N-th root of a number

In this article, we will learn about the numpy nth root of a number. We are going to learn how to solve the nth root of any number. In python, we use numpy.sqrt() to find the of a number. And we use numpy.cbrt() to find the third root, the cube of a number. But If we want the 5th root or 6th root of a number. What will we do? How to find this root? You will get the solutions for all these questions at the end of the article.

If B is equal to A multiplied n times, then B’s nth root would be equal to A, which is the nth root. The number in a smaller subscript tells which root it is so. Suppose we want the square root of 16. Then we can easily say two times of 4 will give 16. Likewise, if we want the cube root of 8, we say three times of 2 will give 27. Now we go for 32 roots of 5; then we have to find out what number multiplied by itself gives us 32. This is the common idea to get the nth root of any number.

How numpy. power is to get the root?

We are going to use to get the root of a number. It is useful to calculate the power of a number. Here we are using it to find the root. Many of you are confused about how it is possible? We know that 2⁵ = 32. Now, if we want the root, we can interchange it as 32^1/5. Now we will get the root as 2. This is the change we will make in power() to get the root of a number.

What is numpy. power()?

An element in the second array is raised to the power of the element in the first array. The limit of both arrays must be the same and positive numbers.

Syntax

numpy.power(x1, x2, /, out=None, *, where=True, casting='same_kind', order='K', dtype=None, subok=True[, signature, extobj]) = <ufunc 'power'>

Parameters

• x1: base element
• x2: power element
• out: the result will be stored
• where: If the value is True calculate the univerfunction. Otherwise, leave the value.
• **kwargs: Allows to pass keyword variable length of the argument

Return

The bases in x1 are raised to the exponents in x2. This is a scalar if both x1 and x2 are scalars.

The change we are going to do in the power function

To calculate the power:

np.power(element1, element2)

To calculate the root:

np.power(element1, (1/element2))

Example 1

import numpy as np
def nthroot(a,n):
    return np.power(a,(1/n))
a=81
n=7
result=nthroot(a,n)
print(f'The {n}th root of {a} is:',result)

• Import numpy as np.
• Create a function named nthroot.
• The parameters of the function are a and n. a is the number. n is to represent which root it is for.
• Inside a function using np.power() to calculate the root.
• Giving the values of a and n.
• Printing the result

The 7th root of 81 is: 1.873444004574479

7 times of final result will give the value 81.

Example 2

import numpy as np
a = int(input("Enter a number:"))
n =int(input("Enter which root do you want:"))
result=(np.power(a,(1/n)))
print(f'The {n}th root of {a} is:',result)

• Import numpy as np
• Get the values of a and n from the user.
• Using np.power() to calculate the root.
• Finally printing the result

Enter a number:81
Enter which root do you want:6
The 6th root of 81 is: 2.080083823051904

6 times of final result will give the value 81.

Example 3

import numpy as np
a=90
n=4
print ("number: ",a)
print("The nth root you gave: ",n)
result=(np.power(a,(1/n)))
print(f'The {n}th root of {a} is:',result)

• Import numpy as np.
• Giving the values of a and n.
• Using np.power() to calculate the root.
• Printing the result

number:  90
The nth root you gave:  4
The 4th root of 90 is: 3.080070288241023

4 times the final result gives the value 90.

Until we saw how to get a root for a single element, now we will see how to get the root for the array of elements.

