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{{short description|Arithmetic operation, inverse of nth power}}
{{about|nth-roots of real and complex numbers|other uses|Root (disambiguation)#Mathematics}}
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{{More citations needed|date=October 2022}}
In [[mathematics]], an '''{{mvar|n}}th root''' of a [[number]] {{mvar|x}} is a number {{mvar|r}} which, when [[exponentiation|raised to the power]] of {{mvar|n}}, yields&nbsp;{{mvar|x}}: <math display="block">r^n = \underbrace{r \times r \times \dotsb \times r}_{n\text{ factors}} = x.</math>

The [[positive integer]] {{mvar|n}} is called the ''index'' or ''degree'', and the number {{mvar|x}} of which the root is taken is the ''radicand.'' A root of degree 2 is called a ''[[square root]]'' and a root of degree 3, a ''[[cube root]]''. Roots of higher degree are referred by using [[ordinal numeral|ordinal numbers]], as in ''fourth root'', ''twentieth root'', etc.  The computation of an {{mvar|n}}th root is a '''root extraction'''.

For example, {{math|3}} is a square root of {{math|9}}, since {{math|1=3{{sup|2}} = 9}}, and {{math|−3}} is also a square root of {{math|9}}, since {{math|1=(−3){{sup|2}} = 9}}.

The {{mvar|n}}th root of {{mvar|x}} is written as <math>\sqrt[n]{x}</math> using the [[radical symbol]] <math>\sqrt{\phantom x}</math>. The square root is usually written as {{tmath|\sqrt x}}, with the degree omitted. Taking the {{mvar|n}}th root of a number, for fixed {{tmath|n}}, is the [[inverse function#Squaring and square root functions|inverse]] of raising a number to the {{mvar|n}}th power,<ref>{{cite web |url=https://www.nagwa.com/en/explainers/985195836913 |access-date=22 July 2023 |title=Lesson Explainer: nth Roots: Integers}}</ref> and can be written as a [[Fraction (mathematics)|fractional]] exponent:

<math display="block">\sqrt[n]{x} = x^{1/n}.</math>

For a positive real number {{mvar|x}}, <math>\sqrt{x}</math> denotes the positive square root of {{mvar|x}} and <math>\sqrt[n]{x}</math> denotes the positive real {{mvar|n}}th root. A negative real number {{math|−''x''}} has no real-valued square roots, but when {{mvar|x}} is treated as a complex number it has two [[imaginary number|imaginary]] square roots, {{tmath|+i\sqrt x }} and {{tmath|-i\sqrt x }}, where {{mvar|i}} is the [[imaginary unit]].

In general, any non-zero [[complex number]] has {{mvar|n}} distinct complex-valued {{mvar|n}}th roots, equally distributed around a complex circle of constant [[Absolute value#Complex numbers|absolute value]]. (The {{mvar|n}}th root of {{math|0}} is zero with [[multiple root|multiplicity]] {{mvar|n}}, and this circle degenerates to a point.) Extracting the {{mvar|n}}th roots of a complex number {{mvar|x}} can thus be taken to be a [[multivalued function]]. By convention the [[principal value]] of this function, called the '''principal root''' and denoted {{tmath|\sqrt[n]{x} }}, is taken to be the {{mvar|n}}th root with the greatest real part and in the special case when {{mvar|x}} is a negative real number, the one with a positive [[imaginary part]]. The principal root of a positive real number is thus also a positive real number. As a [[function (mathematics)|function]], the principal root is [[continuous function|continuous]] in the whole [[complex plane]], except along the negative real axis.

An unresolved root, especially one using the radical symbol, is sometimes referred to as a '''surd'''<ref>{{cite book |title=New Approach to CBSE Mathematics IX |first=R.K. |last=Bansal |page=25 |year=2006 |isbn=978-81-318-0013-3 |publisher=Laxmi Publications |url=https://books.google.com/books?id=1C4iQNUWLBwC&pg=PA25}}</ref> or a '''radical'''.<ref name=silver>{{cite book|last=Silver|first=Howard A.|title=Algebra and trigonometry|year=1986|publisher=Prentice-Hall|location=Englewood Cliffs, New Jersey|isbn=978-0-13-021270-2|url-access=registration|url=https://archive.org/details/algebratrigonome00silv}}</ref> Any expression containing a radical, whether it is a square root, a cube root, or a higher root, is called a '''''radical expression''''', and if it contains no [[transcendental functions]] or [[transcendental numbers]] it is called an ''[[algebraic expression]]''.

{{Arithmetic operations}}

Roots are used for determining the [[radius of convergence]] of a [[power series]] with the [[root test]]. The {{mvar|n}}th roots of 1 are called [[roots of unity]] and play a fundamental role in various areas of mathematics, such as [[number theory]], [[theory of equations]], and [[Fourier transform]].

