Identify and Simplify Roots

Learning Outcomes

Simplify principal square roots using factorization
Use cube root notation to write cube roots
Simplify cube roots using factorization
Simplify square roots with variables
Determine when a simplified root needs an absolute value
Convert between radical and exponent notation
Use the laws of exponents to simplify expressions with rational exponents
Use rational exponents to simplify radical expressions

We know how to square a number:

$5^2=25$ and $\left(-5\right)^2=25$

Taking a square root is the opposite of squaring so we can make these statements:

5 is the nonngeative square root of 25
-5 is the negative square root of 25

Find the square roots of the following numbers:

We want to find a number whose square is 36. $6^2=36$ therefore, the nonnegative square root of 36 is 6 and the negative square root of 36 is -6
We want to find a number whose square is 81. $9^2=81$ therefore, the nonnegative square root of 81 is 9 and the negative square root of 81 is -9
We want to find a number whose square is -49. When you square a real number, the result is always positive. Stop and think about that for a second. A negative number times itself is positive, and a positive number times itself is positive. Therefore, -49 does not have square roots, there are no real number solutions to this question.
We want to find a number whose square is 0. $0^2=0$ therefore, the nonnegative square root of 0 is 0. We do not assign 0 a sign, so it has only one square root, and that is 0.

The notation that we use to express a square root for any real number, a, is as follows:

Writing a Square Root

The symbol for the square root is called a radical symbol. For a real number, a the square root of a is written as $\sqrt{a}$

The number that is written under the radical symbol is called the radicand.

By definition, the square root symbol, $\sqrt{\hphantom{5}}$ always means to find the nonnegative root, called the principal root.

$\sqrt{-a}$ is not defined, therefore $\sqrt{a}$ is defined for $a>0$

Let’s do an example similar to the example from above, this time using square root notation. Note that using the square root notation means that you are only finding the principal root – the nonnegative root.

Example

Simplify the following square roots:

$\sqrt{16}$
$\sqrt{9}$
$\sqrt{-9}$
$\sqrt{5^2}$

Show Solution

$\sqrt{16}$ . We are looking for a number whose square is 16, so $\sqrt{16}=4$ . We only write the nonnegative root because that is how the root symbol is defined.
$\sqrt{9}$ . We are looking for a number whose square is 9, so $\sqrt{9}=3$ . We only write the nonnegative root because that is how the root symbol is defined.
$\sqrt{-9}$ . We are looking for a number whose square is -9. There are no real numbers whose square is -9, so this radical is not a real number.
$\sqrt{5^2}$ . We are looking for a number whose square is $5^2$ . We already have the number whose square is $5^2$ , it’s 5!

The last problem in the previous example shows us an important relationship between squares and square roots, and we can summarize it as follows:

The square root of a square

For a nonnegative real number, a, $\sqrt{a^2}=a$

In the video that follows, we simplify more square roots using the fact that $\sqrt{a^2}=a$ means finding the principal square root.

What if you are working with a number whose square you do not know right away? We can use factoring and the product rule for square roots to find square roots such as $\sqrt{144}$ , or $\sqrt{225}$ .

The Product Rule for Square Roots

Given that a and b are nonnegative real numbers, $\sqrt{a\cdot{b}}=\sqrt{a}\cdot\sqrt{b}$

In the examples that follow we will bring together these ideas to simplify square roots of numbers that are not obvious at first glance:

square root of a square,
the product rule for square roots
factoring

Example

Simplify $\sqrt{144}$

Show Solution

Determine the prime factors of 144.

$\begin{array}{c}\sqrt{144}\\\\\sqrt{2\cdot 72}\\\\\sqrt{2\cdot 2\cdot 36}\\\\\sqrt{2\cdot 2\cdot 2\cdot 18}\\\\\sqrt{2\cdot 2\cdot 2\cdot 2\cdot 9}\\\\\sqrt{2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3}\end{array}$

Because we are finding a square root, we regroup these factors into squares.

$\sqrt{2^2\cdot 2^2\cdot3^2}$

Now we can use the product rule for square roots and the square root of a square idea to finish finding the square root.