Code

import numpy as np
arr = np.arange(5)
n=4
print ("array is: ", arr)
result = np.power(arr,(1/n))
print(f'The {n}th root of {arr[0]} is:',result[0])
print(f'The {n}th root of {arr[1]} is:',result[1])
print(f'The {n}th root of {arr[2]} is:',result[2])
print(f'The {n}th root of {arr[3]} is:',result[3])
print(f'The {n}th root of {arr[4]} is:',result[4])

• Import numpy as np
• Giving the n value as 4.
• So we are going to calculate the 4th root of the every array element.
• Array values are given from 0 to 4.Printing the 4th root of every array elements.

array is:  [0 1 2 3 4]
The 4th root of 0 is: 0.0
The 4th root of 1 is: 1.0
The 4th root of 2 is: 1.189207115002721
The 4th root of 3 is: 1.3160740129524924
The 4th root of 4 is: 1.4142135623730951

Learn how to calculate the sqrt() and cbrt() using numpy

import numpy as np
square_root=np.sqrt(16)
print("The square root of 16 is:",int(square_root))
cube_root=np.cbrt(27)
print("The cube root of 27 is:",int(cube_root))

• Import numpy as np
• Calculating square root using a built-in function np.sqrt()
• Next calculating the cube root using a buil-in function np.cbrt()
• Printing the result.

The square root of 16 is: 4
The cube root of 27 is: 3

1. Which function in numpy is useful to calculate the nth root of a number?

numpy.power() is useful to calculate the nth root of a number.

2. How to calculate the square root of a number using numpy?

numpy.sqrt() is useful to calculate the square root of a number.

3. How to calculate the cube root of a number using numpy?

cbrt() is useful to calculate the square root of a number.

Conclusion

Here we learned about the numpy nth root of a number briefly. We have used numpy.power() to get the root of a number. We hope this article is beneficial for all. Try to solve all the python programs on your own. Learn python in a python pool.

In Mathematics, the nth root of a number x is a number y which when raised to the power n, obtains x:

Here, n is a positive integer, sometimes known as the degree of the root. A root of degree 2 is known as a square root, whereas the root of degree 3 is known as a . Roots of higher degree are also referred to using ordinary as in fourth root, fifth root, twentieth root, etc. The calculation of the nth root is a root extraction.

(Image will be uploaded soon)

Nth Root Definition

Let n be an integer greater than 1, then y is the nth root of x if and only if yⁿ = x.

Nth Root Symbol

It is a radical symbol used for square roots with a little n to define the nth root.

In order to understand the definition of the nth root more precisely, the student needs to be aware of a few other topics that will play a major role in the understanding of the nth root. These topics are explained briefly below. 

Real numbers are referred to as the combination of rational and irrational numbers. All the arithmetic functions are said to be performed on these numbers and they can also be represented on the number line. 

While, on the other hand, imaginary numbers are those that cannot be expressed on a number line, and are usually used to represent complex numbers. Real numbers can be both positive or negative and are usually denoted using the letter R. the natural numbers, fractions, and decimals fall under this category. 

Irrational numbers refers to the real numbers that cannot be expressed in the form of a fraction. It cannot be denoted in the form of a ratio p/q, where the letters p and q refer to integers and q is not equal to zero. One can say that it is the opposite of the rational numbers. 

Irrational numbers are normally expressed in the form R∖Q. The backward slash refers to the ‘set minus’. It is also often expressed in the form of R-Q, which refers to the difference between a set of real numbers and a set of irrational numbers. 

Complex numbers are referred to as numbers that can be expressed in the form of a + ib. a, b are the real numbers, while i refers to the imaginary numbers. For instance, 2+3i is a complex number where 2 is a real number while 3i denotes the imaginary number.

A square root of the number r can be referred to as x, which when squared, gives the result r. 

It is to be noted that every positive real number possesses two square roots, one that is positive and one that is negative. For instance, the number 25 has two square roots, one is 5 and the other is -5. The positive square root is also denoted as the principal square root.

As the square root of every number is non negative, the negative numbers do not possess a square root. But every negative real number has two imaginary square roots associated with them. For example, the square root of -25 will be 5i and -5i. The i here represents the number whose square is supposed to be -1. 

The cube root of a given number x can be a number r whose cube will be x.

How to find the Nth Root of a Number?

Ans: The nth root of a number can be calculated using the Newton method. Let us understand how to find the nth root of a number, ‘A’ using the Newton method.

