8.2 Simplify Radical Expressions

Learning objectives.

By the end of this section, you will be able to:

  • Use the Product Property to simplify radical expressions
  • Use the Quotient Property to simplify radical expressions

Be Prepared 8.4

Before you get started, take this readiness quiz.

Simplify: x 9 x 4 . x 9 x 4 . If you missed this problem, review Example 5.13 .

Be Prepared 8.5

Simplify: y 3 y 11 . y 3 y 11 . If you missed this problem, review Example 5.13 .

Be Prepared 8.6

Simplify: ( n 2 ) 6 . ( n 2 ) 6 . If you missed this problem, review Example 5.17 .

Use the Product Property to Simplify Radical Expressions

We will simplify radical expressions in a way similar to how we simplified fractions. A fraction is simplified if there are no common factors in the numerator and denominator. To simplify a fraction, we look for any common factors in the numerator and denominator.

A radical expression , a n , a n , is considered simplified if it has no factors of m n . m n . So, to simplify a radical expression, we look for any factors in the radicand that are powers of the index.

Simplified Radical Expression

For real numbers a and m , and n ≥ 2 , n ≥ 2 ,

For example, 5 5 is considered simplified because there are no perfect square factors in 5. But 12 12 is not simplified because 12 has a perfect square factor of 4.

Similarly, 4 3 4 3 is simplified because there are no perfect cube factors in 4. But 24 3 24 3 is not simplified because 24 has a perfect cube factor of 8.

To simplify radical expressions, we will also use some properties of roots. The properties we will use to simplify radical expressions are similar to the properties of exponents. We know that ( a b ) n = a n b n . ( a b ) n = a n b n . The corresponding of Product Property of Roots says that a b n = a n · b n . a b n = a n · b n .

Product Property of n th Roots

If a n a n and b n b n are real numbers, and n ≥ 2 n ≥ 2 is an integer, then

We use the Product Property of Roots to remove all perfect square factors from a square root.

Example 8.13

Simplify square roots using the product property of roots.

Simplify: 98 . 98 .

Try It 8.25

Simplify: 48 . 48 .

Try It 8.26

Simplify: 45 . 45 .

Notice in the previous example that the simplified form of 98 98 is 7 2 , 7 2 , which is the product of an integer and a square root. We always write the integer in front of the square root.

Be careful to write your integer so that it is not confused with the index. The expression 7 2 7 2 is very different from 2 7 . 2 7 .

Simplify a radical expression using the Product Property.

  • Step 1. Find the largest factor in the radicand that is a perfect power of the index. Rewrite the radicand as a product of two factors, using that factor.
  • Step 2. Use the product rule to rewrite the radical as the product of two radicals.
  • Step 3. Simplify the root of the perfect power.

We will apply this method in the next example. It may be helpful to have a table of perfect squares, cubes, and fourth powers.

Example 8.14

Simplify: ⓐ 500 500 ⓑ 16 3 16 3 ⓒ 243 4 . 243 4 .

Try It 8.27

Simplify: ⓐ 288 288 ⓑ 81 3 81 3 ⓒ 64 4 . 64 4 .

Try It 8.28

Simplify: ⓐ 432 432 ⓑ 625 3 625 3 ⓒ 729 4 . 729 4 .

The next example is much like the previous examples, but with variables. Don’t forget to use the absolute value signs when taking an even root of an expression with a variable in the radical.

Example 8.15

Simplify: ⓐ x 3 x 3 ⓑ x 4 3 x 4 3 ⓒ x 7 4 . x 7 4 .

Try It 8.29

Simplify: ⓐ b 5 b 5 ⓑ y 6 4 y 6 4 ⓒ z 5 3 z 5 3

Try It 8.30

Simplify: ⓐ p 9 p 9 ⓑ y 8 5 y 8 5 ⓒ q 13 6 q 13 6

We follow the same procedure when there is a coefficient in the radicand. In the next example, both the constant and the variable have perfect square factors.

Example 8.16

Simplify: ⓐ 72 n 7 72 n 7 ⓑ 24 x 7 3 24 x 7 3 ⓒ 80 y 14 4 . 80 y 14 4 .

Try It 8.31

Simplify: ⓐ 32 y 5 32 y 5 ⓑ 54 p 10 3 54 p 10 3 ⓒ 64 q 10 4 . 64 q 10 4 .

Try It 8.32

Simplify: ⓐ 75 a 9 75 a 9 ⓑ 128 m 11 3 128 m 11 3 ⓒ 162 n 7 4 . 162 n 7 4 .

In the next example, we continue to use the same methods even though there are more than one variable under the radical.

Example 8.17

Simplify: ⓐ 63 u 3 v 5 63 u 3 v 5 ⓑ 40 x 4 y 5 3 40 x 4 y 5 3 ⓒ 48 x 4 y 7 4 . 48 x 4 y 7 4 .

Try It 8.33

Simplify: ⓐ 98 a 7 b 5 98 a 7 b 5 ⓑ 56 x 5 y 4 3 56 x 5 y 4 3 ⓒ 32 x 5 y 8 4 . 32 x 5 y 8 4 .

Try It 8.34

Simplify: ⓐ 180 m 9 n 11 180 m 9 n 11 ⓑ 72 x 6 y 5 3 72 x 6 y 5 3 ⓒ 80 x 7 y 4 4 . 80 x 7 y 4 4 .

Example 8.18

Simplify: ⓐ −27 3 −27 3 ⓑ −16 4 . −16 4 .

Try It 8.35

Simplify: ⓐ −64 3 −64 3 ⓑ −81 4 . −81 4 .

Try It 8.36

Simplify: ⓐ −625 3 −625 3 ⓑ −324 4 . −324 4 .

We have seen how to use the order of operations to simplify some expressions with radicals. In the next example, we have the sum of an integer and a square root. We simplify the square root but cannot add the resulting expression to the integer since one term contains a radical and the other does not. The next example also includes a fraction with a radical in the numerator. Remember that in order to simplify a fraction you need a common factor in the numerator and denominator.

Example 8.19

Simplify: ⓐ 3 + 32 3 + 32 ⓑ 4 − 48 2 . 4 − 48 2 .

The terms cannot be added as one has a radical and the other does not. Trying to add an integer and a radical is like trying to add an integer and a variable. They are not like terms!

Try It 8.37

Simplify: ⓐ 5 + 75 5 + 75 ⓑ 10 − 75 5 10 − 75 5

Try It 8.38

Simplify: ⓐ 2 + 98 2 + 98 ⓑ 6 − 45 3 6 − 45 3

Use the Quotient Property to Simplify Radical Expressions

Whenever you have to simplify a radical expression, the first step you should take is to determine whether the radicand is a perfect power of the index. If not, check the numerator and denominator for any common factors, and remove them. You may find a fraction in which both the numerator and the denominator are perfect powers of the index.

Example 8.20

Simplify: ⓐ 45 80 45 80 ⓑ 16 54 3 16 54 3 ⓒ 5 80 4 . 5 80 4 .

Try It 8.39

Simplify: ⓐ 75 48 75 48 ⓑ 54 250 3 54 250 3 ⓒ 32 162 4 . 32 162 4 .

Try It 8.40

Simplify: ⓐ 98 162 98 162 ⓑ 24 375 3 24 375 3 ⓒ 4 324 4 . 4 324 4 .

In the last example, our first step was to simplify the fraction under the radical by removing common factors. In the next example we will use the Quotient Property to simplify under the radical. We divide the like bases by subtracting their exponents,

Example 8.21

Simplify: ⓐ m 6 m 4 m 6 m 4 ⓑ a 8 a 5 3 a 8 a 5 3 ⓒ a 10 a 2 4 . a 10 a 2 4 .

Try It 8.41

Simplify: ⓐ a 8 a 6 a 8 a 6 ⓑ x 7 x 3 4 x 7 x 3 4 ⓒ y 17 y 5 4 . y 17 y 5 4 .

Try It 8.42

Simplify: ⓐ x 14 x 10 x 14 x 10 ⓑ m 13 m 7 3 m 13 m 7 3 ⓒ n 12 n 2 5 . n 12 n 2 5 .

Remember the Quotient to a Power Property ? It said we could raise a fraction to a power by raising the numerator and denominator to the power separately.

We can use a similar property to simplify a root of a fraction. After removing all common factors from the numerator and denominator, if the fraction is not a perfect power of the index, we simplify the numerator and denominator separately.

Quotient Property of Radical Expressions

If a n a n and b n b n are real numbers, b ≠ 0 , b ≠ 0 , and for any integer n ≥ 2 n ≥ 2 then,

Example 8.22

How to simplify the quotient of radical expressions.

Simplify: 27 m 3 196 . 27 m 3 196 .

Try It 8.43

Simplify: 24 p 3 49 . 24 p 3 49 .

Try It 8.44

Simplify: 48 x 5 100 . 48 x 5 100 .

Simplify a square root using the Quotient Property.

  • Step 1. Simplify the fraction in the radicand, if possible.
  • Step 2. Use the Quotient Property to rewrite the radical as the quotient of two radicals.
  • Step 3. Simplify the radicals in the numerator and the denominator.

Example 8.23

Simplify: ⓐ 45 x 5 y 4 45 x 5 y 4 ⓑ 24 x 7 y 3 3 24 x 7 y 3 3 ⓒ 48 x 10 y 8 4 . 48 x 10 y 8 4 .

Try It 8.45

Simplify: ⓐ 80 m 3 n 6 80 m 3 n 6 ⓑ 108 c 10 d 6 3 108 c 10 d 6 3 ⓒ 80 x 10 y 4 4 . 80 x 10 y 4 4 .

Try It 8.46

Simplify: ⓐ 54 u 7 v 8 54 u 7 v 8 ⓑ 40 r 3 s 6 3 40 r 3 s 6 3 ⓒ 162 m 14 n 12 4 . 162 m 14 n 12 4 .