==History==

{{Main article|Square root#History|Cube root#History}}
An archaic term for the operation of taking ''n''th roots is ''radication''.<ref>{{cite web|url=https://www.merriam-webster.com/dictionary/radication|title=Definition of RADICATION|website=www.merriam-webster.com}}</ref><ref>{{cite web|url=https://en.oxforddictionaries.com/definition/radication|archive-url=https://web.archive.org/web/20180403112348/https://en.oxforddictionaries.com/definition/radication|url-status=dead|archive-date=April 3, 2018|title=radication – Definition of radication in English by Oxford Dictionaries|website=Oxford Dictionaries }}</ref>

==Definition and notation==
[[File:NegativeOne4Root.svg|thumb|The four 4th roots of −1,<br /> none of which are real]]
[[File:NegativeOne3Root.svg|thumb|The three 3rd roots of −1,<br /> one of which is a negative real]]
An  ''{{mvar|n}}th root'' of a number ''x'', where ''n'' is a positive integer, is any of the ''n'' real or complex numbers ''r'' whose ''n''th power is ''x'':

<math display="block">r^n = x.</math>

Every positive [[real number]] ''x'' has a single positive ''n''th root, called the [[principal value|principal ''n''th root]], which is written <math>\sqrt[n]{x}</math>. For ''n'' equal to 2 this is called the principal square root and the  ''n'' is omitted. The ''n''th root can also be represented using [[exponentiation]] as ''x''{{sup|1/n}}.

For even values of ''n'', positive numbers also have a negative ''n''th root, while negative numbers do not have a real ''n''th root.  For odd values of ''n'', every negative number ''x'' has a real negative ''n''th root.  For example, −2 has a real 5th root, <math>\sqrt[5]{-2} = -1.148698354\ldots</math> but −2 does not have any real 6th roots.

Every non-zero number ''x'', real or [[Complex number|complex]], has ''n'' different complex number ''n''th roots.  (In the case ''x'' is real, this count includes any real ''n''th roots.) The only complex root of 0 is 0.

The ''n''th roots of almost all numbers (all integers except the ''n''th powers, and all rationals except the quotients of two ''n''th powers) are [[irrational number|irrational]].  For example,

<math display="block">\sqrt{2} = 1.414213562\ldots</math>

All ''n''th roots of rational numbers are [[algebraic number]]s, and all ''n''th roots of integers are [[algebraic integer]]s.

The term "surd" traces back to [[Al-Khwarizmi]] ({{circa|825}}), who referred to rational and irrational numbers as "audible" and "inaudible", respectively. This later led to the Arabic word {{lang|ar|أصم}} ({{lang|ar-Latn|asamm}}, meaning "deaf" or "dumb") for "irrational number" being translated into Latin as {{lang|la|surdus}} (meaning "deaf" or "mute"). [[Gerard of Cremona]] ({{circa|1150}}), [[Fibonacci]] (1202), and then [[Robert Recorde]] (1551) all used the term to refer to "unresolved irrational roots", that is, expressions of the form <math>\sqrt[n]{r}</math>, in which <math>n</math> and <math>r</math> are integer numerals and the whole expression denotes an irrational number.<ref>{{cite web |url=http://jeff560.tripod.com/s.html |title=Earliest Known Uses of Some of the Words of Mathematics|website=Mathematics Pages |first=Jeff |last=Miller|access-date=2008-11-30}}</ref> Irrational numbers of the form <math>\pm\sqrt{a},</math> where <math>a</math> is rational, are called "pure quadratic surds"; irrational numbers of the form  <math>a \pm\sqrt{b}</math>, where <math>a</math> and <math>b</math> are rational, are called ''[[Quadratic irrational number|mixed quadratic surds]]''.<ref>{{cite book |last=Hardy |first=G. H. |author-link=G. H. Hardy |at=§1.13 "Quadratic Surds" – §1.14, {{pgs|19–23}} |url=https://archive.org/details/coursepuremath00hardrich/page/n36/mode/2up |title=A Course of Pure Mathematics |year=1921 |edition=3rd |publisher=Cambridge}}</ref>

===Square roots===
{{Main article|Square root}}
[[Image:Square-root function.svg|thumb|right|The graph <math>y=\pm \sqrt{x}</math>.]]
A '''square root''' of a number ''x'' is a number ''r'' which, when [[square (algebra)|squared]], becomes ''x'':

<math display="block">r^2 = x.</math>

Every positive real number has two square roots, one positive and one negative.  For example, the two square roots of 25 are 5 and −5.  The positive square root is also known as the '''principal square root''', and is denoted with a radical sign:

<math display="block">\sqrt{25} = 5.</math>

Since the square of every real number is nonnegative, negative numbers do not have real square roots. However, for every negative real number there are two [[imaginary number|imaginary]] square roots.  For example, the square roots of −25 are 5''i'' and −5''i'', where ''[[imaginary unit|i]]'' represents a number whose square is {{math|−1}}.