Start with the initial guess x, and then repeat using the 

When does the Nth Root exist?

In a real 

If n is an even whole number, the nth root of x exists whenever x is positive, and for all x.

If n is an odd whole number, the nth root of x exists for all x.

Things get more complicated in the complex number system.

Every number has a square root, cube root, fourth root, fifth root, and so on.

The fourth root of a number 81 are 3, -3, 3i, -3i, because

3⁴ = 81

-3⁴ = 81

(3i)⁴ = 3⁴ i⁴ = 81

(-3i)⁴ = (-3)⁴ i⁴ = 81

Properties of Nth Root

• Expressing the degree of the nth root in its exponent form as in y¹ makes it easier to manipulate roots and power.

• There is exactly one positive nth root in every positive real number. Hence, the rules of operation with surds including positive radicand x, and y are straightforward within a real number.

• Subtleties can take place while calculating the nth root of a negative or complex number. For example,

Facts to Remember

• The nth root of 0 is 0 for all positive integers n, as 0 is equal to 0.

• The nth root of 1 is known as roots of unity and plays an important role in different areas of  Mathematics such as number theory, the theory of equation, etc.

Simplifying Nth Root

Ans: Let us learn to simplify the nth root through the examples below:

I am looking for an efficient algorithm to find nth root of a number. The answer must be an integer. I have found that newtons method and bisection method are popular methods. Are there any efficient and simple methods for integer output?

asked Dec 22, 2013 at 13:40

20 gold badges71 silver badges104 bronze badges

#include <math.h>
inline int root(int input, int n)
{
  return round(pow(input, 1./n));
}

This works for pretty much the whole integer range (as IEEE754 8-byte doubles can represent the whole 32-bit int range exactly, which are the representations and sizes that are used on pretty much every system). And I doubt any integer based algorithm is faster on non-ancient hardware. Including ARM. Embedded controllers (the microwave washing machine kind) might not have floating point hardware though. But that part of the question was underspecified.

answered Dec 22, 2013 at 13:44

int root(int a, int n) {
    int v = 1, bit, tp, t;
    if (n == 0) return 0; //error: zeroth root is indeterminate!
    if (n == 1) return a;
    tp = iPow(v,n);
    while (tp < a) {    // first power of two such that v**n >= a
        v <<= 1;
        tp = iPow(v,n);
    }
    if (tp == a) return v;  // answer is a power of two
    v >>= 1;
    bit = v >> 1;
    tp = iPow(v, n);    // v is highest power of two such that v**n < a
    while (a > tp) {
        v += bit;       // add bit to value
        t = iPow(v, n);
        if (t > a) v -= bit;    // did we add too much?
        else tp = t;
        if ( (bit >>= 1) == 0) break;
    }
    return v;   // closest integer such that v**n <= a
}
// used by root function...
int iPow(int a, int e) {
    int r = 1;
    if (e == 0) return r;
    while (e != 0) {
        if ((e & 1) == 1) r *= a;
        e >>= 1;
        a *= a;
    }
    return r;
}

answered Sep 12, 2015 at 17:55

1 silver badge2 bronze badges

I question your use of «algorithm» when speaking of C programs. Programs and algorithms are not the same (an algorithm is mathematical; a C program is expected to be implementing some algorithm).

But on current processors (like in recent x86-64 laptops or desktops) the FPU is doing fairly well. I guess (but did not benchmark) that a fast way of computing the n-th root could be,

 inline unsigned root(unsigned x, unsigned n) {
   switch (n) {
     case 0: return 1;
     case 1: return x;
     case 2: return (unsigned)sqrt((double)x);
     case 3: return (unsigned)cbrt((double)x);
     default: return (unsigned) pow (x, 1.0/n);
   }
 }

I am not sure that n-th root of a negative number makes sense in general. So my root function takes some unsigned x and returns some unsigned number.  

