B > 0$, and with $A + \sqrt{B} > 0$, but doesn't work with complex numbers. I ...
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### What is $\lim\limits_{n\to \infty} \sqrt{\frac1{1^2}+\sqrt{\frac1{2^2}+\sqrt{\frac1{3^2}+\cdots+\sqrt{\frac1{n^2}}}}}$? [closed]

By numerical calculations we yields 1.466723564..., from which I can't see any useful properties of that value. The same problem can be found at https://www.quora.com/What-is-the-value-of-sqrt-1-sqrt-...
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### Prove that inequality $\sqrt{2\sqrt{4\sqrt{8…\sqrt{2^n}}}} \leqslant n+1$

Let $n$ be the integer. Prove that $$\sqrt{2\sqrt{4\sqrt{8....\sqrt{2^n}}}} \leqslant n+1$$ SOURCE: BANGLADESH MATH OLYMPIAD I am a new beginner at the infinite radical and sequence. I don't know ...
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### Proof that 2*2/sqrt(2)*2/sqrt(2+sqrt(2))*2/sqrt(2+sqrt(2+sqrt(2)))*… equals PI?

I found this formula that $\pi=2\cdot\frac{2}{\sqrt{2}}\cdot\frac{2}{\sqrt{2+\sqrt{2}}}\cdot\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\cdot...$ I tested it out and it seems to be true, but I don't get why ...
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### How can I simplify $\sqrt{\frac{5+\sqrt{5}}{2}}$?

I've tried to see the root as $\sqrt{\frac{5+\sqrt{5}}{2}} = \sqrt{a}+\sqrt{b},$ but this method doesn't give me something good.
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### How do I simplify $\sqrt {4(2- \sqrt{3})}$ into $\sqrt{6} - \sqrt{2}$

This might be a stupid question, but how do I get from $$\sqrt {4(2- \sqrt{3})}$$ to $$\sqrt{6} - \sqrt{2}$$ It is obvious if you squared both, they both equal $8 - 4 \sqrt{3}$, but I'm wondering how ...
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### Finding rational numbers in an equation with two variables

How should we find two rational numbers $\alpha$, $\beta$ such that $\sqrt{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}$? The answer I got alpha = 1 and betta = 1. If I'm wrong, please correct me. Thank you
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### Tough Irrational Equation highschool

Have been trying to solve this irrational equation for a day but as it seems, i'm not going anywere with it. Can somebody offer me a tip ? Thanks! *Tried a "t" substitution for x squared but it still ...
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