===Cube roots===
{{Main article|Cube root}}
[[Image:cube-root function.svg|thumb|right|The graph <math>y=\sqrt[3]{x}</math>.]]
A '''cube root''' of a number ''x'' is a number ''r'' whose [[cube (algebra)|cube]] is ''x'':

<math display="block">r^3 = x.</math>

Every real number ''x'' has exactly one real cube root, written <math>\sqrt[3]{x}</math>.  For example,

<math display="block">\begin{align}
\sqrt[3]{8} &= 2\\
\sqrt[3]{-8} &= -2.
\end{align}</math>

Every real number has two additional [[complex number|complex]] cube roots.

==Identities and properties==
Expressing the degree of an ''n''th root in its exponent form, as in <math>x^{1/n}</math>, makes it easier to manipulate powers and roots. If <math>a</math> is a [[non-negative number|non-negative real number]],

<math display="block">\sqrt[n]{a^m} = (a^m)^{1/n} = a^{m/n} = (a^{1/n})^m = (\sqrt[n]a)^m.</math>

Every non-negative number has exactly one non-negative real ''n''th root, and so the rules for operations with surds involving non-negative radicands <math>a</math> and <math>b</math> are straightforward within the real numbers:

<math display="block">\begin{align}
           \sqrt[n]{ab} &= \sqrt[n]{a} \sqrt[n]{b} \\
  \sqrt[n]{\frac{a}{b}} &= \frac{\sqrt[n]{a}}{\sqrt[n]{b}}
\end{align}</math>

Subtleties can occur when taking the ''n''th roots of negative or [[complex number]]s. For instance:

<math display="block">\sqrt{-1}\times\sqrt{-1} \neq \sqrt{-1 \times -1} = 1,\quad</math>

but, rather,

<math display="block">\quad\sqrt{-1}\times\sqrt{-1} = i \times i = i^2 = -1.</math>

Since the rule <math>\sqrt[n]{a} \times \sqrt[n]{b} =  \sqrt[n]{ab} </math> strictly holds for non-negative real radicands only, its application leads to the inequality in the first step above.

==Simplified form of a radical expression==

A [[nested radical|non-nested radical expression]] is said to be in '''simplified form''' if no factor of the radicand can be written as a power greater than or equal to the index; there are no fractions inside the radical sign; and there are no radicals in the denominator.<ref>{{cite book|last=McKeague|first=Charles P.|title=Elementary algebra|page=470|year=2011|publisher=Cengage Learning |url=https://books.google.com/books?id=etTbP0rItQ4C&q=editions:q0hGn6PkOxsC|isbn=978-0-8400-6421-9}}</ref>

For example, to write the radical expression <math>\textstyle \sqrt{32/5}</math> in simplified form, we can proceed as follows. First, look for a perfect square under the square root sign and remove it:

<math display="block">
\sqrt{\frac{32}{5}} = \sqrt{\frac{16 \cdot 2}{5}}
= \sqrt{16} \cdot \sqrt{\frac{2}{5}} = 4 \sqrt{\frac{2}{5}}
</math>

Next, there is a fraction under the radical sign, which we change as follows:

<math display="block">4 \sqrt{\frac{2}{5}} = \frac{4 \sqrt{2}}{\sqrt{5}}</math>

Finally, we remove the radical from the denominator as follows:

<math display="block">\frac{4 \sqrt{2}}{\sqrt{5}} = \frac{4 \sqrt{2}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{4 \sqrt{10}}{5} = \frac{4}{5}\sqrt{10}</math>

When there is a denominator involving surds it is always possible to find a factor to multiply both numerator and denominator by to simplify the expression.<ref>{{Cite conference|first1=B. F. |last1=Caviness|first2=R. J. |last2=Fateman|chapter-url=http://www.eecs.berkeley.edu/~fateman/papers/radcan.pdf|chapter=Simplification of Radical Expressions|title=Proceedings of the 1976 ACM Symposium on Symbolic and Algebraic Computation|page=329}}</ref><ref>{{cite journal|last=Richard|first=Zippel|title=Simplification of Expressions Involving Radicals|journal=Journal of Symbolic Computation|volume=1|number=189–210|year=1985|pages=189–210 |doi=10.1016/S0747-7171(85)80014-6}}</ref> For instance using the [[Factorization#Sum/difference of two cubes|factorization of the sum of two cubes]]:

<math display="block">
  \frac{1}{\sqrt[3]{a} + \sqrt[3]{b}} =
  \frac{\sqrt[3]{a^2} - \sqrt[3]{ab} + \sqrt[3]{b^2}}{\left(\sqrt[3]{a} + \sqrt[3]{b}\right)\left(\sqrt[3]{a^2} - \sqrt[3]{ab} + \sqrt[3]{b^2}\right)} =
  \frac{\sqrt[3]{a^2} - \sqrt[3]{ab} + \sqrt[3]{b^2}}{a + b} .
</math>

Simplifying radical expressions involving [[nested radical]]s can be quite difficult. In particular, denesting is not always possible, and when possible, it may involve advanced [[Galois theory]]. Moreover, when complete denesting is impossible, there is no general [[canonical form]] such that the equality of two numbers can be tested by simply looking at their canonical expressions. 

For example, it is not obvious that

<math display="block">\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}.</math>

The above can be derived through:

<math display="block">\sqrt{3 + 2\sqrt{2}} = \sqrt{1 + 2\sqrt{2} + 2} = \sqrt{1^2 + 2\sqrt{2} + \sqrt{2}^2} = \sqrt{\left(1 + \sqrt{2}\right)^2} = 1 + \sqrt{2}</math>

Let <math>r=p/q</math>, with {{mvar|p}} and {{mvar|q}} coprime and positive integers. Then <math>\sqrt[n]r = \sqrt[n]{p}/\sqrt[n]{q}</math> is rational if and only if both <math>\sqrt[n]{p}</math> and <math>\sqrt[n]{q}</math> are integers, which means that both {{mvar|p}} and {{mvar|q}} are ''n''th powers of some integer.

==Infinite series==
The radical or root may be represented by the [[infinite series]]:

<math display="block">(1+x)^\frac{s}{t} = \sum_{n=0}^\infty \frac{\prod_{k=0}^{n-1} (s-kt)}{n!t^n}x^n</math>

with <math>|x|<1</math>. This expression can be derived from the [[binomial series]].

==Computing principal roots==

===Using Newton's method===

The {{mvar|n}}th root of a number {{math|''A''}} can be computed with [[Newton's method]], which starts with an initial guess {{math|''x''<sub>0</sub>}} and then iterates using the [[recurrence relation]]

<math display="block">x_{k+1} = x_k-\frac{x_k^n-A}{nx_k^{n-1}}</math>

until the desired precision is reached. For computational efficiency, the recurrence relation is commonly rewritten 

<math display="block">x_{k+1} = \frac{n-1}{n}\,x_k+\frac{A}{n}\,\frac 1{x_k^{n-1}}.</math>

This allows to have only one [[exponentiation]], and to compute once for all the first factor of each term.

For example, to find the fifth root of 34, we plug in {{math|1=''n'' = 5, ''A'' = 34}} and {{math|1=''x''<sub>0</sub> = 2}} (initial guess). The first 5 iterations are, approximately:

{{block indent|{{math|1=''x''<sub>0</sub> = 2}}}}
{{block indent|{{math|1=''x''<sub>1</sub> = 2.025}}}}
{{block indent|{{math|1=''x''<sub>2</sub> = 2.02439 7...}}}}
{{block indent|{{math|1=''x''<sub>3</sub> = 2.02439 7458...}}}}
{{block indent|{{math|1=''x''<sub>4</sub> = 2.02439 74584 99885 04251 08172...}}}}
{{block indent|{{math|1=''x''<sub>5</sub> = 2.02439 74584 99885 04251 08172 45541 93741 91146 21701 07311 8...}}}}

(All correct digits shown.)

The approximation {{math|''x''<sub>4</sub>}} is accurate to 25 decimal places and {{math|''x''<sub>5</sub>}} is good for 51.

Newton's method can be modified to produce various [[generalized continued fraction#Roots of positive numbers|generalized continued fractions]] for the ''n''th root. For example,

<math display="block">
  \sqrt[n]{z} = \sqrt[n]{x^n+y} = x+\cfrac{y} {nx^{n-1}+\cfrac{(n-1)y} {2x+\cfrac{(n+1)y} {3nx^{n-1}+\cfrac{(2n-1)y} {2x+\cfrac{(2n+1)y} {5nx^{n-1}+\cfrac{(3n-1)y} {2x+\ddots}}}}}}.
</math>

=== Digit-by-digit calculation of principal roots of decimal (base 10) numbers ===
[[Image:PascalForDecimalRoots.svg|right|thumb|[[Pascal's triangle]] showing <math>P(4,1) = 4</math>.]]
Building on the [[Methods of computing square roots#Decimal (base 10)|digit-by-digit calculation of a square root]], it can be seen that the formula used there, <math>x(20p + x) \le c</math>, or <math>x^2 + 20xp \le c</math>, follows a pattern involving Pascal's triangle.  For the ''n''th root of a number <math>P(n,i)</math> is defined as the value of element <math>i</math> in row <math>n</math> of Pascal's Triangle such that <math>P(4,1) = 4</math>, we can rewrite the expression as <math>\sum_{i=0}^{n-1}10^i P(n,i)p^i x^{n-i}</math>.  For convenience, call the result of this expression <math>y</math>.  Using this more general expression, any positive principal root can be computed, digit-by-digit, as follows.