Be sure to simplify the fraction in the radicand first, if possible.

Example 8.24

Simplify: ⓐ 18 p 5 q 7 32 p q 2 18 p 5 q 7 32 p q 2 ⓑ 16 x 5 y 7 54 x 2 y 2 3 16 x 5 y 7 54 x 2 y 2 3 ⓒ 5 a 8 b 6 80 a 3 b 2 4 . 5 a 8 b 6 80 a 3 b 2 4 .

Try It 8.47

Simplify: ⓐ 50 x 5 y 3 72 x 4 y 50 x 5 y 3 72 x 4 y ⓑ 16 x 5 y 7 54 x 2 y 2 3 16 x 5 y 7 54 x 2 y 2 3 ⓒ 5 a 8 b 6 80 a 3 b 2 4 . 5 a 8 b 6 80 a 3 b 2 4 .

Try It 8.48

Simplify: ⓐ 48 m 7 n 2 100 m 5 n 8 48 m 7 n 2 100 m 5 n 8 ⓑ 54 x 7 y 5 250 x 2 y 2 3 54 x 7 y 5 250 x 2 y 2 3 ⓒ 32 a 9 b 7 162 a 3 b 3 4 . 32 a 9 b 7 162 a 3 b 3 4 .

In the next example, there is nothing to simplify in the denominators. Since the index on the radicals is the same, we can use the Quotient Property again, to combine them into one radical. We will then look to see if we can simplify the expression.

Example 8.25

Simplify: ⓐ 48 a 7 3 a 48 a 7 3 a ⓑ −108 3 2 3 −108 3 2 3 ⓒ 96 x 7 4 3 x 2 4 . 96 x 7 4 3 x 2 4 .

Try It 8.49

Simplify: ⓐ 98 z 5 2 z 98 z 5 2 z ⓑ −500 3 2 3 −500 3 2 3 ⓒ 486 m 11 4 3 m 5 4 . 486 m 11 4 3 m 5 4 .

Try It 8.50

Simplify: ⓐ 128 m 9 2 m 128 m 9 2 m ⓑ −192 3 3 3 −192 3 3 3 ⓒ 324 n 7 4 2 n 3 4 . 324 n 7 4 2 n 3 4 .

Access these online resources for additional instruction and practice with simplifying radical expressions.

  • Simplifying Square Root and Cube Root with Variables
  • Express a Radical in Simplified Form-Square and Cube Roots with Variables and Exponents
  • Simplifying Cube Roots

Section 8.2 Exercises

Practice makes perfect.

In the following exercises, use the Product Property to simplify radical expressions.

ⓐ 32 4 32 4 ⓑ 64 5 64 5

ⓐ 625 3 625 3 ⓑ 128 6 128 6

ⓐ 64 4 64 4 ⓑ 256 3 256 3

ⓐ 3125 4 3125 4 ⓑ 81 3 81 3

In the following exercises, simplify using absolute value signs as needed.

ⓐ y 11 y 11 ⓑ r 5 3 r 5 3 ⓒ s 10 4 s 10 4

ⓐ m 13 m 13 ⓑ u 7 5 u 7 5 ⓒ v 11 6 v 11 6

ⓐ n 21 n 21 ⓑ q 8 3 q 8 3 ⓒ n 10 8 n 10 8

ⓐ r 25 r 25 ⓑ p 8 5 p 8 5 ⓒ m 5 4 m 5 4

ⓐ 125 r 13 125 r 13 ⓑ 108 x 5 3 108 x 5 3 ⓒ 48 y 6 4 48 y 6 4

ⓐ 80 s 15 80 s 15 ⓑ 96 a 7 5 96 a 7 5 ⓒ 128 b 7 6 128 b 7 6

ⓐ 242 m 23 242 m 23 ⓑ 405 m 10 4 405 m 10 4 ⓒ 160 n 8 5 160 n 8 5

ⓐ 175 n 13 175 n 13 ⓑ 512 p 5 5 512 p 5 5 ⓒ 324 q 7 4 324 q 7 4

ⓐ 147 m 7 n 11 147 m 7 n 11 ⓑ 48 x 6 y 7 3 48 x 6 y 7 3 ⓒ 32 x 5 y 4 4 32 x 5 y 4 4

ⓐ 96 r 3 s 3 96 r 3 s 3 ⓑ 80 x 7 y 6 3 80 x 7 y 6 3 ⓒ 80 x 8 y 9 4 80 x 8 y 9 4

ⓐ 192 q 3 r 7 192 q 3 r 7 ⓑ 54 m 9 n 10 3 54 m 9 n 10 3 ⓒ 81 a 9 b 8 4 81 a 9 b 8 4

ⓐ 150 m 9 n 3 150 m 9 n 3 ⓑ 81 p 7 q 8 3 81 p 7 q 8 3 ⓒ 162 c 11 d 12 4 162 c 11 d 12 4

ⓐ −864 3 −864 3 ⓑ −256 4 −256 4

ⓐ −486 5 −486 5 ⓑ −64 6 −64 6

ⓐ −32 5 −32 5 ⓑ −1 8 −1 8

ⓐ −8 3 −8 3 ⓑ −16 4 −16 4

ⓐ 5 + 12 5 + 12 ⓑ 10 − 24 2 10 − 24 2

ⓐ 8 + 96 8 + 96 ⓑ 8 − 80 4 8 − 80 4

ⓐ 1 + 45 1 + 45 ⓑ 3 + 90 3 3 + 90 3

ⓐ 3 + 125 3 + 125 ⓑ 15 + 75 5 15 + 75 5

In the following exercises, use the Quotient Property to simplify square roots.

ⓐ 45 80 45 80 ⓑ 8 27 3 8 27 3 ⓒ 1 81 4 1 81 4

ⓐ 72 98 72 98 ⓑ 24 81 3 24 81 3 ⓒ 6 96 4 6 96 4

ⓐ 100 36 100 36 ⓑ 81 375 3 81 375 3 ⓒ 1 256 4 1 256 4

ⓐ 121 16 121 16 ⓑ 16 250 3 16 250 3 ⓒ 32 162 4 32 162 4

ⓐ x 10 x 6 x 10 x 6 ⓑ p 11 p 2 3 p 11 p 2 3 ⓒ q 17 q 13 4 q 17 q 13 4

ⓐ p 20 p 10 p 20 p 10 ⓑ d 12 d 7 5 d 12 d 7 5 ⓒ m 12 m 4 8 m 12 m 4 8

ⓐ y 4 y 8 y 4 y 8 ⓑ u 21 u 11 5 u 21 u 11 5 ⓒ v 30 v 12 6 v 30 v 12 6

ⓐ q 8 q 14 q 8 q 14 ⓑ r 14 r 5 3 r 14 r 5 3 ⓒ c 21 c 9 4 c 21 c 9 4

96 x 7 121 96 x 7 121

108 y 4 49 108 y 4 49

300 m 5 64 300 m 5 64

125 n 7 169 125 n 7 169

98 r 5 100 98 r 5 100

180 s 10 144 180 s 10 144

28 q 6 225 28 q 6 225

150 r 3 256 150 r 3 256

ⓐ 75 r 9 s 8 75 r 9 s 8 ⓑ 54 a 8 b 3 3 54 a 8 b 3 3 ⓒ 64 c 5 d 4 4 64 c 5 d 4 4

ⓐ 72 x 5 y 6 72 x 5 y 6 ⓑ 96 r 11 s 5 5 96 r 11 s 5 5 ⓒ 128 u 7 v 12 6 128 u 7 v 12 6

ⓐ 28 p 7 q 2 28 p 7 q 2 ⓑ 81 s 8 t 3 3 81 s 8 t 3 3 ⓒ 64 p 15 q 12 4 64 p 15 q 12 4

ⓐ 45 r 3 s 10 45 r 3 s 10 ⓑ 625 u 10 v 3 3 625 u 10 v 3 3 ⓒ 729 c 21 d 8 4 729 c 21 d 8 4

ⓐ 32 x 5 y 3 18 x 3 y 32 x 5 y 3 18 x 3 y ⓑ 5 x 6 y 9 40 x 5 y 3 3 5 x 6 y 9 40 x 5 y 3 3 ⓒ 5 a 8 b 6 80 a 3 b 2 4 5 a 8 b 6 80 a 3 b 2 4

ⓐ 75 r 6 s 8 48 r s 4 75 r 6 s 8 48 r s 4 ⓑ 24 x 8 y 4 81 x 2 y 3 24 x 8 y 4 81 x 2 y 3 ⓒ 32 m 9 n 2 162 m n 2 4 32 m 9 n 2 162 m n 2 4

ⓐ 27 p 2 q 108 p 4 q 3 27 p 2 q 108 p 4 q 3 ⓑ 16 c 5 d 7 250 c 2 d 2 3 16 c 5 d 7 250 c 2 d 2 3 ⓒ 2 m 9 n 7 128 m 3 n 6 2 m 9 n 7 128 m 3 n 6

ⓐ 50 r 5 s 2 128 r 2 s 6 50 r 5 s 2 128 r 2 s 6 ⓑ 24 m 9 n 7 375 m 4 n 3 24 m 9 n 7 375 m 4 n 3 ⓒ 81 m 2 n 8 256 m 1 n 2 4 81 m 2 n 8 256 m 1 n 2 4

ⓐ 45 p 9 5 q 2 45 p 9 5 q 2 ⓑ 64 4 2 4 64 4 2 4 ⓒ 128 x 8 5 2 x 2 5 128 x 8 5 2 x 2 5

ⓐ 80 q 5 5 q 80 q 5 5 q ⓑ −625 3 5 3 −625 3 5 3 ⓒ 80 m 7 4 5 m 4 80 m 7 4 5 m 4

ⓐ 50 m 7 2 m 50 m 7 2 m ⓑ 1250 2 3 1250 2 3 ⓒ 486 y 9 2 y 3 4 486 y 9 2 y 3 4

ⓐ 72 n 11 2 n 72 n 11 2 n ⓑ 162 6 3 162 6 3 ⓒ 160 r 10 5 r 3 4 160 r 10 5 r 3 4

Writing Exercises

Explain why x 4 = x 2 . x 4 = x 2 . Then explain why x 16 = x 8 . x 16 = x 8 .

