• 1.3 Radicals and Rational Exponents
  • Introduction to Prerequisites
  • 1.1 Real Numbers: Algebra Essentials
  • 1.2 Exponents and Scientific Notation
  • 1.4 Polynomials
  • 1.5 Factoring Polynomials
  • 1.6 Rational Expressions
  • Key Equations
  • Key Concepts
  • Review Exercises
  • Practice Test
  • Introduction to Equations and Inequalities
  • 2.1 The Rectangular Coordinate Systems and Graphs
  • 2.2 Linear Equations in One Variable
  • 2.3 Models and Applications
  • 2.4 Complex Numbers
  • 2.5 Quadratic Equations
  • 2.6 Other Types of Equations
  • 2.7 Linear Inequalities and Absolute Value Inequalities
  • Introduction to Functions
  • 3.1 Functions and Function Notation
  • 3.2 Domain and Range
  • 3.3 Rates of Change and Behavior of Graphs
  • 3.4 Composition of Functions
  • 3.5 Transformation of Functions
  • 3.6 Absolute Value Functions
  • 3.7 Inverse Functions
  • Introduction to Linear Functions
  • 4.1 Linear Functions
  • 4.2 Modeling with Linear Functions
  • 4.3 Fitting Linear Models to Data
  • Introduction to Polynomial and Rational Functions
  • 5.1 Quadratic Functions
  • 5.2 Power Functions and Polynomial Functions
  • 5.3 Graphs of Polynomial Functions
  • 5.4 Dividing Polynomials
  • 5.5 Zeros of Polynomial Functions
  • 5.6 Rational Functions
  • 5.7 Inverses and Radical Functions
  • 5.8 Modeling Using Variation
  • Introduction to Exponential and Logarithmic Functions
  • 6.1 Exponential Functions
  • 6.2 Graphs of Exponential Functions
  • 6.3 Logarithmic Functions
  • 6.4 Graphs of Logarithmic Functions
  • 6.5 Logarithmic Properties
  • 6.6 Exponential and Logarithmic Equations
  • 6.7 Exponential and Logarithmic Models
  • 6.8 Fitting Exponential Models to Data
  • Introduction to Systems of Equations and Inequalities
  • 7.1 Systems of Linear Equations: Two Variables
  • 7.2 Systems of Linear Equations: Three Variables
  • 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables
  • 7.4 Partial Fractions
  • 7.5 Matrices and Matrix Operations
  • 7.6 Solving Systems with Gaussian Elimination
  • 7.7 Solving Systems with Inverses
  • 7.8 Solving Systems with Cramer's Rule
  • Introduction to Analytic Geometry
  • 8.1 The Ellipse
  • 8.2 The Hyperbola
  • 8.3 The Parabola
  • 8.4 Rotation of Axes
  • 8.5 Conic Sections in Polar Coordinates
  • Introduction to Sequences, Probability and Counting Theory
  • 9.1 Sequences and Their Notations
  • 9.2 Arithmetic Sequences
  • 9.3 Geometric Sequences
  • 9.4 Series and Their Notations
  • 9.5 Counting Principles
  • 9.6 Binomial Theorem
  • 9.7 Probability

Learning Objectives

In this section, you will:

  • Evaluate square roots.
  • Use the product rule to simplify square roots.
  • Use the quotient rule to simplify square roots.
  • Add and subtract square roots.
  • Rationalize denominators.
  • Use rational roots.

A hardware store sells 16-ft ladders and 24-ft ladders. A window is located 12 feet above the ground. A ladder needs to be purchased that will reach the window from a point on the ground 5 feet from the building. To find out the length of ladder needed, we can draw a right triangle as shown in Figure 1 , and use the Pythagorean Theorem.

Now, we need to find out the length that, when squared, is 169, to determine which ladder to choose. In other words, we need to find a square root. In this section, we will investigate methods of finding solutions to problems such as this one.

Evaluating Square Roots

When the square root of a number is squared, the result is the original number. Since 4 2 = 16 , 4 2 = 16 , the square root of 16 16 is 4. 4. The square root function is the inverse of the squaring function just as subtraction is the inverse of addition. To undo squaring, we take the square root.