$\begin{array}{c}\sqrt{2^2\cdot 2^2\cdot3^2}\\\\=\sqrt{2^2}\cdot\sqrt{2^2}\cdot\sqrt{3^2}\\\\=2\cdot3\cdot2\\\\=12\end{array}$

Answer

$\sqrt{144}=12$

Example

Simplify $\sqrt{225}$

Show Solution

First, factor 225:

$\begin{array}{c}\sqrt{225}\\\\=\sqrt{5\cdot45}\\\\=\sqrt{5\cdot5\cdot9}\\\\=\sqrt{5\cdot5\cdot3\cdot3}\end{array}$

Because we are finding a square root, we regroup these factors into squares. Finish simplifying with the product rule for roots, and the square of a square idea.

$\begin{array}{c}\sqrt{5^2\cdot3^2}\\\\=\sqrt{5^2}\cdot\sqrt{3^2}\\\\=5\cdot3=15\end{array}$

Answer

$\sqrt{225}=15$

Caution! The square root of a product rule applies when you have multiplication ONLY under the square root. You cannot apply the rule to sums:

$\sqrt{a+b}\ne\sqrt{a}+\sqrt{b}$

Prove this to yourself with some real numbers: let a = 64 and b = 36, then use the order of operations to simplify each expression.

$\begin{array}{c}\sqrt{64+36}=\sqrt{100}=10\\\\\sqrt{64}+\sqrt{36}=8+6=14\\\\10\ne14\end{array}$

So far, you have seen examples that are perfect squares. That is, each is a number whose square root is an integer. But many radical expressions are not perfect squares. Some of these radicals can still be simplified by finding perfect square factors. The example below illustrates how to factor the radicand, looking for pairs of factors that can be expressed as a square.

Example

Simplify. $\sqrt{63}$

Show Solution

Factor 63

$\sqrt{7\cdot 3\cdot3}$

Regroup factors into squares

$\sqrt{7\cdot3^2}$

Finish simplifying with the product rule for roots, and the square of a square idea.

$\sqrt{7\cdot3^2}\\\\=\sqrt{7}\cdot\sqrt{3^2}\\\\=\sqrt{7}\cdot3$

Since 7 is prime and we can’t write it as a square, it will have to stay under the radical sign. As a matter of convention, we write the constant, 3, in front of the radical. This helps the reader know that the 3 is not under the radical anymore.

$3\cdot \sqrt{7}$

Answer

$\sqrt{63}=3\sqrt{7}$

The final answer $3\sqrt{7}$ may look a bit odd, but it is in simplified form. You can read this as “three radical seven” or “three times the square root of seven.”

In the next example, we take a bit of a shortcut by making use of the common squares we know, instead of using prime factors. It helps to have the squares of the numbers between 0 and 10 fresh in your mind to make simplifying radicals faster.

$0^2=0$
$2^2=4$
$3^2=9$
$4^2=16$
$5^2=25$
$6^2=36$
$7^2=49$
$8^2=64$
$9^2=81$
$10^2=100$

Example

Simplify. $\sqrt{2,000}$

Show Solution

Factor 2,000 to find perfect squares.

$\begin{array}{r}\sqrt{100\cdot 20}\\\\=\sqrt{100\cdot 4\cdot 5}\end{array}$

$100=10^2,4=2^2$

$\begin{array}\sqrt{100\cdot 4\cdot 5}\\\\= \sqrt{10^2\cdot 4^2\cdot 5}\\\\=\sqrt{10^2}\cdot\sqrt{4^2}\cdot\sqrt{5}\\\\=10\cdot4\cdot\sqrt{5}\end{array}$

Multiply.

$20\cdot \sqrt{5}$

Answer

$\sqrt{2,000}=20\sqrt{5}$

In this last video, we show examples of simplifying radicals that are not perfect squares.

Cube Roots

While square roots are probably the most common radical, you can also find the third root, the fifth root, the 10th root, or really any other nth root of a number. Just as the square root is a number that, when squared, gives the radicand, the cube root is a number that, when cubed, gives the radicand.

Find the cube roots of the following numbers:

27
8
-8
0

We want to find a number whose cube is 27. $3\cdot9=27$ and $9=3^2$ , so $3/cdot3/cdot3=3^3=27$
We want to find a number whose cube is 8. $2\cdot2\cdot2=8$ the cube root of 8 is 2.
We want to find a number whose cube is -8. We know 2 is the cube root of 8, so maybe we can try -2. $-2\cdot{-2}\cdot{-2}=-8$ , so the cube root of -8 is -2. This is different from square roots because multiplying three negative numbers together results in a negative number.
We want to find a number whose cube is 0. $0\cdot0\cdot0$ , no matter how many times you multiply $0$ by itself, you will always get $0$ .