Explain why 7 + 9 7 + 9 is not equal to 7 + 9 . 7 + 9 .

Explain how you know that x 10 5 = x 2 . x 10 5 = x 2 .

Explain why −64 4 −64 4 is not a real number but −64 3 −64 3 is.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

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7.1: Roots and Radicals

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Learning Objectives

  • Identify and evaluate square and cube roots.
  • Determine the domain of functions involving square and cube roots.
  • Evaluate \(n\)th roots.
  • Simplify radicals using the product and quotient rules for radicals.

Square and Cube Roots

Recall that a square root 1 of a number is a number that when multiplied by itself yields the original number. For example, \(5\) is a square root of \(25\), because \(5^{2} = 25\). Since \((−5)^{2} = 25\), we can say that \(−5\) is a square root of \(25\) as well. Every positive real number has two square roots, one positive and one negative. For this reason, we use the radical sign \(√\) to denote the principal (nonnegative) square root 2 and a negative sign in front of the radical \(−√\) to denote the negative square root.

\(\begin{aligned} \sqrt { 25 } & = 5 \quad\quad\color{Cerulean} { Positive\: square \:root \:of \: 25} \\ - \sqrt { 25 } & = - 5 \quad\:\color{Cerulean} { Negative \:square\: root \:of\: 25} \end{aligned}\)

Zero is the only real number with one square root.

\(\sqrt { 0 } = 0 \text { because } 0 ^ { 2 } = 0\)

Example \(\PageIndex{1}\):

  • \(\sqrt { 121 }\)
  • \(- \sqrt { 81 }\)
  • \(\sqrt { 121 } = \sqrt { 11 ^ { 2 } } = 11\)
  • \(- \sqrt { 81 } = - \sqrt { 9 ^ { 2 } } = - 9\)

If the radicand 3 , the number inside the radical sign, can be factored as the square of another number, then the square root of the number is apparent. In this case, we have the following property:

\(\sqrt { a ^ { 2 } } = a \quad \text { if } \quad a \geq 0\)

Or more generally,

\(\sqrt { a ^ { 2 } } = | a | \quad \text { if } \quad a \in R\)

The absolute value is important because \(a\) may be a negative number and the radical sign denotes the principal square root. For example,

\(\sqrt { ( - 8 ) ^ { 2 } } = | -8| = 8\)

Make use of the absolute value to ensure a positive result.

Example \(\PageIndex{2}\):

Simplify: \(\sqrt { ( x - 2 ) ^ { 2 } }\).

Here the variable expression \(x − 2\) could be negative, zero, or positive. Since the sign depends on the unknown quantity \(x\), we must ensure that we obtain the principal square root by making use of the absolute value.

\(\sqrt { ( x - 2 ) ^ { 2 } } = | x - 2 |\)

\(| x - 2 |\)

The importance of the use of the absolute value in the previous example is apparent when we evaluate using values that make the radicand negative. For example, when \(x = 1\),

\(\begin{aligned} \sqrt { ( x - 2 ) ^ { 2 } } & = | x - 2 | \\ & = | 1 - 2 | \\ & = | - 1 | \\ & = 1 \end{aligned}\)

Next, consider the square root of a negative number. To determine the square root of \(−25\), you must find a number that when squared results in \(−25\):

\(\sqrt { - 25 } = \color{Cerulean}{?} \color{black}\quad{ \text { or }} \quad( \color{Cerulean}{?}\color{black}{ )} ^ { 2 } = - 25\)

However, any real number squared always results in a positive number. The square root of a negative number is currently left undefined. For now, we will state that \(\sqrt { - 25 }\) is not a real number. Therefore, the square root function 4 given by \(f ( x ) = \sqrt { x }\) is not defined to be a real number if the \(x\)-values are negative. The smallest value in the domain is zero. For example,\(f ( 0 ) = \sqrt { 0 } = 0\) and \(f ( 4 ) = \sqrt { 4 } = 2\). Recall the graph of the square root function.

5a5bec559b23c37c3dc80885e5f0026c.png

The domain and range both consist of real numbers greater than or equal to zero: \([0, ∞)\). To determine the domain of a function involving a square root we look at the radicand and find the values that produce nonnegative results.

Example \(\PageIndex{3}\):

Determine the domain of the function defined by \(f ( x ) = \sqrt { 2 x + 3 }\).

Here the radicand is \(2x + 3\). This expression must be zero or positive. In other words,

\(2 x + 3 \geq 0\)

Solve for \(x\).

\(\begin{aligned} 2 x + 3 & \geq 0 \\ 2 x & \geq - 3 \\ x & \geq - \frac { 3 } { 2 } \end{aligned}\)

Domain: \(\left[ - \frac { 3 } { 2 } , \infty \right)\)

A cube root 5 of a number is a number that when multiplied by itself three times yields the original number. Furthermore, we denote a cube root using the symbol \(\sqrt [ 3 ] { }\), where \(3\) is called the index 6 . For example,

\(\sqrt [ 3 ] { 64 } = 4 , \text { because } 4 ^ { 3 } = 64\)

The product of three equal factors will be positive if the factor is positive and negative if the factor is negative. For this reason, any real number will have only one real cube root. Hence the technicalities associated with the principal root do not apply. For example,

\(\sqrt [ 3 ] { - 64 } = - 4 , \text { because } ( - 4 ) ^ { 3 } = - 64\)

In general, given any real number \(a\), we have the following property:

\(\sqrt [ 3 ] { a ^ { 3 } } = a \quad \text { if } \quad a \in R\)

When simplifying cube roots, look for factors that are perfect cubes.

Example \(\PageIndex{4}\):

  • \(\sqrt [ 3 ] { 8 }\)
  • \(\sqrt [ 3 ] { 0 }\)
  • \(\sqrt [ 3 ] { \frac { 1 } { 27 } }\)
  • \(\sqrt [ 3 ] { - 1 }\)
  • \(\sqrt [ 3 ] { - 125 }\)
  • \(\sqrt [ 3 ] { 8 } = \sqrt [ 3 ] { 2 ^ { 3 } } = 2\)
  • \(\sqrt [ 3 ] { 0 } = \sqrt [ 3 ] { 0 ^ { 3 } } = 0\)
  • \(\sqrt [ 3 ] { \frac { 1 } { 27 } } = \sqrt [ 3 ] { \left( \frac { 1 } { 3 } \right) ^ { 3 } } = \frac { 1 } { 3 }\)
  • \(\sqrt [ 3 ] { - 1 } = \sqrt [ 3 ] { ( - 1 ) ^ { 3 } } = - 1\)
  • \(\sqrt [ 3 ] { - 125 } = \sqrt [ 3 ] { ( - 5 ) ^ { 3 } } = - 5\)

It may be the case that the radicand is not a perfect square or cube. If an integer is not a perfect power of the index, then its root will be irrational. For example, \(\sqrt [ 3 ] { 2 }\) is an irrational number that can be approximated on most calculators using the root button \(\sqrt [ x ] { }\).Depending on the calculator, we typically type in the index prior to pushing the button and then the radicand as follows:

\(3 \quad\sqrt [ x ] {y }\quad2\quad=\)

Therefore, we have

\(\sqrt [ 3 ] { 2 } \approx 1.260 , \quad \text { because } \quad 1.260 ^{\wedge} 3 \approx 2\)

Since cube roots can be negative, zero, or positive we do not make use of any absolute values.

Example \(\PageIndex{5}\):

Simplify: \(\sqrt [ 3 ] { ( y - 7 ) ^ { 3 } }\).

The cube root of a quantity cubed is that quantity.

\(\sqrt [ 3 ] { ( y - 7 ) ^ { 3 } } = y - 7\)

Exercise \(\PageIndex{1}\)

Evaluate: \(\sqrt [ 3 ] { - 1000 }\).

www.youtube.com/v/B06NIs-3gig

Next, consider the cube root function 7 :

\(f ( x ) = \sqrt [ 3 ] { x } \quad\color{Cerulean}{Cube\:root\:function.}\)

Since the cube root could be either negative or positive, we conclude that the domain consists of all real numbers. Sketch the graph by plotting points. Choose some positive and negative values for \(x\), as well as zero, and then calculate the corresponding \(y\)-values.

Plot the points and sketch the graph of the cube root function.

507f092c47524abca6b1b6d100809a5e.png

The graph passes the vertical line test and is indeed a function. In addition, the range consists of all real numbers.

Example \(\PageIndex{6}\):

Given \(g ( x ) = \sqrt [ 3 ] { x + 1 } + 2\), find \(g ( - 9 ) , g ( - 2 ) , g ( - 1 )\), and \(g(0)\). Sketch the graph of \(g\).

Replace \(x\) with the given values.