In general terms, if a a is a positive real number, then the square root of a a is a number that, when multiplied by itself, gives a . a . The square root could be positive or negative because multiplying two negative numbers gives a positive number. The principal square root is the nonnegative number that when multiplied by itself equals a . a . The square root obtained using a calculator is the principal square root.

The principal square root of a a is written as a . a . The symbol is called a radical , the term under the symbol is called the radicand , and the entire expression is called a radical expression .

Principal Square Root

The principal square root of a a is the nonnegative number that, when multiplied by itself, equals a . a . It is written as a radical expression , with a symbol called a radical over the term called the radicand : a . a .

Does 25 = ± 5 ? 25 = ± 5 ?

No. Although both 5 2 5 2 and ( −5 ) 2 ( −5 ) 2 are 25 , 25 , the radical symbol implies only a nonnegative root, the principal square root. The principal square root of 25 is 25 = 5. 25 = 5.

Evaluate each expression.

  • ⓒ 25 + 144 25 + 144
  • ⓓ 49 − 81 49 − 81
  • ⓐ 100 = 10 100 = 10 because 10 2 = 100 10 2 = 100
  • ⓑ 16 = 4 = 2 16 = 4 = 2 because 4 2 = 16 4 2 = 16 and 2 2 = 4 2 2 = 4
  • ⓒ 25 + 144 = 169 = 13 25 + 144 = 169 = 13 because 13 2 = 169 13 2 = 169
  • ⓓ 49 − 81 = 7 − 9 = −2 49 − 81 = 7 − 9 = −2 because 7 2 = 49 7 2 = 49 and 9 2 = 81 9 2 = 81

For 25 + 144 , 25 + 144 , can we find the square roots before adding?

No. 25 + 144 = 5 + 12 = 17. 25 + 144 = 5 + 12 = 17. This is not equivalent to 25 + 144 = 13. 25 + 144 = 13. The order of operations requires us to add the terms in the radicand before finding the square root.

  • ⓒ 25 − 9 25 − 9
  • ⓓ 36 + 121 36 + 121

Using the Product Rule to Simplify Square Roots

To simplify a square root, we rewrite it such that there are no perfect squares in the radicand. There are several properties of square roots that allow us to simplify complicated radical expressions. The first rule we will look at is the product rule for simplifying square roots, which allows us to separate the square root of a product of two numbers into the product of two separate rational expressions. For instance, we can rewrite 15 15 as 3 ⋅ 5 . 3 ⋅ 5 . We can also use the product rule to express the product of multiple radical expressions as a single radical expression.

The Product Rule for Simplifying Square Roots

If a a and b b are nonnegative, the square root of the product a b a b is equal to the product of the square roots of a a and b . b .

Given a square root radical expression, use the product rule to simplify it.

  • Factor any perfect squares from the radicand.
  • Write the radical expression as a product of radical expressions.

Simplify the radical expression.

  • ⓑ 162 a 5 b 4 162 a 5 b 4
  • ⓐ 100 ⋅ 3 Factor perfect square from radicand . 100 ⋅ 3 Write radical expression as product of radical expressions . 10 3 Simplify . 100 ⋅ 3 Factor perfect square from radicand . 100 ⋅ 3 Write radical expression as product of radical expressions . 10 3 Simplify .
  • ⓑ 81 a 4 b 4 ⋅ 2 a Factor perfect square from radicand . 81 a 4 b 4 ⋅ 2 a Write radical expression as product of radical expressions . 9 a 2 b 2 2 a Simplify . 81 a 4 b 4 ⋅ 2 a Factor perfect square from radicand . 81 a 4 b 4 ⋅ 2 a Write radical expression as product of radical expressions . 9 a 2 b 2 2 a Simplify .

Simplify 50 x 2 y 3 z . 50 x 2 y 3 z .

Given the product of multiple radical expressions, use the product rule to combine them into one radical expression.

  • Express the product of multiple radical expressions as a single radical expression.