The cube root of a number is written with a small number 3, called the index, just outside and above the radical symbol. It looks like $\sqrt[3]{{}}$ . This little 3 distinguishes cube roots from square roots which are written without a small number outside and above the radical symbol.

Caution! Be careful to distinguish between $\sqrt[3]{x}$ , the cube root of x, and $3\sqrt{x}$ , three times the square root of x. They may look similar at first, but they lead you to much different expressions!

We can also use factoring to simplify cube roots such as $\sqrt[3]{125}$ . You can read this as “the third root of 125” or “the cube root of 125.” To simplify this expression, look for a number that, when multiplied by itself two times (for a total of three identical factors), equals 125. Let’s factor 125 and find that number.

Example

Simplify. $\sqrt[3]{125}$

Show Solution

125 ends in 5, so you know that 5 is a factor. Expand 125 into

5 \cdot 25

$5\cdot25$ .

$\sqrt[3]{5\cdot 25}$

Factor 25 into 5 and 5.

$\sqrt[3]{5\cdot 5\cdot 5}$

The factors are $5\cdot5\cdot5$ , or $5^{3}$ .

$\sqrt[3]{{{5}^{3}}}$

Answer

$\sqrt[3]{125}=5$

The prime factors of 125 are $5\cdot5\cdot5$ , which can be rewritten as $5^{3}$ . The cube root of a cubed number is the number itself, so $\sqrt[3]{{{5}^{3}}}=5$ . You have found the cube root, the three identical factors that when multiplied together give 125. 125 is known as a perfect cube because its cube root is an integer.

Here’s an example of how to simplify a radical that is not a perfect cube.

Example

Simplify. $\sqrt[3]{32{{m}^{5}}}$

Show Solution

Factor 32 into prime factors.

$\sqrt[3]{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot {{m}^{5}}}$

Since you are looking for the cube root, you need to find factors that appear 3 times under the radical. Rewrite $2\cdot 2\cdot 2$ as ${{2}^{3}}$ .

$\sqrt[3]{{{2}^{3}}\cdot 2\cdot 2\cdot {{m}^{5}}}$

Rewrite ${{m}^{5}}$ as ${{m}^{3}}\cdot {{m}^{2}}$ .

$\sqrt[3]{{{2}^{3}}\cdot 2\cdot 2\cdot {{m}^{3}}\cdot {{m}^{2}}}$

Rewrite the expression as a product of multiple radicals.

$\sqrt[3]{{{2}^{3}}}\cdot \sqrt[3]{2\cdot 2}\cdot \sqrt[3]{{{m}^{3}}}\cdot \sqrt[3]{{{m}^{2}}}$

Simplify and multiply.

$2\cdot \sqrt[3]{4}\cdot m\cdot \sqrt[3]{{{m}^{2}}}$

Answer

$\sqrt[3]{32{{m}^{5}}}=2m\sqrt[3]{4{{m}^{2}}}$

In the example below, we use the following idea:

$\sqrt[3]{{{(-1)}^{3}}}=-1$

to simplify the radical. You do not have to do this, but it may help you recognize cubes more easily when they are nonnegative.

Example

Simplify. $\sqrt[3]{-27{{x}^{4}}{{y}^{3}}}$

Show Solution

Factor the expression into cubes.

Separate the cubed factors into individual radicals.

$\begin{array}{r}\sqrt[3]{-1\cdot 27\cdot {{x}^{4}}\cdot {{y}^{3}}}\\\sqrt[3]{{{(-1)}^{3}}\cdot {{(3)}^{3}}\cdot {{x}^{3}}\cdot x\cdot {{y}^{3}}}\\\sqrt[3]{{{(-1)}^{3}}}\cdot \sqrt[3]{{{(3)}^{3}}}\cdot \sqrt[3]{{{x}^{3}}}\cdot \sqrt[3]{x}\cdot \sqrt[3]{{{y}^{3}}}\end{array}$

Simplify the cube roots.

$-1\cdot 3\cdot x\cdot y\cdot \sqrt[3]{x}$

Answer

$\sqrt[3]{-27{{x}^{4}}{{y}^{3}}}=-3xy\sqrt[3]{x}$

In the video that follows, we show more examples if simplifying cube roots.

