${\sqrt {5-2{\sqrt {5}}\ }}$ which arises in discussing the regular pentagon;

${\sqrt {5+2{\sqrt {6}}\ }},$ or more complicated ones such as:

${\sqrt[{3}]{2+{\sqrt {3}}+{\sqrt[{3}]{4}}\ }}.$ Some nested radicals can be rewritten in a form that is not nested. For example,

${\sqrt {3+2{\sqrt {2}}}}=1+{\sqrt {2}}\,,$ ${\sqrt[{3}]{{\sqrt[{3}]{2}}-1}}={\frac {1-{\sqrt[{3}]{2}}+{\sqrt[{3}]{4}}}{\sqrt[{3}]{9}}}\,.$ Rewriting a nested radical in this way is called denesting. This process is generally considered a difficult problem, although a special class of nested radical can be denested by assuming it denests into a sum of two surds:

${\sqrt {a+b{\sqrt {c}}\ }}={\sqrt {d}}+{\sqrt {e}}.$ Squaring both sides of this equation yields:

$a+b{\sqrt {c}}=d+e+2{\sqrt {de}}.$ This can be solved by using the quadratic formula and setting rational and irrational parts on both sides of the equation equal to each other. The solutions for e and d can be obtained by first equating the rational parts:

$a=d+e,$ which gives

$d=a-e,$ $e=a-d.$ For the irrational parts note that

$b{\sqrt {c}}=2{\sqrt {de}},$ and squaring both sides yields

$b^{2}c=4de.$ By plugging in ae for d one obtains

$b^{2}c=4(a-e)e=4ae-4e^{2}.$ Rearranging terms will give a quadratic equation which can be solved for e:

$4e^{2}-4ae+b^{2}c=0,$ $e={\frac {a\pm {\sqrt {a^{2}-b^{2}c}}}{2}}.$ The solution d is the algebraic conjugate of e. If

$e={\frac {a\pm {\sqrt {a^{2}-b^{2}c}}}{2}},$ then

$d={\frac {a\mp {\sqrt {a^{2}-b^{2}c}}}{2}}.$ However, this approach works for nested radicals of the form ${\sqrt {a+b{\sqrt {c}}\ }}$ if and only if ${\sqrt {a^{2}-b^{2}c}}$ is an integer, in which case the nested radical can be denested into a sum of surds.

In some cases, higher-power radicals may be needed to denest the nested radical.

### Some identities of Ramanujan

Srinivasa Ramanujan demonstrated a number of curious identities involving denesting of radicals. Among them are the following:

${\sqrt[{4}]{\frac {3+2{\sqrt[{4}]{5}}}{3-2{\sqrt[{4}]{5}}}}}={\frac {{\sqrt[{4}]{5}}+1}{{\sqrt[{4}]{5}}-1}}={\tfrac {1}{2}}\left(3+{\sqrt[{4}]{5}}+{\sqrt {5}}+{\sqrt[{4}]{125}}\right),$ ${\sqrt {{\sqrt[{3}]{28}}-{\sqrt[{3}]{27}}}}={\tfrac {1}{3}}\left({\sqrt[{3}]{98}}-{\sqrt[{3}]{28}}-1\right),$ ${\sqrt[{3}]{{\sqrt[{5}]{\frac {32}{5}}}-{\sqrt[{5}]{\frac {27}{5}}}}}={\sqrt[{5}]{\frac {1}{25}}}+{\sqrt[{5}]{\frac {3}{25}}}-{\sqrt[{5}]{\frac {9}{25}}},$ ${\sqrt[{3}]{\ {\sqrt[{3}]{2}}\ -1}}={\sqrt[{3}]{\frac {1}{9}}}-{\sqrt[{3}]{\frac {2}{9}}}+{\sqrt[{3}]{\frac {4}{9}}}.$ Other odd-looking radicals inspired by Ramanujan:

${\sqrt[{4}]{49+20{\sqrt {6}}}}+{\sqrt[{4}]{49-20{\sqrt {6}}}}=2{\sqrt {3}},$ ${\sqrt[{3}]{\left({\sqrt {2}}+{\sqrt {3}}\right)\left(5-{\sqrt {6}}\right)+3\left(2{\sqrt {3}}+3{\sqrt {2}}\right)}}={\sqrt {10-{\frac {13-5{\sqrt {6}}}{5+{\sqrt {6}}}}}}.$ ## Landau's algorithm

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In 1989 Susan Landau introduced the first algorithm for deciding which nested radicals can be denested. Earlier algorithms worked in some cases but not others.