Using the Product Rule to Simplify the Product of Multiple Square Roots

Simplify the radical expression. 12 ⋅ 3 12 ⋅ 3

12 ⋅ 3 Express the product as a single radical expression . 36 Simplify . 6 12 ⋅ 3 Express the product as a single radical expression . 36 Simplify . 6

Simplify 50 x ⋅ 2 x 50 x ⋅ 2 x assuming x > 0. x > 0.

Using the Quotient Rule to Simplify Square Roots

Just as we can rewrite the square root of a product as a product of square roots, so too can we rewrite the square root of a quotient as a quotient of square roots, using the quotient rule for simplifying square roots. It can be helpful to separate the numerator and denominator of a fraction under a radical so that we can take their square roots separately. We can rewrite 5 2 5 2 as 5 2 . 5 2 .

The Quotient Rule for Simplifying Square Roots

The square root of the quotient a b a b is equal to the quotient of the square roots of a a and b , b , where b ≠ 0. b ≠ 0.

Given a radical expression, use the quotient rule to simplify it.

  • Write the radical expression as the quotient of two radical expressions.
  • Simplify the numerator and denominator.

5 36 Write as quotient of two radical expressions . 5 6 Simplify denominator . 5 36 Write as quotient of two radical expressions . 5 6 Simplify denominator .

Simplify 2 x 2 9 y 4 . 2 x 2 9 y 4 .

Using the Quotient Rule to Simplify an Expression with Two Square Roots

234 x 11 y 26 x 7 y 234 x 11 y 26 x 7 y

234 x 11 y 26 x 7 y Combine numerator and denominator into one radical expression . 9 x 4 Simplify fraction . 3 x 2   Simplify square root . 234 x 11 y 26 x 7 y Combine numerator and denominator into one radical expression . 9 x 4 Simplify fraction . 3 x 2   Simplify square root .

Simplify 9 a 5 b 14 3 a 4 b 5 . 9 a 5 b 14 3 a 4 b 5 .

Adding and Subtracting Square Roots

We can add or subtract radical expressions only when they have the same radicand and when they have the same radical type such as square roots. For example, the sum of 2 2 and 3 2 3 2 is 4 2 . 4 2 . However, it is often possible to simplify radical expressions, and that may change the radicand. The radical expression 18 18 can be written with a 2 2 in the radicand, as 3 2 , 3 2 , so 2 + 18 = 2 + 3 2 = 4 2 . 2 + 18 = 2 + 3 2 = 4 2 .

Given a radical expression requiring addition or subtraction of square roots, simplify.

  • Simplify each radical expression.
  • Add or subtract expressions with equal radicands.

Adding Square Roots

Add 5 12 + 2 3 . 5 12 + 2 3 .

We can rewrite 5 12 5 12 as 5 4 · 3 . 5 4 · 3 . According the product rule, this becomes 5 4 3 . 5 4 3 . The square root of 4 4 is 2, so the expression becomes 5 ( 2 ) 3 , 5 ( 2 ) 3 , which is 10 3 . 10 3 . Now the terms have the same radicand so we can add.

10 3 + 2 3 = 12 3 10 3 + 2 3 = 12 3

Add 5 + 6 20 . 5 + 6 20 .

Subtracting Square Roots

Subtract 20 72 a 3 b 4 c − 14 8 a 3 b 4 c . 20 72 a 3 b 4 c − 14 8 a 3 b 4 c .

Factor 9 out of the first term so that both terms have equal radicands.

Subtract 3 80 x − 4 45 x . 3 80 x − 4 45 x .

Rationalizing Denominators

When an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. We can remove radicals from the denominators of fractions using a process called rationalizing the denominator .

We know that multiplying by 1 does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by the form of 1 that will eliminate the radical.

For a denominator containing a single term, multiply by the radical in the denominator over itself. In other words, if the denominator is b c , b c , multiply by c c . c c .

For a denominator containing the sum or difference of a rational and an irrational term, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign of the radical portion of the denominator. If the denominator is a + b c , a + b c , then the conjugate is a − b c . a − b c .

Given an expression with a single square root radical term in the denominator, rationalize the denominator.

  • Multiply the numerator and denominator by the radical in the denominator.

Rationalizing a Denominator Containing a Single Term

Write 2 3 3 10 2 3 3 10 in simplest form.