Write the original number in decimal form. The numbers are written similar to the [[long division]] algorithm, and, as in long division, the root will be written on the line above. Now separate the digits into groups of digits equating to the root being taken, starting from the decimal point and going both left and right. The decimal point of the root will be above the decimal point of the radicand. One digit of the root will appear above each group of digits of the original number.

Beginning with the left-most group of digits, do the following procedure for each group:

# Starting on the left, bring down the most significant (leftmost) group of digits not yet used (if all the digits have been used, write "0" the number of times required to make a group) and write them to the right of the remainder from the previous step (on the first step, there will be no remainder). In other words, multiply the remainder by <math>10^n</math> and add the digits from the next group. This will be the '''current value ''c'''''.
# Find ''p'' and ''x'', as follows:
#* Let <math>p</math> be the '''part of the root found so far''', ignoring any decimal point. (For the first step, <math>p = 0</math> and <math>0^0 = 1</math>).
#* Determine the greatest digit <math>x</math> such that <math>y \le c</math>.
#* Place the digit <math>x</math> as the next digit of the root, i.e., above the group of digits you just brought down. Thus the next ''p'' will be the old ''p'' times 10 plus ''x''.
# Subtract <math>y</math> from <math>c</math> to form a new remainder.
# If the remainder is zero and there are no more digits to bring down, then the algorithm has terminated. Otherwise go back to step 1 for another iteration.

====Examples====
{{MOS|section|date=April 2022}}

'''Find the square root of 152.2756.'''

         <u>  1  2. 3  4 </u>
     <u> </u>  /
      \/  01 52.27 56          (Results)    (Explanations)
    &nbsp;
          01                   x = 1        10{{sup|0}}·1·0{{sup|0}}·'''1'''{{sup|2}} + 10{{sup|1}}·2·0{{sup|1}}·'''1'''{{sup|1}}     ≤      1   <   10{{sup|0}}·1·0{{sup|0}}·2{{sup|2}}   + 10{{sup|1}}·2·0{{sup|1}}·2{{sup|1}}
         <u> 01 </u>                  y = 1        y = 10{{sup|0}}·1·0{{sup|0}}·1{{sup|2}}   + 10{{sup|1}}·2·0{{sup|1}}·1{{sup|1}}   =  1 +    0   =     '''1'''
          00 52                x = 2        10{{sup|0}}·1·1{{sup|0}}·'''2'''{{sup|2}} + 10{{sup|1}}·2·1{{sup|1}}·'''2'''{{sup|1}}     ≤     52   <   10{{sup|0}}·1·1{{sup|0}}·3{{sup|2}}   + 10{{sup|1}}·2·1{{sup|1}}·3{{sup|1}}
         <u> 00 44 </u>               y = 44       y = 10{{sup|0}}·1·1{{sup|0}}·2{{sup|2}}   + 10{{sup|1}}·2·1{{sup|1}}·2{{sup|1}}   =  4 +   40   =    '''44'''
             08 27             x = 3        10{{sup|0}}·1·12{{sup|0}}·'''3'''{{sup|2}} + 10{{sup|1}}·2·12{{sup|1}}·'''3'''{{sup|1}}   ≤    827   <   10{{sup|0}}·1·12{{sup|0}}·4{{sup|2}}  + 10{{sup|1}}·2·12{{sup|1}}·4{{sup|1}}
            <u> 07 29 </u>            y = 729      y = 10{{sup|0}}·1·12{{sup|0}}·3{{sup|2}}  + 10{{sup|1}}·2·12{{sup|1}}·3{{sup|1}}  =  9 +  720   =   '''729'''
                98 56          x = 4        10{{sup|0}}·1·123{{sup|0}}·'''4'''{{sup|2}} + 10{{sup|1}}·2·123{{sup|1}}·'''4'''{{sup|1}} ≤   9856   <   10{{sup|0}}·1·123{{sup|0}}·5{{sup|2}} + 10{{sup|1}}·2·123{{sup|1}}·5{{sup|1}}
               <u> 98 56 </u>         y = 9856     y = 10{{sup|0}}·1·123{{sup|0}}·4{{sup|2}} + 10{{sup|1}}·2·123{{sup|1}}·4{{sup|1}} = 16 + 9840   =  '''9856'''
                00 00

Algorithm terminates: Answer is 12.34

'''Find the cube root of 4192 truncated to the nearest thousandth.'''