answered Dec 22, 2013 at 13:50

Here is an efficient general implementation in C, using a simplified version of the «shifting nth root algorithm» to compute the floor of the nth root of x:

uint64_t iroot(const uint64_t x, const unsigned n)
{
  if ((x == 0) || (n == 0)) return 0;
  if (n == 1) return x;

  uint64_t r = 1;
  for (int s = ((ilog2(x) / n) * n) - n; s >= 0; s -= n)
  {
    r <<= 1;
    r |= (ipow(r|1, n) <= (x >> s));
  }

  return r;
}

It needs this function to compute the nth power of x (using the method of exponentiation by squaring):

uint64_t ipow(uint64_t x, unsigned n)
{
  if (x <= 1) return x;

  uint64_t y = 1;
  for (; n != 0; n >>= 1, x *= x)
    if (n & 1)
      y *= x;
  return y;
}

and this function to compute the floor of base-2 logarithm of x:

int ilog2(uint64_t x)
{
  #if __has_builtin(__builtin_clzll)
    return 63 - ((x != 0) * (int)__builtin_clzll(x)) - ((x == 0) * 64);
  #else
    int y = -(x == 0);
    for (unsigned k = 64 / 2; k != 0; k /= 2)
      if ((x >> k) != 0)
        { x >>= k; y += k; }
    return y;
  #endif
}

Note: This assumes that your compiler understands GCC’s __has_builtin test and that your compiler’s uint64_t type is the same size as an unsigned long long.

answered Oct 26, 2018 at 2:58

Todd Lehman

1 gold badge26 silver badges32 bronze badges

You can try this C function to get the nth_root of an unsigned integer :

unsigned initial_guess_nth_root(unsigned n, unsigned nth){
    unsigned res = 1;
    for(; n >>= 1; ++res);
    return nth ? 1 << (res + nth - 1) / nth : 0 ;
}

// return a number that, when multiplied by itself nth times, makes N.
unsigned nth_root(const unsigned n, const unsigned nth) {
    unsigned a = initial_guess_nth_root(n , nth), b, c, r = nth ? a + (n > 0) : n == 1 ;
    for (; a < r; b = a + (nth - 1) * r, a = b / nth)
        for (r = a, a = n, c = nth - 1; c && (a /= r); --c);
    return r;
}

Example of output :

  24 == (int) pow(15625, 1.0/3)

  25 == nth_root(15625, 3)
   0 == nth_root(0, 0)
   1 == nth_root(1, 0)
   4 == nth_root(4096, 6)
  13 == nth_root(18446744073709551614, 17) // 64-bit 20 digits
  11 == nth_root(340282366920938463463374607431768211454, 37) // 128-bit 39 digits

Here is the github source.

answered Apr 28, 2022 at 15:35

2 silver badges3 bronze badges

What’s the best algorithm to do this myself?

asked Sep 5, 2011 at 13:14

Can you use something like this?

Math.pow(n, 1/root);

Math.pow(25, 1/2) == 5

answered Sep 5, 2011 at 13:16

Digital Plane

7 gold badges56 silver badges59 bronze badges

The nth root of x is the same as x to the power of 1/n. You can simply use Math.pow:

var original = 1000;
var fourthRoot = Math.pow(original, 1/4);
original == Math.pow(fourthRoot, 4); // (ignoring floating-point error)

answered Sep 5, 2011 at 13:16

Note that it does not handle negative nicely — here is a discussion and some code that does

function nthroot(x, n) {
  try {
    var negate = n % 2 == 1 && x < 0;
    if(negate)
      x = -x;
    var possible = Math.pow(x, 1 / n);
    n = Math.pow(possible, n);
    if(Math.abs(x - n) < 1 && (x > 0 == n > 0))
      return negate ? -possible : possible;
  } catch(e){}
}

4 gold badges54 silver badges83 bronze badges

answered Sep 5, 2011 at 13:17

28 gold badges173 silver badges235 bronze badges

You could use

Math.nthroot = function(x,n) {
    //if x is negative function returns NaN
    return this.exp((1/n)*this.log(x));
}
//call using Math.nthroot();

answered Mar 7, 2016 at 11:42

3 silver badges12 bronze badges

For the special cases of square and cubic root, it’s best to use the native functions Math.sqrt and Math.cbrt respectively.