The radical in the denominator is 10 . 10 . So multiply the fraction by 10 10 . 10 10 . Then simplify.

Write 12 3 2 12 3 2 in simplest form.

Given an expression with a radical term and a constant in the denominator, rationalize the denominator.

  • Find the conjugate of the denominator.
  • Multiply the numerator and denominator by the conjugate.
  • Use the distributive property.

Rationalizing a Denominator Containing Two Terms

Write 4 1 + 5 4 1 + 5 in simplest form.

Begin by finding the conjugate of the denominator by writing the denominator and changing the sign. So the conjugate of 1 + 5 1 + 5 is 1 − 5 . 1 − 5 . Then multiply the fraction by 1 − 5 1 − 5 . 1 − 5 1 − 5 .

Write 7 2 + 3 7 2 + 3 in simplest form.

Using Rational Roots

Although square roots are the most common rational roots, we can also find cube roots, 4th roots, 5th roots, and more. Just as the square root function is the inverse of the squaring function, these roots are the inverse of their respective power functions. These functions can be useful when we need to determine the number that, when raised to a certain power, gives a certain number.

Understanding n th Roots

Suppose we know that a 3 = 8. a 3 = 8. We want to find what number raised to the 3rd power is equal to 8. Since 2 3 = 8 , 2 3 = 8 , we say that 2 is the cube root of 8.

The n th root of a a is a number that, when raised to the n th power, gives a . a . For example, −3 −3 is the 5th root of −243 −243 because ( −3 ) 5 = −243. ( −3 ) 5 = −243. If a a is a real number with at least one n th root, then the principal n th root of a a is the number with the same sign as a a that, when raised to the n th power, equals a . a .

The principal n th root of a a is written as a n , a n , where n n is a positive integer greater than or equal to 2. In the radical expression, n n is called the index of the radical.

Principal n n th Root

If a a is a real number with at least one n th root, then the principal n th root of a , a , written as a n , a n , is the number with the same sign as a a that, when raised to the n th power, equals a . a . The index of the radical is n . n .

Simplifying n th Roots

Simplify each of the following:

  • ⓐ −32 5 −32 5
  • ⓑ 4 4 ⋅ 1 , 024 4 4 4 ⋅ 1 , 024 4
  • ⓒ − 8 x 6 125 3 − 8 x 6 125 3
  • ⓓ 8 3 4 − 48 4 8 3 4 − 48 4
  • ⓐ −32 5 = −2 −32 5 = −2 because ( −2 ) 5 = −32 ( −2 ) 5 = −32
  • ⓑ First, express the product as a single radical expression. 4,096 4 = 8 4,096 4 = 8 because 8 4 = 4,096 8 4 = 4,096
  • ⓒ − 8 x 6 3 125 3 Write as quotient of two radical expressions . − 2 x 2 5 Simplify . − 8 x 6 3 125 3 Write as quotient of two radical expressions . − 2 x 2 5 Simplify .
  • ⓓ 8 3 4 − 2 3 4 Simplify to get equal radicands . 6 3 4   Add . 8 3 4 − 2 3 4 Simplify to get equal radicands . 6 3 4   Add .
  • ⓐ −216 3 −216 3
  • ⓑ 3 80 4 5 4 3 80 4 5 4
  • ⓒ 6 9 , 000 3 + 7 576 3 6 9 , 000 3 + 7 576 3

Using Rational Exponents

Radical expressions can also be written without using the radical symbol. We can use rational (fractional) exponents. The index must be a positive integer. If the index n n is even, then a a cannot be negative.

We can also have rational exponents with numerators other than 1. In these cases, the exponent must be a fraction in lowest terms. We raise the base to a power and take an n th root. The numerator tells us the power and the denominator tells us the root.

All of the properties of exponents that we learned for integer exponents also hold for rational exponents.

  • Rational Exponents

Rational exponents are another way to express principal n th roots. The general form for converting between a radical expression with a radical symbol and one with a rational exponent is

Given an expression with a rational exponent, write the expression as a radical.

  • Determine the power by looking at the numerator of the exponent.
  • Determine the root by looking at the denominator of the exponent.
  • Using the base as the radicand, raise the radicand to the power and use the root as the index.

