${\displaystyle {\sqrt {5-2{\sqrt {5}}\ }}}$

which arises in discussing the regular pentagon;

${\displaystyle {\sqrt {5+2{\sqrt {6}}\ }},}$

or more complicated ones such as:

${\displaystyle {\sqrt[{3}]{2+{\sqrt {3}}+{\sqrt[{3}]{4}}\ }}.}$

Some nested radicals can be rewritten in a form that is not nested. For example,

${\displaystyle {\sqrt {3+2{\sqrt {2}}}}=1+{\sqrt {2}}\,,}$
${\displaystyle {\sqrt[{3}]{{\sqrt[{3}]{2}}-1}}={\frac {1-{\sqrt[{3}]{2}}+{\sqrt[{3}]{4}}}{\sqrt[{3}]{9}}}\,.}$

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:

${\displaystyle {\sqrt {a+b{\sqrt {c}}\ }}={\sqrt {d}}+{\sqrt {e}}.}$

Squaring both sides of this equation yields:

${\displaystyle a+b{\sqrt {c}}=d+e+2{\sqrt {de}}.}$

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:

${\displaystyle a=d+e,}$

which gives

${\displaystyle d=a-e,}$
${\displaystyle e=a-d.}$

For the irrational parts note that

${\displaystyle b{\sqrt {c}}=2{\sqrt {de}},}$

and squaring both sides yields

${\displaystyle b^{2}c=4de.}$

By plugging in ae for d one obtains

${\displaystyle b^{2}c=4(a-e)e=4ae-4e^{2}.}$

Rearranging terms will give a quadratic equation which can be solved for e:

${\displaystyle 4e^{2}-4ae+b^{2}c=0,}$
${\displaystyle e={\frac {a\pm {\sqrt {a^{2}-b^{2}c}}}{2}}.}$

The solution d is the algebraic conjugate of e. If

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

then

${\displaystyle d={\frac {a\mp {\sqrt {a^{2}-b^{2}c}}}{2}}.}$

However, this approach works for nested radicals of the form ${\displaystyle {\sqrt {a+b{\sqrt {c}}\ }}}$ if and only if ${\displaystyle {\sqrt {a^{2}-b^{2}c}}}$ 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:[1]

${\displaystyle {\sqrt[{4}]{\frac {3+2{\sqrt[{4}]{5}}}{3-2{\sqrt[{4}]{5}}}}}={\frac {{\sqrt[{4}]{5}}+1}{{\sqrt[{4}]{5}}-1}}={\tfrac {1}{2}}\left(3+{\sqrt[{4}]{5}}+{\sqrt {5}}+{\sqrt[{4}]{125}}\right),}$
${\displaystyle {\sqrt {{\sqrt[{3}]{28}}-{\sqrt[{3}]{27}}}}={\tfrac {1}{3}}\left({\sqrt[{3}]{98}}-{\sqrt[{3}]{28}}-1\right),}$
${\displaystyle {\sqrt[{3}]{{\sqrt[{5}]{\frac {32}{5}}}-{\sqrt[{5}]{\frac {27}{5}}}}}={\sqrt[{5}]{\frac {1}{25}}}+{\sqrt[{5}]{\frac {3}{25}}}-{\sqrt[{5}]{\frac {9}{25}}},}$
${\displaystyle {\sqrt[{3}]{\ {\sqrt[{3}]{2}}\ -1}}={\sqrt[{3}]{\frac {1}{9}}}-{\sqrt[{3}]{\frac {2}{9}}}+{\sqrt[{3}]{\frac {4}{9}}}.}$ [2]

Other odd-looking radicals inspired by Ramanujan:

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

## Landau's algorithm

{{#invoke:main|main}}

In 1989 Susan Landau introduced the first algorithm for deciding which nested radicals can be denested.[3] Earlier algorithms worked in some cases but not others.

### Square roots

Under certain conditions infinitely nested square roots such as

${\displaystyle x={\sqrt {2+{\sqrt {2+{\sqrt {2+{\sqrt {2+\cdots }}}}}}}}}$

represent rational numbers. This rational number can be found by realizing that x also appears under the radical sign, which gives the equation

${\displaystyle x={\sqrt {2+x}}.}$

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:

${\displaystyle {\sqrt {n+{\sqrt {n+{\sqrt {n+{\sqrt {n+\cdots }}}}}}}}={\tfrac {1}{2}}\left(1+{\sqrt {1+4n}}\right)}$

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

${\displaystyle {\sqrt {n-{\sqrt {n-{\sqrt {n-{\sqrt {n-\cdots }}}}}}}}={\tfrac {1}{2}}\left(-1+{\sqrt {1+4n}}\right).}$

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

${\displaystyle n=x^{2}+x.\,}$

Ramanujan posed this problem to the 'Journal of Indian Mathematical Society':

${\displaystyle ?={\sqrt {1+2{\sqrt {1+3{\sqrt {1+\cdots }}}}}}.\,}$

This can be solved by noting a more general formulation:

${\displaystyle ?={\sqrt {ax+(n+a)^{2}+x{\sqrt {a(x+n)+(n+a)^{2}+(x+n){\sqrt {\mathrm {\cdots } }}}}}}\,}$

Setting this to F(x) and squaring both sides gives us:

${\displaystyle F(x)^{2}=ax+(n+a)^{2}+x{\sqrt {a(x+n)+(n+a)^{2}+(x+n){\sqrt {\mathrm {\cdots } }}}}\,}$

Which can be simplified to:

${\displaystyle F(x)^{2}=ax+(n+a)^{2}+xF(x+n)\,}$

It can then be shown that:

${\displaystyle F(x)={x+n+a}\,}$

So, setting a =0, n = 1, and x = 2:

${\displaystyle 3={\sqrt {1+2{\sqrt {1+3{\sqrt {1+\cdots }}}}}}.\,}$

Ramanujan stated this radical in his lost notebook

${\displaystyle {\sqrt {5+{\sqrt {5+{\sqrt {5-{\sqrt {5+{\sqrt {5+{\sqrt {5+{\sqrt {5-\cdots }}}}}}}}}}}}}}={\frac {2+{\sqrt {5}}+{\sqrt {15-6{\sqrt {5}}}}}{2}}}$

(The repeating pattern of the signs is ${\displaystyle (+,+,-,+)}$

### Cube roots

In certain cases, infinitely nested cube roots such as

${\displaystyle x={\sqrt[{3}]{6+{\sqrt[{3}]{6+{\sqrt[{3}]{6+{\sqrt[{3}]{6+\cdots }}}}}}}}}$

can represent rational numbers as well. Again, by realizing that the whole expression appears inside itself, we are left with the equation

${\displaystyle x={\sqrt[{3}]{6+x}}.}$

If we solve this equation, we find that x = 2. More generally, we find that

${\displaystyle {\sqrt[{3}]{n+{\sqrt[{3}]{n+{\sqrt[{3}]{n+{\sqrt[{3}]{n+\cdots }}}}}}}}}$

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

${\displaystyle {\sqrt[{3}]{n-{\sqrt[{3}]{n-{\sqrt[{3}]{n-{\sqrt[{3}]{n-\cdots }}}}}}}}}$

as the real root of the equation x3 + x − n = 0 for all n and x where n > 0 and |x| ≥ 1.