*n*th Root

How to find the *n*th root of a number: examples and their solutions.

## How to Solve

To solve an *n*th root,

make the power of *n*,

then cancel the *n*th root and the power of *n*.

For an even root (*n* = 2, 4, 6, ...),

if the signs of both sides are not the same,

use the absolute value signs

to make the signs of both sides the same.

Absolute value

## Example 1: Simplify ^{4}√81

^{4}√81 is a fourth root.

So make 81 the power of fourth.

81 = 3^{4}

Cancel the fourth root and the power of fourth.

Then (given) = 3.

^{4}√81 is an even root.

So check if

the signs of ^{4}√81 and 3 are the same.^{4}√81 is (+).

3 is (+).

So 3 is the answer.

## Example 2: Simplify ^{5}√-32

^{5}√-32 is a fifth root.

So make -32 the power of fifth.

-32 = (-2)^{5}

Cancel the fifth root and the power of fifth.

Then (given) = -2.

^{5}√-32 is not an even root.

So you don't have to check the signs of both sides.

So -2 is the answer.

## Example 3: Simplify √(-7)^{2}

Cancel the square and the square.

Then (given) = -7.

√(-7)^{2} is an even root.

So check if

the signs of √(-7)^{2} and -7 are the same.

√(-7)^{2} is (+).

-7 is (-).

So, to make -7 (+),

change -7 to |-7|.

|-7| = 7

So 7 is the answer.

## Example 4: Simplify √-7^{2}

For an even root,

if the radicand is (-),

then that radical is not a real number.

(Radicand: the number inside the radical sign)

The radicand of √-7^{2} is -7^{2}: (-).

So √-7^{2} is not a real number.

## Example 5: Simplify √*x*^{2}

Cancel the square and the square.

Then (given) = *x*.

√*x*^{2} is an even root.

So check if

the signs of √*x*^{2} and *x* are always the same.

In most cases,

when the variable is (-),

both sides can have different signs.

So check for the case

when the variable is (-).

If *x* is (-),

√*x*^{2} is (+).*x* (brown) is (-).

The signs should be the same.

So, to make *x* (+),

change *x* to |*x*|.

So |*x*| is the answer.

## Example 6: Simplify √*x*^{4}

To cancel the square root,

make a square.*x*^{4} = (*x*^{2})^{2}

Power of a power

Cancel the square and the square.

Then (given) = *x*^{2}.

√*x*^{4} is an even root.

So check if

the signs of √*x*^{4} and *x*^{2} are always the same

when the variable is (-).

If *x* is (-),

√*x*^{4} is (+).*x*^{2} is (+).

The signs are the same.

So *x*^{2} is the answer.

## Example 7: Simplify ^{5}√*x*^{15}

To cancel the cube root,

make a cube.*x*^{15} = (*x*^{3})^{5}

Cancel the fifth root and the power of fifth.

Then (given) = *x*^{3}.

^{5}√*x*^{15} is not an even root.

So you don't have to check the signs of both sides.

So *x*^{3} is the answer.

## Example 8: Simplify ^{4}√16*x*^{12}*y*^{8}

^{4}√16*x*^{12}*y*^{8} is a fourth root.

So change the radicand to the powers of fourth.

16 = 2^{4}*x*^{12} = (*x*^{3})^{4}*y*^{8} = (*y*^{2})^{4}

Cancel the fourth root and the powers of fourth.

Then (given) = 2⋅*x*^{3}⋅*y*^{2}.

^{4}√16*x*^{12}*y*^{8} is an even root.

So check if

the signs of ^{4}√16*x*^{12}*y*^{8} and 2⋅*x*^{3}⋅*y*^{2}

are always the same.

1. If *x* is (-),^{4}√16*x*^{12}*y*^{8} is (+).

2⋅*x*^{3}⋅*y*^{2} is,

2⋅(-)^{3}⋅*y*^{2} = (+)⋅(-)⋅(+),

(-).

So change *x*^{3} to |*x*^{3}|.

2. If *y* is (-),^{4}√16*x*^{12}*y*^{8} is (+).

2⋅|*x*^{3}|⋅*y*^{2} is,

2⋅|*x*^{3}|⋅(-)^{2} = (+)⋅(+)⋅(+),

(+).

So just leave *y*^{2}.

So 2|*x*^{3}|*y*^{2} is the answer.

## Example 9: Simplify ^{3}√27*x*^{12}*y*^{15}

^{3}√27*x*^{12}*y*^{15} is a cube root.

So change the radicand to the cubes.

27 = 3^{3}

(*x* - 8)^{12} = ((*x* - 8)^{4})^{3}*y*^{15} = (*y*^{5})^{3}

Cancel the cube root and the cubes.

Then (given) = 3⋅(*x* - 8)^{4}⋅*y*^{5}.

^{3}√27*x*^{12}*y*^{15} is not an even root.

So you don't have to check the signs of both sides.

So 3(*x* - 8)^{4}*y*^{5} is the answer.