Writing Rational Exponents as Radicals

Write 343 2 3 343 2 3 as a radical. Simplify.

The 2 tells us the power and the 3 tells us the root.

343 2 3 = ( 343 3 ) 2 = 343 2 3 343 2 3 = ( 343 3 ) 2 = 343 2 3

We know that 343 3 = 7 343 3 = 7 because 7 3 = 343. 7 3 = 343. Because the cube root is easy to find, it is easiest to find the cube root before squaring for this problem. In general, it is easier to find the root first and then raise it to a power.

343 2 3 = ( 343 3 ) 2 = 7 2 = 49 343 2 3 = ( 343 3 ) 2 = 7 2 = 49

Write 9 5 2 9 5 2 as a radical. Simplify.

Writing Radicals as Rational Exponents

Write 4 a 2 7 4 a 2 7 using a rational exponent.

The power is 2 and the root is 7, so the rational exponent will be 2 7 . 2 7 . We get 4 a 2 7 . 4 a 2 7 . Using properties of exponents, we get 4 a 2 7 = 4 a −2 7 . 4 a 2 7 = 4 a −2 7 .

Write x ( 5 y ) 9 x ( 5 y ) 9 using a rational exponent.

Simplifying Rational Exponents

  • ⓐ 5 ( 2 x 3 4 ) ( 3 x 1 5 ) 5 ( 2 x 3 4 ) ( 3 x 1 5 )
  • ⓑ ( 16 9 ) − 1 2 ( 16 9 ) − 1 2

ⓐ 30 x 3 4 x 1 5 Multiply the coefficients . 30 x 3 4 + 1 5 Use properties of exponents . 30 x 19 20 Simplify . 30 x 3 4 x 1 5 Multiply the coefficients . 30 x 3 4 + 1 5 Use properties of exponents . 30 x 19 20 Simplify .

ⓑ ( 9 16 ) 1 2    Use definition of negative exponents . 9 16    Rewrite as a radical . 9 16    Use the quotient rule . 3 4    Simplify . ( 9 16 ) 1 2    Use definition of negative exponents . 9 16    Rewrite as a radical . 9 16    Use the quotient rule . 3 4    Simplify .

Simplify ( 8 x ) 1 3 ( 14 x 6 5 ) . ( 8 x ) 1 3 ( 14 x 6 5 ) .

Access these online resources for additional instruction and practice with radicals and rational exponents.

  • Simplify Radicals
  • Rationalize Denominator

1.3 Section Exercises

What does it mean when a radical does not have an index? Is the expression equal to the radicand? Explain.

Where would radicals come in the order of operations? Explain why.

Every number will have two square roots. What is the principal square root?

Can a radical with a negative radicand have a real square root? Why or why not?

For the following exercises, simplify each expression.

4 ( 9 + 16 ) 4 ( 9 + 16 )

289 − 121 289 − 121

27 64 27 64

169 + 144 169 + 144

18 162 18 162

14 6 − 6 24 14 6 − 6 24

15 5 + 7 45 15 5 + 7 45

96 100 96 100

( 42 ) ( 30 ) ( 42 ) ( 30 )