As of ES7, the exponentiation operator ** can be used to calculate the nth root as the ¹/_nth power of a non-negative base:

let root1 = Math.PI ** (1 / 3); // cube root of π

let root2 = 81 ** 0.25;         // 4th root of 81

This doesn’t work with negative bases, though.

let root3 = (-32) ** 5;         // NaN

answered Sep 17, 2016 at 16:28

GOTO 0

21 gold badges124 silver badges155 bronze badges

The n-th root of x is a number r such that r to the power of 1/n is x.

In real numbers, there are some subcases:

• There are two solutions (same value with opposite sign) when x is positive and r is even.
• There is one positive solution when x is positive and r is odd.
• There is one negative solution when x is negative and r is odd.
• There is no solution when x is negative and r is even.

Since Math.pow doesn’t like a negative base with a non-integer exponent, you can use

function nthRoot(x, n) {
  if(x < 0 && n%2 != 1) return NaN; // Not well defined
  return (x < 0 ? -1 : 1) * Math.pow(Math.abs(x), 1/n);
}

nthRoot(+4, 2); // 2 (the positive is chosen, but -2 is a solution too)
nthRoot(+8, 3); // 2 (this is the only solution)
nthRoot(-8, 3); // -2 (this is the only solution)
nthRoot(-4, 2); // NaN (there is no solution)

answered Feb 29, 2016 at 1:01

62 gold badges430 silver badges508 bronze badges

Well, I know this is an old question. But, based on SwiftNinjaPro’s answer, I simplified the function and fixed some NaN issues. Note: This function used ES6 feature, arrow function and template strings, and exponentation. So, it might not work in older browsers:

Math.numberRoot = (x, n) => {
  return (((x > 1 || x < -1) && n == 0) ? Infinity : ((x > 0 || x < 0) && n == 0) ? 1 : (x < 0 && n % 2 == 0) ? `${((x < 0 ? -x : x) ** (1 / n))}${"i"}` : (n == 3 && x < 0) ? -Math.cbrt(-x) : (x < 0) ? -((x < 0 ? -x : x) ** (1 / n)) : (n == 3 && x > 0 ? Math.cbrt(x) : (x < 0 ? -x : x) ** (1 / n)));
};

Math.numberRoot(-64, 3); // Returns -4

Example (Imaginary number result):

Math.numberRoot(-729, 6); // Returns a string containing "3i".

answered Sep 3, 2019 at 18:13

Here’s a function that tries to return the imaginary number. It also checks for a few common things first, ex: if getting square root of 0 or 1, or getting 0th root of number x

function root(x, n){
        if(x == 1){
          return 1;
        }else if(x == 0 && n > 0){
          return 0;
        }else if(x == 0 && n < 0){
          return Infinity;
        }else if(n == 1){
          return x;
        }else if(n == 0 && x > 1){
          return Infinity;
        }else if(n == 0 && x == 1){
          return 1;
        }else if(n == 0 && x < 1 && x > -1){
          return 0;
        }else if(n == 0){
          return NaN;
        }
        var result = false;
        var num = x;
        var neg = false;
        if(num < 0){
            //not using Math.abs because I need the function to remember if the number was positive or negative
            num = num*-1;
            neg = true;
        }
        if(n == 2){
            //better to use square root if we can
            result = Math.sqrt(num);
        }else if(n == 3){
            //better to use cube root if we can
            result = Math.cbrt(num);
        }else if(n > 3){
            //the method Digital Plane suggested
            result = Math.pow(num, 1/n);
        }else if(n < 0){
            //the method Digital Plane suggested
            result = Math.pow(num, 1/n);
        }
        if(neg && n == 2){
            //if square root, you can just add the imaginary number "i=√-1" to a string answer
            //you should check if the functions return value contains i, before continuing any calculations
            result += 'i';
        }else if(neg && n % 2 !== 0 && n > 0){
            //if the nth root is an odd number, you don't get an imaginary number
            //neg*neg=pos, but neg*neg*neg=neg
            //so you can simply make an odd nth root of a negative number, a negative number
            result = result*-1;
        }else if(neg){
            //if the nth root is an even number that is not 2, things get more complex
            //if someone wants to calculate this further, they can
            //i'm just going to stop at *n√-1 (times the nth root of -1)
            //you should also check if the functions return value contains * or √, before continuing any calculations
            result += '*'+n+√+'-1';
        }
        return result;
    }