We can also sketch the graph using the following translations:

\(\begin{array} { l } { y = \sqrt [ 3 ] { x } \quad\quad\quad\quad \color{Cerulean} { Basic\: cube \:root\: function } } \\ { y = \sqrt [ 3 ] { x + 1 } \quad \quad\:\color{Cerulean} { Horizontal\: shift\: left\: 1\: unit } } \\ { y = \sqrt [ 3 ] { x + 1 } + 2 \:\:\:\color{Cerulean} { Vertical\: shift\: up\: 2\: units } } \end{array}\)

c7808cb98ec1b47125d20c368e032309.png

\(n\)th Roots

For any integer \(n ≥ 2\), we define an \(n\)th root 8 of a positive real number as a number that when raised to the \(n\)th power yields the original number. Given any nonnegative real number \(a\), we have the following property:

\(\sqrt [ n ] { a ^ { n } } = a , \quad \text { if } \quad a \geq 0\)

Here n is called the index and \(a^{n}\) is called the radicand. Furthermore, we can refer to the entire expression \(\sqrt [ n ] { A }\) as a radical 9 . When the index is an integer greater than or equal to \(4\), we say “fourth root,” “fifth root,” and so on. The \(n\)th root of any number is apparent if we can write the radicand with an exponent equal to the index.

Example \(\PageIndex{7}\):

  • \(\sqrt [ 4 ] { 81 }\)
  • \(\sqrt [ 5 ] { 32 }\)
  • \(\sqrt [ 7 ] { 1 }\)
  • \(\sqrt [ 4 ] { \frac { 1 } { 16 } }\)
  • \(\sqrt [ 4 ] { 81 } = \sqrt [ 4 ] { 3 ^ { 4 } } = 3\)
  • \(\sqrt [ 5 ] { 32 } = \sqrt [ 5 ] { 2 ^ { 5 } } = 2\)
  • \(\sqrt [ 7 ] { 1 } = \sqrt [ 7 ] { 1 ^ { 7 } } = 1\)
  • \(\sqrt [ 4 ] { \frac { 1 } { 16 } } = \sqrt [ 4 ] { \left( \frac { 1 } { 2 } \right) ^ { 4 } } = \frac { 1 } { 2 }\)

If the index is \(n = 2\), then the radical indicates a square root and it is customary to write the radical without the index; \(\sqrt [ 2 ] { a } = \sqrt { a }\).

We have already taken care to define the principal square root of a real number. At this point, we extend this idea to nth roots when n is even. For example, \(3\) is a fourth root of \(81\), because \(3^{4} = 81\). And since \((−3)^{4} = 81\), we can say that \(−3\) is a fourth root of \(81\) as well. Hence we use the radical sign \(\sqrt [ n ] { }\) to denote the principal (nonnegative) \(n\)th root 10 when \(n\) is even. In this case, for any real number \(a\), we use the following property:

\(\sqrt [ n ] { a ^ { n } } = | a | \quad \color{Cerulean} { When\:n\:is\:even } \)

For example,

\(\begin{aligned} \sqrt [ 4 ] { 81 } & = \sqrt [ 4 ] { 3 ^ { 4 } } \quad\quad= |3| \:\:\:\:\:= 3 \\ \sqrt [ 4 ] { 81 } & = \sqrt [ 4 ] { ( - 3 ) ^ { 4 } } \:\:= | - 3 | = 3 \end{aligned}\)

The negative \(n\)th root, when \(n\) is even, will be denoted using a negative sign in front of the radical \(- \sqrt [ n ] { }\).

\(- \sqrt [ 4 ] { 81 } = - \sqrt [ 4 ] { 3 ^ { 4 } } = - 3\)

We have seen that the square root of a negative number is not real because any real number that is squared will result in a positive number. In fact, a similar problem arises for any even index:

\(\sqrt [ 4 ] { - 81 } =\color{Cerulean}{ ?} \quad \color{black}{\text { or }} \quad (\color{Cerulean}{ ?}\color{black}{ )} ^ { 4 } = - 81\)

We can see that a fourth root of \(−81\) is not a real number because the fourth power of any real number is always positive.

\(\left. \begin{array} { l } { \sqrt { - 4 } } \\ { \sqrt [ 4 ] { - 81 } } \\ { \sqrt [ 6 ] { - 64 } } \end{array} \right\} \quad \color{Cerulean}{These\:radicals\:are\:not\:real\:numbers.}\)

You are encouraged to try all of these on a calculator. What does it say?

Example \(\PageIndex{8}\):

  • \(\sqrt [ 4 ] { ( - 10 ) ^ { 4 } }\)
  • \(\sqrt [ 4 ] { - 10 ^ { 4 } }\)
  • \(\sqrt [ 6 ] { ( 2 y + 1 ) ^ { 6 } }\)

Since the indices are even, use absolute values to ensure nonnegative results.

  • \(\sqrt [ 4 ] { ( - 10 ) ^ { 4 } } = | - 10 | = 10\)
  • \(\sqrt [ 4 ] { - 10 ^ { 4 } } = \sqrt [ 4 ] { - 10,000 }\) is not a real number.
  • \(\sqrt [ 6 ] { ( 2 y + 1 ) ^ { 6 } } = | 2 y + 1 |\)

When the index \(n\) is odd, the same problems do not occur. The product of an odd number of positive factors is positive and the product of an odd number of negative factors is negative. Hence when the index \(n\) is odd, there is only one real \(n\)th root for any real number \(a\). And we have the following property:

\(\sqrt [ n ] { a ^ { n } } = a \quad \color{Cerulean} { When \: n\:is\:odd}\)

Example \(\PageIndex{9}\):

  • \(\sqrt [ 5 ] { ( - 10 ) ^ { 5 } }\)
  • \(\sqrt [ 5 ] { - 32 }\)
  • \(\sqrt [ 7 ] { ( 2 y + 1 ) ^ { 7 } }\)

Since the indices are odd, the absolute value is not used.

  • \(\sqrt [ 5 ] { ( - 10 ) ^ { 5 } } = - 10\)
  • \(\sqrt [ 5 ] { - 32 } = \sqrt [ 5 ] { ( - 2 ) ^ { 5 } } = - 2\)
  • \(\sqrt [ 7 ] { ( 2 y + 1 ) ^ { 7 } } = 2 y + 1\)

In summary, for any real number \(a\) we have,

\(\begin{aligned} \sqrt [ n ] { a^ { n } } & = | a | \color{Cerulean}\:\:\: { When \: n\: is\: even } \\ \sqrt [ n ] {a^ { n } } & = a \quad\: \color{Cerulean} { When \: n\: is\: odd } \end{aligned}\)

When \(n\) is odd , the \(n\)th root is positive or negative depending on the sign of the radicand.

\(\begin{aligned} \sqrt [ 3 ] { 27 } & = \sqrt [ 3 ] { 3 ^ { 3 } } = 3 \\ \sqrt [ 3 ] { - 27 } & = \sqrt [ 3 ] { ( - 3 ) ^ { 3 } } = - 3 \end{aligned}\)

When \(n\) is even , the \(n\)th root is positive or not real depending on the sign of the radicand.

\(\begin{aligned} \sqrt [ 4 ] { 16 } & = \sqrt [ 4 ] { 2 ^ { 4 } } \quad\:\:= 2 \\ \sqrt [ 4 ] { 16 } & = \sqrt [ 4 ] { ( - 2 ) ^ { 4 } } = | - 2| = 2 \\ \sqrt [ 4 ] { - 16 } & \quad\color{Cerulean} { Not \:a \:real\: number } \end{aligned}\)

Exercise \(\PageIndex{2}\)

Simplify: \(- 8 \sqrt [ 5 ] { - 32 }\).

www.youtube.com/v/Ik1xXgq18f0

Simplifying Radicals

It will not always be the case that the radicand is a perfect power of the given index. If it is not, then we use the product rule for radicals 11 and the quotient rule for radicals 12 to simplify them. Given real numbers \(\sqrt [ n ] { A }\) and \(\sqrt [ n ] { B }\),

A radical is simplified 13 if it does not contain any factors that can be written as perfect powers of the index.

Example \(\PageIndex{10}\):

Simplify: \(\sqrt { 150 }\).

Here \(150\) can be written as \(2 \cdot 3 \cdot 5 ^ { 2 }\).

\(\begin{aligned} \sqrt { 150 } & = \sqrt { 2 \cdot 3 \cdot 5 ^ { 2 } }\quad\quad \color{Cerulean} { Apply\: the\: product \:rule\: for\: radicals.} \\ & = \sqrt { 2 \cdot 3 } \cdot \sqrt { 5 ^ { 2 } }\quad\: \color{Cerulean} { Simplify. } \\ & = \sqrt { 6 } \cdot 5 \\ & = 5 \sqrt { 6 } \end{aligned}\)

We can verify our answer on a calculator:

\(\sqrt { 150 } \approx 12.25 \quad\text { and }\quad 5 \sqrt { 6 } \approx 12.25\)

Also, it is worth noting that

\(12.25 ^ { 2 } \approx 150\)

\(5 \sqrt { 6 }\)

\(5 \sqrt { 6 }\) is the exact answer and \(12.25\) is an approximate answer. We present exact answers unless told otherwise.

Example \(\PageIndex{11}\):

Simplify: \(\sqrt [ 3 ] { 160 }\).

Use the prime factorization of \(160\) to find the largest perfect cube factor:

\(\begin{aligned} 160 & = 2 ^ { 5 } \cdot 5 \\ & = \color{Cerulean}{2 ^ { 3} }\color{black}{ \cdot} 2 ^ { 2 } \cdot 5 \end{aligned}\)

Replace the radicand with this factorization and then apply the product rule for radicals.

\(\begin{aligned} \sqrt [ 3 ] { 160 } & = \sqrt [ 3 ] { 2 ^ { 3 } \cdot 2 ^ { 2 } \cdot 5 } \quad\quad\color{Cerulean} { Apply\:the\: product\: rule\: for\: radicals.} \\ & = \sqrt [ 3 ] { 2 ^ { 3 } } \cdot \sqrt [ 3 ] { 2 ^ { 2 } \cdot 5 }\quad \color{Cerulean} { Simplify. } \\ & = 2 \cdot \sqrt [ 3 ] { 20 } \end{aligned}\)

We can verify our answer on a calculator.

