*n*th Root of a Number

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

## Example 1

To solve the *n*th root,

change the radicand to *n*th power form.

Then the *n*th root and the *n*th power are cancelled.

And the base can get out from the radical sign.

Simplifying a radical (part 1)

## Example 2

**1.**, **3.**: The radicand of an even number root cannot be (-).

So it cannot be a real number.

There are no *x* values that satisfy*x*^{2} = -49 or *x*^{4} = -81.**2.**, **4.**: The radicand of an odd number root can be (-).

So it can be simplified just like (+).

(-5)^{3} = -125, (-2)^{5} = -32.

## Example 3

Simplifying an even number root with *x*:

The signs of both 'given' and 'answer'

should be the same,

even if *x* is (-).

√*x*^{2} = *x*

Put *x* = -1:

√(-1)^{2} ≠ -1

(1 ≠ -1)

So add an absolute value sign

to make both sides equal:

√*x*^{2} = |*x*|.

√*x*^{4} = *x*^{2}

Put *x* = -1:

√(-1)^{4} = (-1)^{2}

(1 = 1)

So you don't have to add any absolute value signs:*x*^{2} is the answer.

√*x*^{6} = *x*^{3}

Put *x* = -1:

√(-1)^{6} ≠ (-1)^{3}

(1 ≠ -1)

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

^{4}√16*x*^{12}*y*^{8} = 2⋅*x*^{3}⋅*y*^{2}

Put *x* = *y* = -1:^{4}√16⋅(-1)^{12}⋅(-1)^{8} ≠ 2⋅(-1)^{3}⋅(-1)^{2}

(1 ≠ -1)*x*^{3} makes (-).

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

## Example 4

Simplifying an odd number root with *x*:

You don't need to think of the signs

of the 'given' and the 'answer',

because the radicand of the odd number root

can be (-).