answered Jul 24, 2019 at 20:29

I have written an algorithm but it is slow when you need many numbers after the point:

NRoot(orginal, nthRoot, base, numbersAfterPoint);

The function returns a string.

var original = 1000;
var fourthRoot = NRoot(original, 4, 10, 32);
console.log(fourthRoot); 
//5.62341325190349080394951039776481

answered Aug 7, 2020 at 3:10

The ultra short version: const nthroot=(x,root)=>x**(1/root);

answered Mar 9 at 16:33

2 silver badges9 bronze badges

Input : A = 81
        N = 4
Output : 3 
3^4 = 81

according to newton’s method
x(K+1) = x(K) – f(x) / f’(x)        
here    f(x)  = x^(N) – A
so    f’(x) = N*x^(N - 1)
and     x(K) denoted the value of x at Kth iteration
putting the values and simplifying we get,
x(K + 1) = (1 / N) * ((N - 1) * x(K) + A / x(K) ^ (N - 1))

N^th root of unity is the root of unity when taken which on taking to the power n gives the value 1. N^th root of any number is defined as the number which on taking to the power of n results in the original number. For example, if we take n^th root of any number say a and the result is b, and then b is raised to n power will get a.

We define n^th root of any number as suppose we take n^th power of any number ‘a’

aⁿ = b

then, n^th root of ‘b’ is ‘a’ we represent this as,

ⁿ√b = a

We can also check for n^th root of unity as,

zⁿ = 1

then, n^th root of ‘1‘ is ‘z‘ we represent this as,

ⁿ√1 = z

In this article, we will learn about, n^th root of any number, n^th root o unity, and others in detail.

What is Nth Root?

We define n^th root of any number as the number which when taken to n^th power results in the original number, i.e.on multiplying the n^th of any number n times we get the original number.

We can mathematically express this as if n^th root of x is y then,

ⁿ√x = y

⇒ yⁿ = x

This can also be represented as,

Example: Find the third root of 8

We know that,

2³ = 2×2×2 = 8

³√8 =  ³√(2×2×2) = 2

Thus, the third root of 8 is 2

N^th Root Symbol

The n^th root of any number is represented using the symbol, ⁿ√x, here, we find the n^th root of x. 

In the expression ⁿ√x, x is called the radicand of the term and n is the index of the term. We can also represent this as the exponent of x in fraction, i.e.

ⁿ√x = (x)^1/n

Nth root of Unity

N^th root of unity is the specific case of n^th root of any number we define n^th root of unity as the number which when multiplying n times gives n. As we know for real numbers if any number is multiplied by 1 we get, the number itself, and 1 when multiplied by itself a finite number of times results in 1. Thus, in the case of real numbers the n^th root of unity is always 1 but, if we consider complex number things gets more interesting when we consider the complex number “i”, “ω” and other roots of unity comes to play.

Thus, we have various roots of unity. The solution to the equation,

zⁿ = 1

gives the n^th root of unity.

We can easily solve this equation using complex numbers.