12 3 − 4 75 12 3 − 4 75

4 225 4 225

405 324 405 324

360 361 360 361

5 1 + 3 5 1 + 3

8 1 − 17 8 1 − 17

128 3 + 3 2 3 128 3 + 3 2 3

−32 243 5 −32 243 5

15 125 4 5 4 15 125 4 5 4

3 −432 3 + 16 3 3 −432 3 + 16 3

400 x 4 400 x 4

4 y 2 4 y 2

( 144 p 2 q 6 ) 1 2 ( 144 p 2 q 6 ) 1 2

m 5 2 289 m 5 2 289

9 3 m 2 + 27 9 3 m 2 + 27

3 a b 2 − b a 3 a b 2 − b a

4 2 n 16 n 4 4 2 n 16 n 4

225 x 3 49 x 225 x 3 49 x

3 44 z + 99 z 3 44 z + 99 z

50 y 8 50 y 8

490 b c 2 490 b c 2

32 14 d 32 14 d

q 3 2 63 p q 3 2 63 p

8 1 − 3 x 8 1 − 3 x

20 121 d 4 20 121 d 4

w 3 2 32 − w 3 2 50 w 3 2 32 − w 3 2 50

108 x 4 + 27 x 4 108 x 4 + 27 x 4

12 x 2 + 2 3 12 x 2 + 2 3

147 k 3 147 k 3

125 n 10 125 n 10

42 q 36 q 3 42 q 36 q 3

81 m 361 m 2 81 m 361 m 2

72 c − 2 2 c 72 c − 2 2 c

144 324 d 2 144 324 d 2

24 x 6 3 + 81 x 6 3 24 x 6 3 + 81 x 6 3

162 x 6 16 x 4 4 162 x 6 16 x 4 4

64 y 3 64 y 3

128 z 3 3 − −16 z 3 3 128 z 3 3 − −16 z 3 3

1,024 c 10 5 1,024 c 10 5

Real-World Applications

A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So the length of the guy wire can be found by evaluating 90,000 + 160,000 . 90,000 + 160,000 . What is the length of the guy wire?

A car accelerates at a rate of 6 − 4 t m/s 2 6 − 4 t m/s 2 where t is the time in seconds after the car moves from rest. Simplify the expression.

8 − 16 4 − 2 − 2 1 2 8 − 16 4 − 2 − 2 1 2

4 3 2 − 16 3 2 8 1 3 4 3 2 − 16 3 2 8 1 3

m n 3 a 2 c −3 ⋅ a −7 n −2 m 2 c 4 m n 3 a 2 c −3 ⋅ a −7 n −2 m 2 c 4

a a − c a a − c

x 64 y + 4 y 128 y x 64 y + 4 y 128 y

( 250 x 2 100 b 3 ) ( 7 b 125 x ) ( 250 x 2 100 b 3 ) ( 7 b 125 x )

64 3 + 256 4 64 + 256 64 3 + 256 4 64 + 256

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Laws of Exponents and Radicals

Laws of Exponents (Index Law) 1. $x^n = x \cdot x \cdot x ... \, (n \text{ factors})$

2. $x^m \cdot x^n = x^{m + n}$

3. $(x^m)^n = x^{mn}$

4. $(xyz)^n = x^n \, y^n \, z^n$

5. $\dfrac{x^m}{x^n} = x^{m - n}$

6. $\left( \dfrac{x}{y} \right)^n = \dfrac{x^n}{y^n}$

7. $x^{-n} = \dfrac{1}{x^n}$   and   $\dfrac{1}{x^{-n}} = x^n$

8. $x^0 = 1$,   provided   $x \ne 0$.

9. $(x^m)^{1/n} = (x^{1/n})^m = x^{m/n}$

10. $x^{m/n} = \sqrt[n]{x^m}$

11. If   $x^m = x^n$,   then   $m = n$   provided   $x \ne 0$.  

Properties of Radicals 1. $\sqrt[n]{x} = x^{1/n}$

2. $\sqrt[n]{x^m} = \left( \sqrt[n]{x} \right)^m = x^{m/n}$

3. $\sqrt[n]{x} \, \sqrt[n]{y} = \sqrt[n]{xy}$

4. $\dfrac{\sqrt[n]{x}}{\sqrt[n]{y}} = \sqrt[n]{\dfrac{x}{y}}$

5. $\sqrt[n]{x} \, \sqrt[m]{x} = \sqrt[mn]{x^{m + n}}$

6. $\dfrac{\sqrt[n]{x}}{\sqrt[m]{x}} = \sqrt[mn]{x^{m - n}}$

7. $\left( \sqrt[n]{x} \right)^n = x$  

  • 01 - Solution to Radical Equations
  • 02 - Solution to Radical Equations
  • 03 - Solved Problems Involving Exponents and Radicals
  • 04 - Solution of Radical Equation
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  • Special Products and Factoring
  • Arithmetic, geometric, and harmonic progressions
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  • Verbal Problems in Algebra
  • Probability and Statistics

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Mathematics LibreTexts

5.4: Multiplying and Dividing Radical Expressions

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  • Page ID 6264

Learning Objectives

  • Multiply radical expressions.
  • Divide radical expressions.
  • Rationalize the denominator.