\(\sqrt [ 3 ] { 160 } \approx 5.43 \text { and } 2 \sqrt [ 3 ] { 20 } \approx 5.43\)

\(2 \sqrt [ 3 ] { 20 }\)

Example \(\PageIndex{12}\):

Simplify: \(\sqrt [ 5 ] { - 320 }\).

Here we note that the index is odd and the radicand is negative; hence the result will be negative. We can factor the radicand as follows:

\(- 320 = - 1 \cdot 32 \cdot 10 = ( - 1 ) ^ { 5 } \cdot ( 2 ) ^ { 5 } \cdot 10\)

Then simplify:

\(\begin{aligned} \sqrt [ 5 ] { - 320 } & = \sqrt [ 5 ] { ( - 1 ) ^ { 5 } \cdot ( 2 ) ^ { 5 } \cdot 10 } \quad\quad\quad\color{Cerulean} { Apply\: the\: product\: rule\: for\: radicals.} \\ & = \sqrt [ 5 ] { ( - 1 ) ^ { 5 } } \cdot \sqrt [ 5 ] { ( 2 ) ^ { 5 } } \cdot \sqrt [ 5 ] { 10 }\quad \color{Cerulean} { Simplify. } \\ &= -1\cdot2\cdot \sqrt[5]{10} \\ &=-2\cdot \sqrt[5]{10}\end{aligned}\)

\(- 2 \sqrt [ 5 ] { 10 }\)

Example \(\PageIndex{13}\):

Simplify: \(\sqrt [ 3 ] { - \frac { 8 } { 64 } }\).

In this case, consider the equivalent fraction with \(−8 = (−2)^{3}\) in the numerator and \(64 = 4^{3}\) in the denominator and then simplify.

\(\begin{aligned} \sqrt [ 3 ] { - \frac { 8 } { 64 } } & = \sqrt [ 3 ] { \frac { - 8 } { 64 } } \quad\quad\quad\color{Cerulean} { Apply\: the\: quotient \:rule\: for\: radicals.} \\ & = \frac { \sqrt [ 3 ] { ( - 2 ) ^ { 3 } } } { \sqrt [ 3 ] { 4 ^ { 3 } } }\quad\:\:\: \color{Cerulean} { Simplify. } \\ & = \frac { - 2 } { 4 } \\ & = - \frac { 1 } { 2 } \end{aligned}\)

\(-\frac{1}{2}\)

Exercise \(\PageIndex{3}\)

Simplify: \(\sqrt [ 4 ] { \frac { 80 } { 81 } }\)

\(\frac { 2 \sqrt [ 4 ] { 5 } } { 3 }\)

www.youtube.com/v/8CwbDBFO2FQ

Key Takeaways

  • To simplify a square root, look for the largest perfect square factor of the radicand and then apply the product or quotient rule for radicals.
  • To simplify a cube root, look for the largest perfect cube factor of the radicand and then apply the product or quotient rule for radicals.
  • When working with nth roots, \(n\) determines the definition that applies. We use \(\sqrt [ n ] { a ^ { n } } = a _ { 1 }\) when \(n\) is odd and \(\sqrt [ n ] { a ^ { n } } = | a | \) when \(n\) is even.
  • To simplify \(n\)th roots, look for the factors that have a power that is equal to the index \(n\) and then apply the product or quotient rule for radicals. Typically, the process is streamlined if you work with the prime factorization of the radicand.

Exercise \(\PageIndex{4}\)

  • \(\sqrt { 36 }\)
  • \(\sqrt { 100 }\)
  • \(\sqrt { \frac { 4 } { 9 } }\)
  • \(\sqrt { \frac { 1 } { 64 } }\)
  • \(- \sqrt { 16 }\)
  • \(- \sqrt { 1 }\)
  • \(\sqrt { ( - 5 ) ^ { 2 } }\)
  • \(\sqrt { ( - 1 ) ^ { 2 } }\)
  • \(\sqrt { - 4 }\)
  • \(\sqrt { - 5 ^ { 2 } }\)
  • \(- \sqrt { ( - 3 ) ^ { 2 } }\)
  • \(- \sqrt { ( - 4 ) ^ { 2 } }\)
  • \(\sqrt { x ^ { 2 } }\)
  • \(\sqrt { ( - x ) ^ { 2 } }\)
  • \(\sqrt { ( x - 5 ) ^ { 2 } }\)
  • \(\sqrt { ( 2 x - 1 ) ^ { 2 } }\)
  • \(\sqrt [ 3 ] { 64 }\)
  • \(\sqrt [ 3 ] { 216 }\)
  • \(\sqrt [ 3 ] { - 216 }\)
  • \(\sqrt [ 3 ] { - 64 }\)
  • \(\sqrt [ 3 ] { - 8 }\)
  • \(\sqrt [ 3 ] { 1 }\)
  • \(- \sqrt [ 3 ] { ( - 2 ) ^ { 3 } }\)
  • \(- \sqrt [ 3 ] { ( - 7 ) ^ { 3 } }\)
  • \(\sqrt [ 3 ] { \frac { 1 } { 8 } }\)
  • \(\sqrt [ 3 ] { \frac { 8 } { 27 } }\)
  • \(\sqrt [ 3 ] { ( - y ) ^ { 3 } }\)
  • \(- \sqrt [ 3 ] { y ^ { 3 } }\)
  • \(\sqrt [ 3 ] { ( y - 8 ) ^ { 3 } }\)
  • \(\sqrt [ 3 ] { ( 2 x - 3 ) ^ { 3 } }\)

3. \(\frac{2}{3}\)

5. \(−4\)

9. Not a real number

11. \(−3\)

13. \(|x|\)

15. \(|x − 5|\)

19. \(−6\)

21. \(−2\)

25. \(\frac{1}{2}\)

27. \(−y\)

29. \(y − 8\)

Exercise \(\PageIndex{5}\)

Determine the domain of the given function.

  • \(g ( x ) = \sqrt { x + 5 }\)
  • \(g ( x ) = \sqrt { x - 2 }\)
  • \(f ( x ) = \sqrt { 5 x + 1 }\)
  • \(f ( x ) = \sqrt { 3 x + 4 }\)
  • \(g ( x ) = \sqrt { - x + 1 }\)
  • \(g ( x ) = \sqrt { - x - 3 }\)
  • \(h ( x ) = \sqrt { 5 - x }\)
  • \(h ( x ) = \sqrt { 2 - 3 x }\)
  • \(g ( x ) = \sqrt [ 3 ] { x + 4 }\)
  • \(g ( x ) = \sqrt [ 3 ] { x - 3 }\)

1. \([ - 5 , \infty )\)

3. \(\left[ - \frac { 1 } { 5 } , \infty \right)\)

5. \(( - \infty , 1 ]\)

7. \(( - \infty , 5 ]\)

9. \(( - \infty , \infty )\)

Exercise \(\PageIndex{6}\)

Evaluate given the function definition.

  • Given \(f ( x ) = \sqrt { x - 1 }\), find \(f ( 1 ) , f ( 2 )\), and \(f ( 5 )\)
  • Given \(f ( x ) = \sqrt { x + 5 }\), find \(f ( - 5 ) , f ( - 1 )\), and \(f ( 20 )\)
  • Given \(f ( x ) = \sqrt { x } + 3\), find \(f ( 0 ) , f ( 1 )\), and \(f(16)\)
  • Given \(f ( x ) = \sqrt { x } - 5\), find \(f ( 0 ) , f ( 1 )\), and \(f(25)\)
  • Given \(g ( x ) = \sqrt [ 3 ] { x }\), find \(g ( - 1 ) , g ( 0 )\), and \(g(1)\)
  • Given \(g ( x ) = \sqrt [ 3 ] { x } - 2\) find \(g ( - 1 ) , g ( 0 )\), and \(g(8)\)
  • Given \(g ( x ) = \sqrt [ 3 ] { x + 7 }\), find \(g ( - 15 ) , g ( - 7 )\), and \(g(20)\)
  • Given \(g ( x ) = \sqrt [ 3 ] { x - 1 } + 2\), find \(g ( 0 ) , g ( 2 ) \), and \(g(9)\)

1. \(f ( 1 ) = 0 ; f ( 2 ) = 1 ; f ( 5 ) = 2\)

3. \(f ( 0 ) = 3 ; f ( 1 ) = 4 ; f ( 16 ) = 7\)

5. \(g ( - 1 ) = - 1 ; g ( 0 ) = 0 ; g ( 1 ) = 1\)

7. \(g ( - 15 ) = - 2 ; g ( - 7 ) = 0 ; g ( 20 ) = 3\)

Exercise \(\PageIndex{7}\)

Sketch the graph of the given function and give its domain and range.