How to Find n^th Root of Unity?

N^th root of unity can be easily found by finding the solution to the equation,

zⁿ = 1

In polar form, we write this equation as,

zⁿ = cos 0 + i sin 0

In general form,

zⁿ = cos (0+2mπ) + i sin (0+2mπ)                    (where m∈N)

Taking n^th root on both sides,

Using DeMoivre’s Theorem

This can be represented in Euler Form,

z = e^(i2mπ/n)

This is the n^th root of unity for m ∈ N

N^th Root of Unity in Complex Numbers

We know that the general form of a complex number is x + iy.

Comparing the n^th root of unity to a general complex number we get,

Squaring and adding (i) and (ii) we get,

x² + y² = cos²(2mπ/n) + sin²(2mπ/n) = 1

The above equation is the equation of the circle with the centre at origin (0,0) and radius 1.

If we represent the complex number as ω,

ω =  e^(i2mπ/n)

Taking power n both sides,

(ω)ⁿ =  e^(i2mπ/n)n

⇒ ωⁿ =  e^i2mπ

This gives the n^th root of unity taking n ≥ 0, we get the root of unity as,

These roots can be represented in a unit circle in a complex plane as,

Properties of Nth Root of unity

• N^th root of unity is the points on the circumference of the circle in the complex plane with the center on the origin and a radius of 1 unit.
• 1 and -1 are the 2nd roots of unity.
• 1, -1. i, and -1 are the 4th roots of unity.
• i = √(-1) and i² = -1, i⁴ = 1
• 1, ω = (-1/2 + i√(3)/2), ω² =(-1/2 – i√(3)/2) are the three roots of the unity.
• Multiplying two imaginary cube roots of unity results in 1.
• The sum of all three cube roots of unity, i.e.

1 + ω + ω² = 0

• All the n^th roots of unity are in GP i.e. 1, ω², ω³, ω⁴, …., ω^n-1, are in GP
• The product of all the roots of unity is,

Solved Examples on Nth Root

Example 1: Find ⁵√(-243)

As we know,

(-3)⁵ = -243

Thus,⁵√(-243) =  (-3)⁵ 

Thus, ⁵√(-243) = -3.

Example 2: Simplify ⁴√(16x⁸)

As 16 = 2×2×2×2 = 2⁴

Thus, ⁴√(16x⁸) = ⁴√(2⁴x⁸)

⇒  ⁴√(16x⁸) = 2^4/4x^8/2

⇒  ⁴√(16x⁸) = 2x²

Example 3: Simplify ⁴√(-x²⁴)

⇒ ⁴√(-x²⁴) = (-1)^1/4 ×(x²⁴)^1/4

⇒ ⁴√(-x²⁴) = ix⁸

FAQs on Nth Root

Q1: What is the n^th root?

The n^th root of unity is the number which when multiplied n times gives the original number. If the n^th root of unity is x^1/n = b, then

bⁿ = 1

Q2: What is a Square Root?

The square root of the number is the number which when multiplied by itself gives the original number. It is represented by the symbol √, Now the square root of (x) is represented as, √(x). For example, the square root of 4 is,

√(4) =√(2×2) = 2

Q3: What is a Cube Root?

The cube root of the number is the number which when multiplied by itself thrice gives the original number. It is represented by the symbol ³√, Now the cube root of (x) is represented as, ³√(x). For example, the cube root of 8 is,

√(8) = √(2×2×2) = 2

Q4: What are the Cube Roots of Unity?

Cube root of unity is the number which when multiplied with itself three times gives results as 1. We can represent this as,

³√(1) = 1, ω, ω²

Here, 1, ω, and ω² are the cube root of unity.

Q5: What is the fourth root of unity?

Fourth root of unity is the number which when multiplied with itself four times gives results as 1. We can represent this as,

⁴√(1) = 1, -1, i, -i

Here, 1, -1, i, and -i are the fourth roots of the unity.