### Square roots

Under certain conditions infinitely nested square roots such as

$x={\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+\cdots }}}}}}}}$ represent rational numbers. This rational number can be found by realizing that x also appears under the radical sign, which gives the equation

$x={\sqrt {2+x}}.$ If we solve this equation, we find that x = 2 (the second solution x = −1 doesn't apply, under the convention that the positive square root is meant). This approach can also be used to show that generally, if n > 0, then:

${\sqrt {n+{\sqrt {n+{\sqrt {n+{\sqrt {n+\cdots }}}}}}}}={\tfrac {1}{2}}\left(1+{\sqrt {1+4n}}\right)$ and is the real root of the equation x2 − x − n = 0. For n = 1, this root is the golden ratio φ, approximately equal to 1.618. The same procedure also works to get that

${\sqrt {n-{\sqrt {n-{\sqrt {n-{\sqrt {n-\cdots }}}}}}}}={\tfrac {1}{2}}\left(-1+{\sqrt {1+4n}}\right).$ and is the real root of the equation x2 + x − n = 0. For n = 1, this root is the reciprocal of the golden ratio Φ, which is equal to φ − 1. This method will give a rational x value for all values of n such that

$n=x^{2}+x.\,$ Ramanujan posed this problem to the 'Journal of Indian Mathematical Society':

$?={\sqrt {1+2{\sqrt {1+3{\sqrt {1+\cdots }}}}}}.\,$ This can be solved by noting a more general formulation:

$?={\sqrt {ax+(n+a)^{2}+x{\sqrt {a(x+n)+(n+a)^{2}+(x+n){\sqrt {\mathrm {\cdots } }}}}}}\,$ Setting this to F(x) and squaring both sides gives us:

$F(x)^{2}=ax+(n+a)^{2}+x{\sqrt {a(x+n)+(n+a)^{2}+(x+n){\sqrt {\mathrm {\cdots } }}}}\,$ Which can be simplified to:

$F(x)^{2}=ax+(n+a)^{2}+xF(x+n)\,$ It can then be shown that:

$F(x)={x+n+a}\,$ So, setting a =0, n = 1, and x = 2:

$3={\sqrt {1+2{\sqrt {1+3{\sqrt {1+\cdots }}}}}}.\,$ Ramanujan stated this radical in his lost notebook

${\sqrt {5+{\sqrt {5+{\sqrt {5-{\sqrt {5+{\sqrt {5+{\sqrt {5+{\sqrt {5-\cdots }}}}}}}}}}}}}}={\frac {2+{\sqrt {5}}+{\sqrt {15-6{\sqrt {5}}}}}{2}}$ ### Cube roots

In certain cases, infinitely nested cube roots such as

$x={\sqrt[{3}]{6+{\sqrt[{3}]{6+{\sqrt[{3}]{6+{\sqrt[{3}]{6+\cdots }}}}}}}}$ can represent rational numbers as well. Again, by realizing that the whole expression appears inside itself, we are left with the equation

$x={\sqrt[{3}]{6+x}}.$ If we solve this equation, we find that x = 2. More generally, we find that

${\sqrt[{3}]{n+{\sqrt[{3}]{n+{\sqrt[{3}]{n+{\sqrt[{3}]{n+\cdots }}}}}}}}$ is the real root of the equation x3 − x − n = 0 for all n > 0. For n = 1, this root is the plastic number ρ, approximately equal to 1.3247.

The same procedure also works to get

${\sqrt[{3}]{n-{\sqrt[{3}]{n-{\sqrt[{3}]{n-{\sqrt[{3}]{n-\cdots }}}}}}}}$ as the real root of the equation x3 + x − n = 0 for all n and x where n > 0 and |x| ≥ 1.