You could check your answer by performing the inverse operation. If you are right, when you cube $-3xy\sqrt[3]{x}$ you should get $-27{{x}^{4}}{{y}^{3}}$ .

$\begin{array}{l}\left( -3xy\sqrt[3]{x} \right)\left( -3xy\sqrt[3]{x} \right)\left( -3xy\sqrt[3]{x} \right)\\-3\cdot -3\cdot -3\cdot x\cdot x\cdot x\cdot y\cdot y\cdot y\cdot \sqrt[3]{x}\cdot \sqrt[3]{x}\cdot \sqrt[3]{x}\\-27\cdot {{x}^{3}}\cdot {{y}^{3}}\cdot \sqrt[3]{{{x}^{3}}}\\-27{{x}^{3}}{{y}^{3}}\cdot x\\-27{{x}^{4}}{{y}^{3}}\end{array}$

You can find the odd root of a negative number, but you cannot find the even root of a negative number. This means you can simplify the radicals $\sqrt[3]{-81},\ \sqrt[5]{-64}$ , and $\sqrt[7]{-2187}$ , but you cannot simplify the radicals $\sqrt[{}]{-100},\ \sqrt[4]{-16}$ , or $\sqrt[6]{-2,500}$ .

Let’s look at another example.

Example

Simplify. $\sqrt[3]{-24{{a}^{5}}}$

Show Solution

Factor

- 24

$−24$ to find perfect cubes. Here,

- 1

$−1$ and 8 are the perfect cubes.

$\sqrt[3]{-1\cdot 8\cdot 3\cdot {{a}^{5}}}$

Factor variables. You are looking for cube exponents, so you factor $a^{5}$ into $a^{3}$ and $a^{2}$ .

$\sqrt[3]{{{(-1)}^{3}}\cdot {{2}^{3}}\cdot 3\cdot {{a}^{3}}\cdot {{a}^{2}}}$

Separate the factors into individual radicals.

$\sqrt[3]{{{(-1)}^{3}}}\cdot \sqrt[3]{{{2}^{3}}}\cdot \sqrt[3]{{{a}^{3}}}\cdot \sqrt[3]{3\cdot {{a}^{2}}}$

Simplify, using the property $\sqrt[3]{{{x}^{3}}}=x$ .

$-1\cdot 2\cdot a\cdot \sqrt[3]{3\cdot {{a}^{2}}}$

This is the simplest form of this expression; all cubes have been pulled out of the radical expression.

$-2a\sqrt[3]{3{{a}^{2}}}$

Answer

$\sqrt[3]{-24{{a}^{5}}}=-2a\sqrt[3]{3{{a}^{2}}}$

The steps to consider when simplifying a radical are outlined below.

Simplifying a radical

When working with exponents and radicals:

If n is odd, $\sqrt[n]{{{x}^{n}}}=x$ .
If n is even, $\sqrt[n]{{{x}^{n}}}=\left| x \right|$ . (The absolute value accounts for the fact that if x is negative and raised to an even power, that number will be positive, as will the nth principal root of that number.)

Example

Simplify. $\sqrt{100{{x}^{2}}{{y}^{4}}}$

Show Solution

Separate factors; look for squared numbers and variables. Factor 100 into

10 \cdot 10

$10\cdot10$ .

$\sqrt{10\cdot 10\cdot {{x}^{2}}\cdot {{y}^{4}}}$

Factor $y^{4}$ into $\left(y^{2}\right)^{2}$ .

$\sqrt{10\cdot 10\cdot {{x}^{2}}\cdot {{({{y}^{2}})}^{2}}}$

Separate the squared factors into individual radicals.

$\sqrt{{{10}^{2}}}\cdot \sqrt{{{x}^{2}}}\cdot \sqrt{{{({{y}^{2}})}^{2}}}$

Take the square root of each radical . Since you do not know whether x is positive or negative, use $\left|x\right|$ to account for both possibilities, thereby guaranteeing that your answer will be positive.

$10\cdot\left|x\right|\cdot{y}^{2}$

Simplify and multiply.

$10\left|x\right|y^{2}$

Answer

$\sqrt{100{{x}^{2}}{{y}^{4}}}=10\left| x \right|{{y}^{2}}$

You can check your answer by squaring it to be sure it equals $100{{x}^{2}}{{y}^{4}}$ .