  • \(f ( x ) = \sqrt { x + 9 }\)
  • \(f ( x ) = \sqrt { x - 3 }\)
  • \(f ( x ) = \sqrt { x - 1 } + 2\)
  • \(f ( x ) = \sqrt { x + 1 } + 3\)
  • \(g ( x ) = \sqrt [ 3 ] { x - 1 }\)
  • \(g ( x ) = \sqrt [ 3 ] { x + 1 }\)
  • \(g ( x ) = \sqrt [ 3 ] { x } - 4\)
  • \(g ( x ) = \sqrt [ 3 ] { x } + 5\)
  • \(g ( x ) = \sqrt [ 3 ] { x + 2 } - 1\)
  • \(g ( x ) = \sqrt [ 3 ] { x - 2 } + 3\)
  • \(f ( x ) = - \sqrt [ 3 ] { x }\)
  • \(f ( x ) = - \sqrt [ 3 ] { x - 1 }\)

1. Domain: \([ - 9 , \infty )\); range: \([ 0 , \infty )\)

d41e03bbc0db0d0bbb412151227f80cf.png

3. Domain: \([ 1 , \infty )\); range: \([ 2 , \infty )\)

326c85b7ef2b6468784f06604da2e200.png

5. Domain: \(\mathbb { R }\); range; \(\mathbb { R }\)

9da27c50cbdbe8f826496fa49c70e01a.png

7. Domain: \(\mathbb { R }\); range; \(\mathbb { R }\)

e17bdac87837ac3587ddf019eb7226ba.png

9. Domain: \(\mathbb { R }\); range; \(\mathbb { R }\)

b84dad2e3ddd99f6bf8eca4ab9612020.png

11. Domain: \(\mathbb { R }\); range; \(\mathbb { R }\)

b7cb2d25f3f876f0c587f47f97e8fa68.png

Exercise \(\PageIndex{8}\)

  • \(\sqrt [ 4 ] { 64 }\)
  • \(\sqrt [ 4 ] { 16 }\)
  • \(\sqrt [ 4 ] { 625 }\)
  • \(\sqrt [ 4 ] { 1 }\)
  • \(\sqrt [ 4 ] { 256 }\)
  • \(\sqrt [ 4 ] { 10,000 }\)
  • \(\sqrt [ 5 ] { 243 }\)
  • \(\sqrt [ 5 ] { 100,000 }\)
  • \(\sqrt [ 5 ] { \frac { 1 } { 32 } }\)
  • \(\sqrt [ 5 ] { \frac { 1 } { 243 } }\)
  • \(- \sqrt [ 4 ] { 16 }\)
  • \(- \sqrt [ 6 ] { 1 }\)
  • \(\sqrt [ 5 ] { - 1 }\)
  • \(\sqrt { - 1 }\)
  • \(\sqrt [ 4 ] { - 16 }\)
  • \(- 6 \sqrt [ 3 ] { - 27 }\)
  • \(- 5 \sqrt [ 3 ] { - 8 }\)
  • \(2 \sqrt [ 3 ] { - 1,000 }\)
  • \(7 \sqrt [ 5 ] { - 243 }\)
  • \(6 \sqrt [ 4 ] { - 16 }\)
  • \(12 \sqrt [ 6 ] { - 64 }\)
  • \(\sqrt [ 3 ] { \frac { 25 } { 16 } }\)
  • \(6 \sqrt { \frac { 16 } { 9 } }\)
  • \(5 \sqrt [ 3 ] { \frac { 27 } { 125 } }\)
  • \(7 \sqrt [ 5 ] { \frac { 32 } { 7 ^ { 5 } } }\)
  • \(- 5 \sqrt [ 3 ] { \frac { 8 } { 27 } }\)
  • \(- 8 \sqrt [ 4 ] { \frac { 625 } { 16 } }\)
  • \(2 \sqrt [ 5 ] { 100,000 }\)
  • \(2 \sqrt [ 7 ] { 128 }\)

9. \(\frac{1}{2}\)

11. \(−2\)

13. \(−2\)

15. Not a real number

19. \(−20\)

21. Not a real number

23. \(\frac{15}{4}\)

27. \(−\frac{10}{3}\)

Exercise \(\PageIndex{9}\)

  • \(\sqrt { 96 }\)
  • \(\sqrt { 500 }\)
  • \(\sqrt { 480 }\)
  • \(\sqrt { 450 }\)
  • \(\sqrt { 320 }\)
  • \(\sqrt { 216 }\)
  • \(5 \sqrt { 112 }\)
  • \(10 \sqrt { 135 }\)
  • \(- 2 \sqrt { 240 }\)
  • \(- 3 \sqrt { 162 }\)
  • \(\sqrt { \frac { 150 } { 49 } }\)
  • \(\sqrt { \frac { 200 } { 9 } }\)
  • \(\sqrt { \frac { 675 } { 121 } }\)
  • \(\sqrt { \frac { 192 } { 81 } }\)
  • \(\sqrt [ 3 ] { 54 }\)
  • \(\sqrt [ 3 ] { 24 }\)
  • \(\sqrt [ 3 ] { 48 }\)
  • \(\sqrt [ 3 ] { 81 }\)
  • \(\sqrt [ 3 ] { 40 }\)
  • \(\sqrt [ 3 ] { 120 }\)
  • \(\sqrt [ 3 ] { 162 }\)
  • \(\sqrt [ 3 ] { 500 }\)
  • \(\sqrt [ 3 ] { \frac { 54 } { 125 } }\)
  • \(\sqrt [ 3 ] { \frac { 40 } { 343 } }\)
  • \(5 \sqrt [ 3 ] { - 48 }\)
  • \(2 \sqrt [ 3 ] { - 108 }\)
  • \(8 \sqrt [ 4 ] { 96 }\)
  • \(7 \sqrt [ 4 ] { 162 }\)
  • \(\sqrt [ 5 ] { 160 }\)
  • \(\sqrt [ 5 ] { 486 }\)
  • \(\sqrt [ 5 ] { \frac { 224 } { 243 } }\)
  • \(\sqrt [ 5 ] { \frac { 5 } { 32 } }\)
  • \(\sqrt [ 5 ] { - \frac { 1 } { 32 } }\)
  • \(\sqrt [ 6 ] { - \frac { 1 } { 64 } }\)

1. \(4 \sqrt { 6 }\)

3. \(4 \sqrt { 30 }\)

5. \(8 \sqrt { 5 }\)

7. \(20 \sqrt { 7 }\)

9. \(- 8 \sqrt { 15 }\)

11. \(\frac { 5 \sqrt { 6 } } { 7 }\)

13. \(\frac { 15 \sqrt { 3 } } { 11 }\)

15. \(3 \sqrt [ 3 ] { 2 }\)

17. \(2 \sqrt [ 3 ] { 6 }\)

19. \(2 \sqrt [ 3 ] { 5 }\)

21. \(3 \sqrt [ 3 ] { 6 }\)

23. \(\frac { 3 \sqrt [ 3 ] { 2 } } { 5 }\)

25. \(- 10 \sqrt [ 3 ] { 6 }\)

27. \(16 \sqrt [ 4 ] { 6 }\)

29. \(2 \sqrt [ 5 ] { 5 }\)

31. \(\frac { 2 \sqrt [ 5 ] { 7 } } { 3 }\)

33. \(- \frac { 1 } { 2 }\)

Exercise \(\PageIndex{10}\)

Simplify. Give the exact answer and the approximate answer rounded to the nearest hundredth.

  • \(\sqrt { 60 }\)
  • \(\sqrt { 600 }\)
  • \(\sqrt { \frac { 96 } { 49 } }\)
  • \(\sqrt { \frac { 192 } { 25 } }\)
  • \(\sqrt [ 3 ] { 240 }\)
  • \(\sqrt [ 3 ] { 320 }\)
  • \(\sqrt [ 3 ] { \frac { 288 } { 125 } }\)
  • \(\sqrt [ 3 ] { \frac { 625 } { 8 } }\)
  • \(\sqrt [ 4 ] { 486 }\)
  • \(\sqrt [ 5 ] { 288 }\)

1. \(2 \sqrt { 15 } ; 7.75\)

3. \(\frac { 4 \sqrt { 6 } } { 7 } ; 1.40\)

5. \(2 \sqrt [ 3 ] { 30 } ; 6.21\)

7. \(\frac { 2 \sqrt [ 3 ] { 36 } } { 5 } ; 1.32\)

9. \(3 \sqrt [ 4 ] { 6 } ; 4.70\)

Exercise \(\PageIndex{11}\)

Rewrite the following as a radical expression with coeffecient \(1\).

  • \(2 \sqrt { 15 }\)
  • \(3 \sqrt { 7 }\)
  • \(5 \sqrt { 10 }\)
  • \(10 \sqrt { 3 }\)
  • \(2 \sqrt [ 3 ] { 7 }\)
  • \(3 \sqrt [ 3 ] { 6 }\)
  • \(2 \sqrt [ 4 ] { 5 }\)
  • \(3\sqrt [ 4 ] { 2 }\)
  • Each side of a square has a length that is equal to the square root of the square’s area. If the area of a square is \(72\) square units, find the length of each of its sides.
  • Each edge of a cube has a length that is equal to the cube root of the cube’s volume. If the volume of a cube is \(375\) cubic units, find the length of each of its edges.
  • The current \(I\) measured in amperes is given by the formula \(I = \sqrt { \frac { P } { R } }\) where \(P\) is the power usage measured in watts and \(R\) is the resistance measured in ohms. If a \(100\) watt light bulb has \(160\) ohms of resistance, find the current needed. (Round to the nearest hundredth of an ampere.)
  • The time in seconds an object is in free fall is given by the formula \(t = \frac { \sqrt { s } } { 4 }\) where \(s\) represents the distance in feet the object has fallen. How long will it take an object to fall to the ground from the top of an \(8\)-foot stepladder? (Round to the nearest tenth of a second.)

1. \(\sqrt { 60 }\)

3. \(\sqrt { 250 }\)

5. \(\sqrt [ 3 ] { 56 }\)

7. \(\sqrt [ 4 ] { 80 }\)

9. \(6 \sqrt { 2 }\) units

11. \(0.79\) ampere

Exercise \(\PageIndex{12}\)

  • Explain why there are two real square roots for any positive real number and one real cube root for any real number.
  • What is the square root of \(1\) and what is the cube root of \(1\)? Explain why.
  • Explain why \(\sqrt { - 1 }\) is not a real number and why \(\sqrt [ 3 ] { - 1 }\) is a real number.
  • Research and discuss the methods used for calculating square roots before the common use of electronic calculators.

1. Answer may vary

3. Answer may vary

1 A number that when multiplied by itself yields the original number.

2 The positive square root of a positive real number, denoted with the symbol \(√\).

3 The expression \(A\) within a radical sign, \(\sqrt [ n ] { A }\).

4 The function defined by \(f ( x ) = \sqrt { x }\).

5 A number that when used as a factor with itself three times yields the original number, denoted with the symbol \(\sqrt [ 3 ] { }\).