      <u>   1   6.  1   2   4</u>
  <u>3</u>  /
   \/  004 192.000 000 000          (Results)    (Explanations)
    &nbsp;
       004                          x = 1        10{{sup|0}}·1·0{{sup|0}}·'''1'''{{sup|3}}    +  10{{sup|1}}·3·0{{sup|1}}·'''1'''{{sup|2}}   + 10{{sup|2}}·3·0{{sup|2}}·'''1'''{{sup|1}}    ≤          4  <  10{{sup|0}}·1·0{{sup|0}}·2{{sup|3}}     + 10{{sup|1}}·3·0{{sup|1}}·2{{sup|2}}    + 10{{sup|2}}·3·0{{sup|2}}·2{{sup|1}}
      <u> 001 </u>                         y = 1        y = 10{{sup|0}}·1·0{{sup|0}}·1{{sup|3}}   + 10{{sup|1}}·3·0{{sup|1}}·1{{sup|2}}   + 10{{sup|2}}·3·0{{sup|2}}·1{{sup|1}}   =   1 +      0 +          0   =          '''1'''
       003 192                      x = 6        10{{sup|0}}·1·1{{sup|0}}·'''6'''{{sup|3}}    +  10{{sup|1}}·3·1{{sup|1}}·'''6'''{{sup|2}}   + 10{{sup|2}}·3·1{{sup|2}}·'''6'''{{sup|1}}    ≤       3192  <  10{{sup|0}}·1·1{{sup|0}}·7{{sup|3}}     + 10{{sup|1}}·3·1{{sup|1}}·7{{sup|2}}    + 10{{sup|2}}·3·1{{sup|2}}·7{{sup|1}}
      <u> 003 096 </u>                     y = 3096     y = 10{{sup|0}}·1·1{{sup|0}}·6{{sup|3}}   + 10{{sup|1}}·3·1{{sup|1}}·6{{sup|2}}   + 10{{sup|2}}·3·1{{sup|2}}·6{{sup|1}}   = 216 +  1,080 +      1,800   =      '''3,096'''
           096 000                  x = 1        10{{sup|0}}·1·16{{sup|0}}·'''1'''{{sup|3}}   + 10{{sup|1}}·3·16{{sup|1}}·'''1'''{{sup|2}}   + 10{{sup|2}}·3·16{{sup|2}}·'''1'''{{sup|1}}   ≤      96000  <  10{{sup|0}}·1·16{{sup|0}}·2{{sup|3}}   + 10{{sup|1}}·3·16{{sup|1}}·2{{sup|2}}   + 10{{sup|2}}·3·16{{sup|2}}·2{{sup|1}}
          <u> 077 281 </u>                 y = 77281    y = 10{{sup|0}}·1·16{{sup|0}}·1{{sup|3}}  + 10{{sup|1}}·3·16{{sup|1}}·1{{sup|2}}  + 10{{sup|2}}·3·16{{sup|2}}·1{{sup|1}}  =   1 +    480 +     76,800   =     '''77,281'''
           018 719 000              x = 2        10{{sup|0}}·1·161{{sup|0}}·'''2'''{{sup|3}}  + 10{{sup|1}}·3·161{{sup|1}}·'''2'''{{sup|2}}  + 10{{sup|2}}·3·161{{sup|2}}·'''2'''{{sup|1}}  ≤   18719000  <  10{{sup|0}}·1·161{{sup|0}}·3{{sup|3}}  + 10{{sup|1}}·3·161{{sup|1}}·3{{sup|2}}  + 10{{sup|2}}·3·161{{sup|2}}·3{{sup|1}}
              <u> 015 571 928 </u>         y = 15571928 y = 10{{sup|0}}·1·161{{sup|0}}·2{{sup|3}} + 10{{sup|1}}·3·161{{sup|1}}·2{{sup|2}} + 10{{sup|2}}·3·161{{sup|2}}·2{{sup|1}} =   8 + 19,320 + 15,552,600   = '''15,571,928'''
               003 147 072 000      x = 4        10{{sup|0}}·1·1612{{sup|0}}·'''4'''{{sup|3}} + 10{{sup|1}}·3·1612{{sup|1}}·'''4'''{{sup|2}} + 10{{sup|2}}·3·1612{{sup|2}}·'''4'''{{sup|1}} ≤ 3147072000  <  10{{sup|0}}·1·1612{{sup|0}}·5{{sup|3}} + 10{{sup|1}}·3·1612{{sup|1}}·5{{sup|2}} + 10{{sup|2}}·3·1612{{sup|2}}·5{{sup|1}}

The desired precision is achieved. The cube root of 4192 is 16.124...