In the last video, we share examples of finding cube roots with negative radicands.

Simplify Square Roots with Variables

Radical expressions are expressions that contain radicals. Radical expressions come in many forms, from simple and familiar, such as $\sqrt{16}$ , to quite complicated, as in $\sqrt[3]{250{{x}^{4}}y}$ . Using factoring, you can simplify these radical expressions, too.

Simplifying Square Roots

Radical expressions will sometimes include variables as well as numbers. Consider the expression $\sqrt{9{{x}^{6}}}$ . Simplifying a radical expression with variables is not as straightforward as the examples we have already shown with integers.

Consider the expression $\sqrt{{{x}^{2}}}$ . This looks like it should be equal to x, right? Let’s test some values for x and see what happens.

In the chart below, look along each row and determine whether the value of x is the same as the value of $\sqrt{{{x}^{2}}}$ . Where are they equal? Where are they not equal?

After doing that for each row, look again and determine whether the value of $\sqrt{{{x}^{2}}}$ is the same as the value of $\left|x\right|$ .

$x$	$x^{2}$	$\sqrt{x^{2}}$	$\left\|x\right\|$
$−5$	25	5	5
$−2$	4	2	2
0	0	0	0
6	36	6	6
10	100	10	10

Notice—in cases where x is a negative number, $\sqrt{x^{2}}\neq{x}$ ! (This happens because the process of squaring the number loses the negative sign, since a negative times a negative is a positive.) However, in all cases $\sqrt{x^{2}}=\left|x\right|$ . You need to consider this fact when simplifying radicals that contain variables, because by definition $\sqrt{x^{2}}$ is always nonnegative.

Taking the Square Root of a Radical Expression

When finding the square root of an expression that contains variables raised to a power, consider that $\sqrt{x^{2}}=\left|x\right|$ .

Examples: $\sqrt{9x^{2}}=3\left|x\right|$ , and $\sqrt{16{{x}^{2}}{{y}^{2}}}=4\left|xy\right|$

Let’s try it.
The goal is to find factors under the radical that are perfect squares so that you can take their square root.

Example

Simplify. $\sqrt{9{{x}^{6}}}$

Show Solution

Factor to find identical pairs.

$\sqrt{3\cdot 3\cdot {{x}^{3}}\cdot {{x}^{3}}}$

Rewrite the pairs as perfect squares, note how we use the power rule for exponents to simplify $x^6$ into a square: ${x^3}^2$

$\sqrt{{{3}^{2}}\cdot {{\left( {{x}^{3}} \right)}^{2}}}$

Separate into individual radicals.

$\sqrt{{{3}^{2}}}\cdot \sqrt{{{\left( {{x}^{3}} \right)}^{2}}}$

Simplify, using the rule that $\sqrt{{{x}^{2}}}=\left|x\right|$ .

$3\left|{{x}^{3}}\right|$

Answer

$\sqrt{9{{x}^{6}}}=3\left|{{x}^{3}}\right|$

Variable factors with even exponents can be written as squares. In the example above, ${{x}^{6}}={{x}^{3}}\cdot{{x}^{3}}={\left|x^3\right|}^{2}$ and

${{y}^{4}}={{y}^{2}}\cdot{{y}^{2}}={\left(|y^2\right|)}^{2}$ .

Let’s try to simplify another radical expression.

Example

Simplify. $\sqrt{49{{x}^{10}}{{y}^{8}}}$

Show Solution

Look for squared numbers and variables. Factor 49 into

7 \cdot 7

$7\cdot7$ ,

x^{10}

$x^{10}$ into

x^{5} \cdot x^{5}

$x^{5}\cdot{x}^{5}$ , and

y^{8}

$y^{8}$ into

y^{4} \cdot y^{4}

$y^{4}\cdot{y}^{4}$ .

$\sqrt{7\cdot 7\cdot {{x}^{5}}\cdot{{x}^{5}}\cdot{{y}^{4}}\cdot{{y}^{4}}}$

Rewrite the pairs as squares.

$\sqrt{{{7}^{2}}\cdot{{({{x}^{5}})}^{2}}\cdot{{({{y}^{4}})}^{2}}}$

Separate the squared factors into individual radicals.

$\sqrt{7^2}\cdot\sqrt{({x^5})^2}\cdot\sqrt{({y^4})^2}$

Take the square root of each radical using the rule that $\sqrt{{{x}^{2}}}=\left|x\right|$ .