6 The positive integer \(n\) in the notation \(\sqrt [ n ] { }\) that is used to indicate an nth root.

7 The function defined by \(f ( x ) = \sqrt [ 3 ] { x }\).

8 A number that when raised to the \(n\)th power \((n ≥ 2)\) yields the original number.

9 Used when referring to an expression of the form \(\sqrt [ n ] { A }\).

10 The positive \(n\)th root when \(n\) is even.

11 Given real numbers \(\sqrt [ n ] { A }\) and \(\sqrt [ n ] { B }\),\(\sqrt [ n ] { A \cdot B } = \sqrt [ n ] { A } \cdot \sqrt [ n ] { B }\).

12 Given real numbers \(\sqrt [ n ] { A }\) and \(\sqrt [ n ] { B }\),\(\sqrt [ n ] { \frac { A } { B } } = \frac { \sqrt [ n ] { A } } { \sqrt [ n ] { B } }\) where \(B ≠ 0\).

13 A radical where the radicand does not consist of any factors that can be written as perfect powers of the index.

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Simplifying Radical Expressions

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Basics on the topic Simplifying Radical Expressions

For a radical expression to be in the simplest form, three conditions must be met: 1. The radicand contains no factor greater than 1 that is a perfect square. 2. There is no fraction under the radical sign. 3. There is no radical in the denominator of a fraction. The procedure used to remove a radical from a denominator is called rationalizing the denominator.

A knowledge of perfect squares and the product property of square roots can be very helpful in simplifying radical expressions. The property states that the root of the product of two terms is equal to the product of the root of each term.

To use the product property of square roots to simplify a radical expression, first write the radicand as the product of a perfect square and a factor that does not contain a perfect square. Then use the product property of square roots to write the expression as a product. Finally, simplify the radical expression.

Another useful property for simplifying radical expressions is the quotient property of square roots; it is used to divide radical expressions. The property states that the root of a rational number is equal to the root of the numerator divided by the root of the denominator.

In a similar fashion, the quotient property can be used to simplify a radical expression by first writing the expression as the root of the numerator divided by the root of the denominator, then simplifying the root in the numerator and the root in the denominator, and finally simplifying the resulting radical.

Expressions and Equations Work with radicals and integer exponents.

CCSS.MATH.CONTENT.8.EE.A.2

Transcript Simplifying Radical Expressions

Long, long ago, somewhere in the deep blue of the Caribbean Sea. Two rival pirate captains were madly in love, and as you can imagine, this was quite a complicated situation. To keep their dalliance a secret, they pretended to be mortal enemies, and to prove this, they constantly pretended to try to overpower the other. Captain Bonny hoisted her sails due east while Calico Jack headed south. Bonny had the wind at her back, so she traveled twice as fast as Jack. Only two hours later and with a heavy heart, she wondered how far she was from her paramour’s sloop.

Simplifying radical expressions

Being a math whiz, the lovelorn captain used simplified radical expressions to figure out the distance. Let’s look at her calculations: Bonny travelled a distance equal to 4x, and Jack travelled a distance equal to 2x. Notice the right angle?

Because the points are in the shape of a right triangle, she used the Pythagorean Theorem to solve for the unknown length . Remember, the Pythagorean Theorem is: a² + b² = c² and 'c' is the hypotenuse , which is the longest side . The hypotenuse is always located opposite the right angle. Now, by subbing in the lengths she knows, Bonny can calculate the unknown length by finding the square root of 20 x².

The notation of roots

Let’s investigate the proper notation for roots . I bet you always wondered about this. The small number is called the index , and it indicates the root . A root of 2 indicates the second root , or the square root ; if no number is present, the second root is assumed. The radical is the boxy shape, and the radicand is the number under the box. Understanding and using the correct terms for math can make calculations easier!

Two Properties to solve problems

There are two properties that will help us to solve problems that include square roots. First, let’s review the Product Property of Square Roots :

The square root of the product of 'ab' is equal to the square root of 'a' times the square root of 'b'.

Let’s take a look at an example: what is the square root of 27? First, factor out any perfect squares. In case you forgot, a perfect square is a number that is the square of a rational whole number . Perfect squares are 4, 9, 16, 25, and so on. Back to root 27. Using the Product Property of Square Roots, we can factor out the perfect square of 9. Then, just do the math. The end result is 3 times the square root of 3, or simply, three root three.

The second property is the Quotient Property of Square Roots . The square root of the fraction 'a/b' is equal to the square root of 'a' divided by the square root of 'b'. Let’s sub in some numbers to make this easier to understand. The square root of the fraction 49 over 81 is equal to the square root of 49 over the square root of 81 or 7 over 9.

Back to Captain Bonny. How did she simplify the radical expression of the square root of 20 x²? First, she looked for perfect square factors of 20, oh there’s one: 4! x² is obviously a perfect square, so we can group these two terms together. The rest of the calculations are easier than firing a cannon. The square root of 20x² can be simplified to 2x times the square root of 5.

And speaking of firing a cannon. Is this for real? Oh, so that was her plan all along!

Simplifying Radical Expressions exercise

Determine the distance $c$..

You can use the Pythagorean theorem to determine the missing length.

Don't forget to simplify your answer.

You can simplify your answer using the product property of roots.

The distance the ships have traveled and the distance between them form a right angle. We can use the Pythagorean theorem to find the distance between the two ships. The distance between them is the hypotenuse. We have

$a^2 + b^2 = c^2$.

Substituting in known values,

$(2x)^2 + (4x)^2 = c^2$,

and simplifying gives

$4x^2 + 16x^2 = c^2$

$20x^2 = c^2$.

Taking the root of both sides gives

$\sqrt{20x^2} = c$.

Now we have an expression for $c$, and we must simplify it as much as possible. We can do this using the product property of roots:

$c = \sqrt{20x^2}$.

We can start by separating the radicand (the term the root operator is applied to) into a product of two terms, one of which contains only perfect squares:

$c = \sqrt{4x^2 \times 5}$.

We know that the product property of roots states that

$\sqrt{ab} = \sqrt{a} \times \sqrt{b}$.

We can apply this to our solution, to arrive at

$c = \sqrt{4x^2} \times \sqrt{5}$.

We can then simplify this expression to a final answer:

$c = 2x \times \sqrt{5}$.

Explain how to multiply and divide radical expressions.

The name of the property includes either "product" or "quotient", which gives you a hint about what operation it involves.

The properties don't use addition or subtraction.

The product property of roots is used to simplify terms that involve multiplication.

The property states that the root of the product of two terms is equal to the product of the root of each term:

The quotient property of roots is used to simplify terms that involve division.

The property states that the root of a rational number is equal to the root of the numerator divided by the root of the denominator:

$\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$.

Decide which radical expressions can be simplified.

Look for factor pairs of the radicand, where one pair is a perfect square.

Keep in mind that $4$, $9$, $16$, $25$, and $x^2$ are all perfect squares.

All of these radical terms involve multiplication, not division. Therefore we should try to simplify each expression using the product property of roots. A radical term can be simplified using this property if its radicand has a factor that is also a perfect square.

Let's look for factors of each radicand, keeping in mind that $4$, $9$, $16$, $25$, and $x^2$ are all perfect squares.

For the first radical term, $\sqrt{32x^2}$, the radicand $32x^2$ has the factor $16x^2$. The square root of this factor is $4x$, so it is a perfect square. Therefore this radical term can be factored. We can prove this by factoring it:

$\sqrt{32x^2}$

$= \sqrt{16x^2 \times 2}$

$= \sqrt{16x^2} \times \sqrt{2}$

$= 4x \times \sqrt{2}$ ✓

Similarly, for the other radical expressions...

  • $\sqrt{14}$ the radicand $14$ has no factors which are a perfect square. ✗
  • $\sqrt{18x}$ the radicand $18x$ has the factor $9$, which is a perfect square. ✓
  • $\sqrt{50}$ the radicand $50$ has the factor $25$, which is a perfect square. ✓

Determine the missing length.

It may help to assign a variable to the side length of one of the squares.

You can find the side length of the square using the area of the square, and the fact that the area is equal to the square of the side length.

Don't forget to simplify radical terms.

Jack labels the side length of a square as $s$. He therefore knows that $x = 2s$.

He can also say that the area of the square, $a$, is $a = s^2$.

He knows that the area of the square is $32$ units, so he substitutes this in to get $32 = s^2$.

He applies opposite operations to isolate $s$: $\sqrt{32} = s$.

Now he must simplify the radical to find the value of $s$:

$s = \sqrt{32}$

$s = \sqrt{16 \times 2}$.

Using the product property of roots, he gets

$s = \sqrt{16} \times \sqrt{2}$.

Simplifying gives

$s = 4 \times \sqrt{2}$.

Jack can now find the missing value $x$:

$x = 2 \times 4 \times \sqrt{2}$

$x = 8 \sqrt{2}$.

Label the parts of a radical expression.

Pay close attention to spelling.

The number that indicates what root is being taken is called the index.

The small number is called the index. It indicates which root is being taken. It is equal to two in the case of a square root.

The boxy shape that looks a bit like a check mark is called the radical.

The large number, inside the radical, is called the radicand.

Simplify the radical expressions.

The different expressions sometimes share common factors. It may help to start by simplifying each radical expression without looking at the answers.

Each expression can be simplified using the product property of roots.

Each expression can be simplified using the product property of roots. Let's simplify each radical expression.

For $\sqrt{32}$, $16$ is both a factor of the radicand, and is a perfect square. Let's factor this number out of the radicand: $\sqrt{16 \times 2}$. Now we can used the product property of roots to simplify the radical expression: $\sqrt{16} \times \sqrt{2}$. This simplifies to $4 \times \sqrt{2}$.