===Logarithmic calculation===

The principal ''n''th root of a positive number can be computed using [[logarithm]]s. Starting from the equation that defines ''r'' as an ''n''th root of ''x'', namely <math>r^n=x,</math> with ''x'' positive and therefore its principal root ''r'' also positive, one takes logarithms of both sides (any [[logarithm#Particular bases|base of the logarithm]] will do) to obtain

<math display="block">n \log_b r = \log_b x \quad \quad \text{hence} \quad \quad \log_b r = \frac{\log_b x}{n}.</math>

The root ''r'' is recovered from this by taking the [[antilog]]:

<math display="block">r = b^{\frac{1}{n}\log_b x}.</math>

(Note: That formula shows ''b'' raised to the power of the result of the division, not ''b'' multiplied by the result of the division.)

For the case in which ''x'' is negative and ''n'' is odd, there is one real root ''r'' which is also negative. This can be found by first multiplying both sides of the defining equation by −1 to obtain <math>|r|^n = |x|,</math> then proceeding as before to find |''r''|, and using {{nowrap|''r'' {{=}} −{{!}}''r''{{!}}}}.

==Geometric constructibility==

The [[ancient Greek mathematicians]] knew how to [[compass-and-straightedge construction|use compass and straightedge]] to construct a length equal to the square root of a given length, when an auxiliary line of unit length is given. In 1837 [[Pierre Wantzel]] proved that an ''n''th root of a given length cannot be constructed if ''n'' is not a power of 2.<ref>{{Citation|first = [[Monsieur|M.]] L.|last = Wantzel|title = Recherches sur les moyens de reconnaître si un Problème de Géométrie peut se résoudre avec la règle et le compas |journal = Journal de Mathématiques Pures et Appliquées|year = 1837|volume = 1|issue = 2|pages = 366–372|url = http://visualiseur.bnf.fr/ConsulterElementNum?O=NUMM-16381&Deb=374&Fin=380&E=PDF}}</ref>

==Complex roots==
Every [[complex number]] other than 0 has ''n'' different ''n''th roots.

===Square roots===
[[Image:Imaginary2Root.svg|thumb|right|The square roots of '''''i''''']]
The two square roots of a complex number are always negatives of each other. For example, the square roots of {{math|−4}} are {{math|2''i''}} and {{math|−2''i''}}, and the square roots of {{math|''i''}} are

<math display="block">\tfrac{1}{\sqrt{2}}(1 + i) \quad\text{and}\quad -\tfrac{1}{\sqrt{2}}(1 + i).</math>

If we express a complex number in [[polar form]], then the square root can be obtained by taking the square root of the radius and halving the angle:

<math display="block">\sqrt{re^{i\theta}} = \pm\sqrt{r} \cdot e^{i\theta/2}.</math>

A ''principal'' root of a complex number may be chosen in various ways, for example

<math display="block">\sqrt{re^{i\theta}} = \sqrt{r} \cdot e^{i\theta/2}</math>

which introduces a [[branch cut]] in the [[complex plane]] along the [[positive real axis]] with the condition {{math|0&nbsp;≤&nbsp;''θ''&nbsp;<&nbsp;2{{pi}}}}, or along the negative real axis with {{math|−{{pi}}&nbsp;<&nbsp;''θ''&nbsp;≤&nbsp;{{pi}}}}.

Using the first(last) branch cut the principal square root <math>\scriptstyle \sqrt z</math> maps <math>\scriptstyle z</math> to the half plane with non-negative imaginary(real) part. The last branch cut is presupposed in mathematical software like [[Matlab]] or [[Scilab]].

===Roots of unity===
{{Main article|Root of unity}}
[[File:3rd roots of unity.svg|thumb|right|The three 3rd roots of 1]]

The number 1 has ''n'' different ''n''th roots in the complex plane, namely

<math display="block">1,\;\omega,\;\omega^2,\;\ldots,\;\omega^{n-1},</math>

where

<math display="block">\omega = e^\frac{2\pi i}{n} = \cos\left(\frac{2\pi}{n}\right) + i\sin\left(\frac{2\pi}{n}\right).</math>

These roots are evenly spaced around the [[unit circle]] in the complex plane, at angles which are multiples of <math>2\pi/n</math>.  For example, the square roots of unity are 1 and −1, and the fourth roots of unity are 1, <math>i</math>, −1, and <math>-i</math>.