$7\cdot\left|{{x}^{5}}\right|\cdot{{y}^{4}}$

Multiply.

$7\left|{{x}^{5}}\right|{{y}^{4}}$

Answer

$\sqrt{49{{x}^{10}}{{y}^{8}}}=7\left|{{x}^{5}}\right|{{y}^{4}}$

You find that the square root of $49{{x}^{10}}{{y}^{8}}$ is $7\left|{{x}^{5}}\right|{{y}^{4}}$ . In order to check this calculation, you could square $7\left|{{x}^{5}}\right|{{y}^{4}}$ , hoping to arrive at $49{{x}^{10}}{{y}^{8}}$ . And, in fact, you would get this expression if you evaluated ${\left({7\left|{{x}^{5}}\right|{{y}^{4}}}\right)^{2}}$ .

In the video that follows we show several examples of simplifying radicals with variables.

Example

Simplify. $\sqrt{{{a}^{3}}{{b}^{5}}{{c}^{2}}}$

Show Solution

Factor to find variables with even exponents.

$\sqrt{{{a}^{2}}\cdot a\cdot {{b}^{4}}\cdot{b}\cdot{{c}^{2}}}$

Rewrite $b^{4}$ as $\left(b^{2}\right)^{2}$ .

$\sqrt{{{a}^{2}}\cdot a\cdot {{({{b}^{2}})}^{2}}\cdot{ b}\cdot{{c}^{2}}}$

Separate the squared factors into individual radicals.

$\sqrt{{{a}^{2}}}\cdot\sqrt{{{({{b}^{2}})}^{2}}}\cdot\sqrt{{{c}^{2}}}\cdot \sqrt{a\cdot b}$

Take the square root of each radical. Remember that $\sqrt{{{a}^{2}}}=\left| a \right|$ .

$\left| a \right|\cdot {{b}^{2}}\cdot\left|{c}\right|\cdot\sqrt{a\cdot b}$

Simplify and multiply. The entire quantity $a{{b}^{2}}c$ can be enclosed in the absolute value sign because $b^2$ will be positive anyway.

$\left| a{{b}^{2}}c \right|\sqrt{ab}$

Answer

$\sqrt{{{a}^{3}}{{b}^{5}}{{c}^{2}}}=\left| a{{b}^{2}}c\right|\sqrt{ab}$

In the next section, we will explore cube roots, and use the methods we have shown here to simplify them. Cube roots are unique from square roots in that it is possible to have a negative number under the root, such as $\sqrt[3]{-125}$ .

Rational Exponents

Roots can also be expressed as fractional exponents. The square root of a number can be written with a radical symbol or by raising the number to the $\frac{1}{2}$ power. This is illustrated in the table below.

Exponent Form	Root Form	Root of a Square	Simplified
${{25}^{\frac{1}{2}}}$	$\sqrt{25}$	$\sqrt{{{5}^{2}}}$	5
${{16}^{\frac{1}{2}}}$	$\sqrt{16}$	$\sqrt{{{4}^{2}}}$	4
${{100}^{\frac{1}{2}}}$	$\sqrt{100}$	$\sqrt{{{10}^{2}}}$	10

Use the example below to familiarize yourself with the different ways to write square roots.

Example

Fill in the missing cells in the table.

Exponent Form	Root Form	Root of a Square
${{36}^{\frac{1}{2}}}$
	$\sqrt{81}$
		$\sqrt{{{12}^{2}}}$

Show Solution

Exponent Form	Root Form	Root of a Square	Simplified
${{36}^{\frac{1}{2}}}$	$\sqrt{36}$	$\sqrt{{{6}^{2}}}$	6
${{81}^{\frac{1}{2}}}$	$\sqrt{81}$	$\sqrt{{{9}^{2}}}$	9
${{144}^{\frac{1}{2}}}$	$\sqrt{144}$	$\sqrt{{{12}^{2}}}$	12

In the following video, we show another example of filling in a table to connect the different notation used for roots.

We can extend the concept of writing $\sqrt{x}=x^{\frac{1}{2}}$ to cube roots. Remember, cubing a number raises it to the power of three. Notice that in these examples, the denominator of the rational exponent is the number 3.