Similarly, we can simplify $\sqrt{18}$ into $3 \times \sqrt{2}$.

The radical expression $\sqrt{12}$ becomes $2 \times \sqrt{3}$.

And the expression $\sqrt{48}$ simplifies to $4 \times \sqrt{3}$.

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Solving Radical Equations

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Rationalize the Denominator

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The Distance Formula

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The Midpoint Formula

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Adding and Subtracting Radical Expressions

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Multiplying Radical Expressions

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Dividing Radical Expressions

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  1. Simplifying radicals

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  2. Simplifying Radicals Expressions Worksheet Answers

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  4. Simplifying Radicals Worksheet 20 Answers

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VIDEO

  1. Simplifying Radical Expressions

  2. 11-3 Simplifying Radical Expressions Homework Q's p.493

  3. How to Simplify Radicals #shorts

  4. Solving for x with RADICALS

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COMMENTS

  1. Simplifying Radicals Worksheet and Answer Key

    Objective. Students will practice simplifying radicals. This sheet focuses on Algebra 1 problems using real numbers. (No Algebraic expressions) The worksheet has model problems worked out, step by step. 25 scaffolded questions that start out relatively easy and end with some real challenges.

  2. PDF Simplifying Radicals Name: Answer Key Answers

    Simplifying Radicals Name: Answer Key Math www.CommonCoreSheets.com 2 1-10 95 90 85 80 75 70 65 60 55 50 11-20 45 40 35 30 25 20 15 10 5 0 Simplify each radical. 1) 84

  3. PDF 10.1 Simplifying Radical Expressions

    Communicate Your Answer 3. How can you multiply square roots? 4. Give an example of multiplying square roots that is different from the examples in Exploration 1. 5. Write an algebraic rule for the product of square roots. REASONING ABSTRACTLY To be profi cient in math, you need to recognize and use counterexamples. Simplifying Radical Expressions

  4. 5.2: Simplifying Radical Expressions

    Answer: \(3x\) Example \(\PageIndex{2}\): ... Key Takeaways. To simplify a radical expression, look for factors of the radicand with powers that match the index. If found, they can be simplified by applying the product and quotient rules for radicals, as well as the property \(\sqrt [ n ] { a ^ { n } } = a\), where \(a\) is nonnegative ...

  5. PDF Simplifying Radical Expressions Date Period

    Simplifying Radical Expressions Date_____ Period____ Simplify. 1) 125 n 2) 216 v 3) 512 k2 4) 512 m3 5) 216 k4 6) 100 v3 7) 80 p3 8) 45 p2 9) 147 m3n3 10) 200 m4n 11) 75 x2y 12) 64 m3n3 13) 16 u4v3 14) 28 x3y3-1- ©s n220 D1b2S kKRumtUa c LSgoqfMtywta1rme0 pL qL 9CY. f H VArl qlV 0r 8i rg OhAtas H yr3e 2sUeGrXvQejd d.Q j 8M BaWdWes 8wYist Dh9 ...

  6. 8.2 Simplify Radical Expressions

    Step 1. Find the largest factor in the radicand that is a perfect power of the index. Rewrite the radicand as a product of two factors, using that factor. Step 2. Use the product rule to rewrite the radical as the product of two radicals. Step 3. Simplify the root of the perfect power.

  7. PDF Central Bucks School District / Homepage

    Created Date: 4/20/2017 9:23:26 AM

  8. 11.1 Simplifying Radicals

    11.1 Simplifying Square Roots. Need a tutor? Click this link and get your first session free! A1.1.6 Simplify numerical expressions, including those involving radicals and absolute value. Packet: 11.1_simplifying_radicals.pdf: File Size: 1254 kb: File Type: pdf: Download File. Practice Solutions: a1_unit_11.1_prac_ans.pdf:

  9. PDF 7.1R Simplifying Radicals 020316

    LO: I can simplify radical expressions including adding, subtracting, multiplying, dividing and rationalizing denominators. DO NOW On the back of this packet (1) calculator Simplifying Radicals: Finding hidden perfect squares and taking their root. Simplify each expression by factoring to find perfect squares and then taking their root.

  10. PDF Simplifying Radicals Date Period

    Simplifying Radicals Date_____ Period____ Simplify. Use absolute value signs when necessary. 1) 24 2) 3 1000 3) 3 −162 4) 512 5) 4 128 n8 6) 98 k 7) 5 224 r7 8) 3 24 m3 9 ... Simplify n 3 ⋅ 2n ⋅ x2n yn + 3 2x2y n 3y3 Create your own worksheets like this one with Infinite Algebra 2. Free trial available at KutaSoftware.com.

  11. 8.2: Simplifying Radical Expressions

    For this reason, we will use the following property for the rest of the section: n√an = a, if a ≥ 0 n th root. When simplifying radical expressions, look for factors with powers that match the index. Example 8.2.3. Simplify: √18x3y4. Solution: Begin by determining the square factors of 18, x3, and y4.

  12. PDF 7.1 Simplifying Radicals

    Simplifying Radicals: Finding hidden perfect squares and taking their root. Simplify each expression by factoring to find perfect squares and then taking their root. Add or subtract radicals by simplifying each term and then combining like terms. (a) Multiply numbers that are BOTH OUTSIDE the radical.

  13. Radicals Review -- Free Worksheet (pdf) and Answer Key. 47 scaffolded

    All of your worksheets are now here on Mathwarehouse.com. Please update your bookmarks! These worksheets focus on the topics typically covered in Algebra I (no variables). Radical Worksheets: Dividing Radicals. Simplify Radicals Worksheet. Adding Radicals. Mulitplying Radicals Worksheet. Radicals Review (Mixed review worksheet on radicals and ...

  14. PDF 1-Simplifying Square Roots

    Simplifying Square Roots Date_____ Period____ Simplify. 1) 96 2) 216 3) 98 4) 18 5) 72 6) 144 7) 45 8) 175 9) 343 10) 12 11) 10 96 12) 9 245-1- ©J 52e0 J1w1L oKGu7t 5ay YSWojf AtgwCaHrzeh cL HLVCj.2 h WAWlhl F Nr9iZgPh ktGsM or IehsUesr Lve1d9. a P 1Mja0dGe3 lw 1iTtyh L uI Knofni rn 9i QtUeL gGzeko ZmKe1tPr6ys. 4 Worksheet by Kuta Software LLC ...

  15. Chapter 10

    Algebra 1 answers to Chapter 10 - Radical Expressions and Equations - 10-2 Simplifying Radicals - Practice and Problem-Solving Exercises - Page 610 17 including work step by step written by community members like you. Textbook Authors: Hall, Prentice, ISBN-10: 0133500403, ISBN-13: 978--13350-040-0, Publisher: Prentice Hall

  16. Algebra Skill

    Simplifying Radicals. ©B Y240f1U1P qKbuDtaaD BSkoFfjtKw5aCrLez 1LyLxCa.X A WArlulH Qr1iegahqtMsy SrEeKs1e0rEvyeGdW.s. Write each expression in simplest radical form. 1) 48. 3) 12. 5) 36. 7) 125. 9) 18. 11) 50.

  17. Unit 6: Radicals

    6.2 Simplifying Radicals. A3 Notes. ... 6.3 Multiplying and Dividing Radicals. A3 Notes. A4 Notes. HW. HW Key. 12/5 6.4 Binomial Radical Expressions. A3 Notes. A4 Notes. HW. HW Key. 12/7 6.5 Solving Radical Equations. A3 Notes. A4 Notes. HW. HW Key. 12/11 Unit 6 Review. Unit 6 Review. Review answer key. Powered by Create your own unique website ...

  18. 7.1: Roots and Radicals

    Key Takeaways. To simplify a square root, look for the largest perfect square factor of the radicand and then apply the product or quotient rule for radicals. ... Simplify. Give the exact answer and the approximate answer rounded to the nearest hundredth. \(\sqrt { 60 }\) ... Explain why there are two real square roots for any positive real ...

  19. Simplifying Radical Expressions

    Perfect squares are 4, 9, 16, 25, and so on. Back to root 27. Using the Product Property of Square Roots, we can factor out the perfect square of 9. Then, just do the math. The end result is 3 times the square root of 3, or simply, three root three. The second property is the Quotient Property of Square Roots. The square root of the fraction 'a ...

  20. Simplifying Radicals

    Created by. aurbaniak25 Teacher. Use prime factorization. Pairs of factors break out of the radical symbol (jail) and prime numbers are left inside. If you have multiple numbers on the inside or outside just multiply them. Ex. square root of 20 has 2 *2*5. Pair of 2's means the 2 breaks out and 5 is on the inside, so 2_/5 is the answer.

  21. PDF Unit 6

    35) '-t -3E) a-I 1b - + wasm Agebta). Name: Unit 6: Radical Functons Homæork 7: Graphing Radial Functions ** This a 2-page document ** For quætions 1-2: Describe the bansformaäons from the parent function. strdch 21 x-axis , 3. The square root parent funcöon is translated so that it has an endpoint located at (4, -1), then vertically ...

  22. Simplifying radically question

    Name: _____ Unit 6: Radical Functions Date: _____ Bell: _____ Homework 1: nth Roots; Simplifying Radicals Directions: Give ALL possible roots to the radicals below. 1. 225 2. 3 512 3. 3 − 64 4. 4 1 296, Directions: Simplify the radicals below. 5. 5 150 6. − 8 63 7. 3 875 8. − 3 2 54 9. − 4 3 567 10. 4

  23. Solved Name: Unit 6: Radical Functions Date: Bell: Homework

    Question: Name: Unit 6: Radical Functions Date: Bell: Homework 1: nth Roots; Simplifying Radicals. Simplfy the radicals below. question 5-19 please help with steps. Show transcribed image text. There's just one step to solve this. Expert-verified.