===''n''th roots===
{{visualisation_complex_number_roots.svg}}
Every complex number has ''n'' different ''n''th roots in the complex plane.  These are

<math display="block">\eta,\;\eta\omega,\;\eta\omega^2,\;\ldots,\;\eta\omega^{n-1},</math>

where ''η'' is a single ''n''th root, and 1,&nbsp;''ω'',&nbsp;''ω''{{sup|2}},&nbsp;...&nbsp;''ω''{{sup|''n''−1}} are the ''n''th roots of unity.  For example, the four different fourth roots of 2 are

<math display="block">\sqrt[4]{2},\quad i\sqrt[4]{2},\quad -\sqrt[4]{2},\quad\text{and}\quad -i\sqrt[4]{2}.</math>

In [[polar form]], a single ''n''th root may be found by the formula

<math display="block">\sqrt[n]{re^{i\theta}} = \sqrt[n]{r} \cdot e^{i\theta/n}.</math>

Here ''r'' is the magnitude (the modulus, also called the [[absolute value]]) of the number whose root is to be taken; if the number can be written as ''a+bi'' then <math>r=\sqrt{a^2+b^2}</math>. Also, <math>\theta</math> is the angle formed as one pivots on the origin counterclockwise from the positive horizontal axis to a ray going from the origin to the number; it has the properties that <math>\cos \theta = a/r,</math>  <math> \sin \theta = b/r,</math> and <math> \tan \theta = b/a.</math>

Thus finding ''n''th roots in the complex plane can be segmented into two steps. First, the magnitude of all the ''n''th roots is the ''n''th root of the magnitude of the original number. Second, the angle between the positive horizontal axis and a ray from the origin to one of the ''n''th roots is <math>\theta / n</math>, where <math>\theta</math> is the angle defined in the same way for the number whose root is being taken. Furthermore, all ''n'' of the ''n''th roots are at equally spaced angles from each other.

If ''n'' is even, a complex number's ''n''th roots, of which there are an even number, come in [[additive inverse]] pairs, so that if a number ''r''<sub>1</sub> is one of the ''n''th roots then ''r''<sub>2</sub> =  −''r''<sub>1</sub> is another. This is because raising the latter's coefficient −1 to the ''n''th power for even ''n'' yields 1: that is, (−''r''<sub>1</sub>){{sup|''n''}} = (−1){{sup|''n''}} × ''r''<sub>1</sub>{{sup|''n''}} = ''r''<sub>1</sub>{{sup|''n''}}.

As with square roots, the formula above does not define a [[continuous function]] over the entire complex plane, but instead has a [[branch cut]] at points where ''θ''&nbsp;/&nbsp;''n'' is discontinuous.

==Solving polynomials==
{{see also|Root-finding algorithms}}

It was once [[conjecture]]d that all [[polynomial equation]]s could be [[Algebraic solution|solved algebraically]] (that is, that all roots of a [[polynomial]] could be expressed in terms of a finite number of radicals and [[elementary arithmetic|elementary operations]]). However, while this is true for third degree polynomials ([[cubic function|cubics]]) and fourth degree polynomials ([[quartic function|quartics]]), the [[Abel–Ruffini theorem]] (1824) shows that this is not true in general when the degree is 5 or greater. For example, the solutions of the equation

<math display="block">x^5 = x + 1</math>

cannot be expressed in terms of radicals. (''cf.'' [[quintic equation]])

== Proof of irrationality for non-perfect ''n''th power ''x'' ==
Assume that <math>\sqrt[n]{x}</math> is rational. That is, it can be reduced to a fraction <math>\frac{a}{b}</math>, where {{mvar|a}} and {{mvar|b}} are integers without a common factor.

This means that <math>x = \frac{a^n}{b^n}</math>.

Since ''x'' is an integer, <math>a^n</math>and <math>b^n</math>must share a common factor if <math>b \neq 1</math>. This means that if <math>b \neq 1</math>, <math>\frac{a^n}{b^n}</math> is not in simplest form. Thus ''b'' should equal 1.

Since <math>1^n = 1</math> and <math>\frac{n}{1} = n</math>, <math>\frac{a^n}{b^n} = a^n</math>.

This means that <math>x = a^n</math> and thus, <math>\sqrt[n]{x} = a</math>. This implies that <math>\sqrt[n]{x}</math> is an integer. Since {{mvar|x}} is not a perfect {{mvar|n}}th power, this is impossible. Thus <math>\sqrt[n]{x}</math> is irrational.

==See also==
* [[Geometric mean]]
* [[Twelfth root of two]]

==References==
{{Reflist|30em}}

== External links ==
{{Wiktionary|surd}}
{{Wiktionary|radical}}

{{Hyperoperations}}
{{DISPLAYTITLE:{{mvar|n}}th root}}

[[Category:Elementary algebra]]
[[Category:Operations on numbers]]