Radical Form	Exponent Form	Integer
$\sqrt[3]{8}$	${{8}^{\tfrac{1}{3}}}$	2
$\sqrt[3]{8}$	${{125}^{\tfrac{1}{3}}}$	5
$\sqrt[3]{1000}$	${{1000}^{\tfrac{1}{3}}}$	10

These examples help us model a relationship between radicals and rational exponents: namely, that the nth root of a number can be written as either $\sqrt[n]{x}$ or ${{x}^{\frac{1}{n}}}$ .

Radical Form	Exponent Form
$\sqrt{x}$	${{x}^{\tfrac{1}{2}}}$
$\sqrt[3]{x}$	${{x}^{\tfrac{1}{3}}}$
$\sqrt[4]{x}$	${{x}^{\tfrac{1}{4}}}$
…	…
$\sqrt[n]{x}$	${{x}^{\tfrac{1}{n}}}$

Convert Between Radical and Exponent Notation

When faced with an expression containing a rational exponent, you can rewrite it using a radical. In the table above, notice how the denominator of the rational exponent determines the index of the root. So, an exponent of $\frac{1}{2}$ translates to the square root, an exponent of $\frac{1}{5}$ translates to the fifth root or $\sqrt[5]{{\hphantom{5}}}$ , and $\frac{1}{8}$ translates to the eighth root or $\sqrt[8]{{\hphantom{5}}}$ .

Example

Write $\sqrt[3]{81}$ as an expression with a rational exponent.

Show Solution

The radical form

\sqrt[4]{}

$\sqrt[4]{{}}$ can be rewritten as the exponent

\frac{1}{4}

$\frac{1}{4}$ . Remove the radical and place the exponent next to the base.

${{81}^{\frac{1}{3}}}$

Answer

$\sqrt[3]{81}={{81}^{\frac{1}{3}}}$

Example

Express ${{(2x)}^{^{\frac{1}{3}}}}$ in radical form.

Show Solution

Rewrite the expression with the fractional exponent as a radical. The denominator of the fraction determines the root, in this case the cube root.

$\sqrt[3]{2x}$

The parentheses in ${{\left( 2x \right)}^{\frac{1}{3}}}$ indicate that the exponent refers to everything within the parentheses.

Answer

${{(2x)}^{^{\frac{1}{3}}}}=\sqrt[3]{2x}$

Remember that exponents only refer to the quantity immediately to their left unless a grouping symbol is used. The example below looks very similar to the previous example with one important difference—there are no parentheses! Look what happens.

Example

Express $2{{x}^{^{\frac{1}{2}}}}$ in radical form.

Show Solution

Rewrite the expression with the fractional exponent as a radical. The denominator of the fraction determines the root, in this case the cube root.

$2\sqrt{x}$

The exponent refers only to the part of the expression immediately to the left of the exponent, in this case x, but not the 2.

Answer

$2{{x}^{^{\frac{1}{2}}}}=2\sqrt{x}$

The next example is intended to help you practice placing a rational exponent on the appropriate terms in an expression that is written in radical form

Example

Express $4\sqrt[3]{xy}$ with rational exponents.

Show Solution

Rewrite the radical using a rational exponent. The root determines the fraction. In this case, the index of the radical is 3, so the rational exponent will be

\frac{1}{3}

$\frac{1}{3}$ .

$4{{(xy)}^{\frac{1}{3}}}$

Since 4 is outside the radical, it is not included in the grouping symbol and the exponent does not refer to it.

Answer

$4\sqrt[3]{xy}=4{{(xy)}^{\frac{1}{3}}}$

In the next video, we show examples of converting between radical and exponent form.

When converting from radical to rational exponent notation, the degree of the root becomes the denominator of the exponent. If you start with a square root, you will have an exponent of $\frac{1}{2}$ on the expression in the radical (the radicand). On the other hand, if you start with an exponent of $\frac{1}{3}$ you will use a cube root. The following statement summarizes this idea.

Writing Fractional Exponents

Any radical in the form $\sqrt[n]{a}$ can be written using a fractional exponent in the form $a^{\frac{1}{n}}$ .

Simplifying Radical Expressions Using Rational Exponents and the Laws of Exponents

Let’s explore some radical expressions now and see how to simplify them. Let’s start by simplifying this expression, $\sqrt[3]{{{a}^{6}}}$ .

One method of simplifying this expression is to factor and pull out groups of $a^{3}$ , as shown below in this example.

Example

Simplify. $\sqrt[3]{{{a}^{6}}}$

Show Solution

Rewrite by factoring out cubes.

$\sqrt[3]{{{a}^{3}}\cdot {{a}^{3}}}$

Write each factor under its own radical and simplify.

$\begin{array}{r}\sqrt[3]{{{a}^{3}}}\cdot \sqrt[3]{{{a}^{3}}}\\a\cdot{a}\end{array}$

Answer

$\sqrt[3]{{{a}^{6}}}={{a}^{2}}$

You can also simplify this expression by thinking about the radical as an expression with a rational exponent, and using the principle that any radical in the form $\sqrt[n]{{{a}^{x}}}$ can be written using a fractional exponent in the form ${{a}^{\tfrac{x}{n}}}$ .

Example

Simplify. $\sqrt[3]{{{a}^{6}}}$

Show Solution

Rewrite the radical using a rational exponent.

${{a}^{\frac{6}{3}}}$

Simplify the exponent.

${{a}^{2}}$

Answer

$\sqrt[3]{{{a}^{6}}}={{a}^{2}}$

Note that rational exponents are subject to all of the same rules as other exponents when they appear in algebraic expressions.

Both simplification methods gave the same result, $a^{2}$ . Depending on the context of the problem, it may be easier to use one method or the other, but for now, you’ll note that you were able to simplify this expression more quickly using rational exponents than when using the “pull-out” method.

Let’s try another example.

Example

Simplify. $\sqrt[4]{81{{x}^{8}}{{y}^{3}}}$

Show Solution

Rewrite the radical using rational exponents.

${{(81{{x}^{8}}{{y}^{3}})}^{\frac{1}{4}}}$

Use the rules of exponents to simplify the expression.

$\begin{array}{r}{{81}^{\frac{1}{4}}}\cdot {{x}^{\frac{8}{4}}}\cdot {{y}^{\frac{3}{4}}}\\{{(3\cdot 3\cdot 3\cdot 3)}^{\frac{1}{4}}}{{x}^{2}}{{y}^{\frac{3}{4}}}\\{{({{3}^{4}})}^{\frac{1}{4}}}{{x}^{2}}{{y}^{\frac{3}{4}}}\\3{{x}^{2}}{{y}^{\frac{3}{4}}}\end{array}$

Change the expression with the rational exponent back to radical form.

$3{{x}^{2}}\sqrt[4]{{{y}^{3}}}$

Answer

$\sqrt[4]{81{{x}^{8}}{{y}^{3}}}=3{{x}^{2}}\sqrt[4]{{{y}^{3}}}$

Again, the alternative method is to work on simplifying under the radical by using factoring. For the example you just solved, it looks like this.

Example

Simplify. $\sqrt[4]{81{{x}^{8}}{{y}^{3}}}$

Show Solution

Rewrite the expression.

$\sqrt[4]{81}\cdot \sqrt[4]{{{x}^{8}}}\cdot \sqrt[4]{{{y}^{3}}}$

Factor each radicand.

$\sqrt[4]{3\cdot 3\cdot 3\cdot 3}\cdot \sqrt[4]{{{x}^{2}}\cdot {{x}^{2}}\cdot {{x}^{2}}\cdot {{x}^{2}}}\cdot \sqrt[4]{{{y}^{3}}}$

Simplify.

$\begin{array}{r}\sqrt[4]{{{3}^{4}}}\cdot \sqrt[4]{{{({{x}^{2}})}^{4}}}\cdot \sqrt[4]{{{y}^{3}}}\\3\cdot {{x}^{2}}\cdot \sqrt[4]{{{y}^{3}}}\end{array}$

Answer

$\sqrt[4]{81x^{8}y^{3}}=3x^{2}\sqrt[4]{y^{3}}$

The following video shows more examples of how to simplify a radical expression using rational exponents.

Module 11QR: Statistical Thinking

Learning Outcomes

Writing a Square Root

Example

The square root of a square

The Product Rule for Square Roots

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Cube Roots

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Simplifying a radical

Example

Answer

Simplify Square Roots with Variables

Simplifying Square Roots

Taking the Square Root of a Radical Expression

Example

Answer

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Rational Exponents

Example

Convert Between Radical and Exponent Notation

Example

Answer

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Writing Fractional Exponents

Simplifying Radical Expressions Using Rational Exponents and the Laws of Exponents

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