Multiplying Radical Expressions

When multiplying radical expressions with the same index, we use the product rule for radicals. Given real numbers \(\sqrt [ n ] { A }\) and \(\sqrt [ n ] { B }\),

\(\sqrt [ n ] { A } \cdot \sqrt [ n ] { B } = \sqrt [ n ] { A \cdot B }\)\

Example \(\PageIndex{1}\):

Multiply: \(\sqrt [ 3 ] { 12 } \cdot \sqrt [ 3 ] { 6 }\).

Apply the product rule for radicals, and then simplify.

\(\begin{aligned} \sqrt [ 3 ] { 12 } \cdot \sqrt [ 3 ] { 6 } & = \sqrt [ 3 ] { 12 \cdot 6 }\quad \color{Cerulean} { Multiply\: the\: radicands. } \\ & = \sqrt [ 3 ] { 72 } \quad\quad\:\color{Cerulean} { Simplify. } \\ & = \sqrt [ 3 ] { 2 ^ { 3 } \cdot 3 ^ { 2 } } \\ & = 2 \sqrt [ 3 ] { {3 } ^ { 2 }} \\ & = 2 \sqrt [ 3 ] { 9 } \end{aligned}\)

\(2 \sqrt [ 3 ] { 9 }\)

Often, there will be coefficients in front of the radicals.

Example \(\PageIndex{2}\):

Multiply: \(3 \sqrt { 6 } \cdot 5 \sqrt { 2 }\)

Using the product rule for radicals and the fact that multiplication is commutative, we can multiply the coefficients and the radicands as follows.

\(\begin{aligned} 3 \sqrt { 6 } \cdot 5 \sqrt { 2 } & = \color{Cerulean}{3 \cdot 5}\color{black}{ \cdot}\color{OliveGreen}{ \sqrt { 6 } \cdot \sqrt { 2} }\quad\color{Cerulean}{Multiplication\:is\:commutative.} \\ & = 15 \cdot \sqrt { 12 } \quad\quad\quad\:\color{Cerulean}{Multiply\:the\:coefficients\:and\:the\:radicands.} \\ & = 15 \sqrt { 4 \cdot 3 } \quad\quad\quad\:\color{Cerulean}{Simplify.} \\ & = 15 \cdot 2 \cdot \sqrt { 3 } \\ & = 30 \sqrt { 3 } \end{aligned}\)

Typically, the first step involving the application of the commutative property is not shown.

\(30 \sqrt { 3 }\)

Example \(\PageIndex{3}\):

Multiply: \(- 3 \sqrt [ 3 ] { 4 y ^ { 2 } } \cdot 5 \sqrt [ 3 ] { 16 y }\).

\(\begin{aligned} - 3 \sqrt [ 3 ] { 4 y ^ { 2 } } \cdot 5 \sqrt [ 3 ] { 16 y } & = - 15 \sqrt [ 3 ] { 64 y ^ { 3 } }\quad\color{Cerulean}{Multiply\:the\:coefficients\:and\:then\:multipy\:the\:rest.} \\ & = - 15 \sqrt [ 3 ] { 4 ^ { 3 } y ^ { 3 } }\quad\color{Cerulean}{Simplify.} \\ & = - 15 \cdot 4 y \\ & = - 60 y \end{aligned}\)

Use the distributive property when multiplying rational expressions with more than one term.

Example \(\PageIndex{4}\):

Multiply: \(5 \sqrt { 2 x } ( 3 \sqrt { x } - \sqrt { 2 x } )\).

Apply the distributive property and multiply each term by \(5 \sqrt { 2 x }\).

\(\begin{aligned} 5 \sqrt { 2 x } ( 3 \sqrt { x } - \sqrt { 2 x } ) & = \color{Cerulean}{5 \sqrt { 2 x } }\color{black}{\cdot} 3 \sqrt { x } - \color{Cerulean}{5 \sqrt { 2 x }}\color{black}{ \cdot} \sqrt { 2 x } \quad\color{Cerulean}{Distribute.}\\ & = 15 \sqrt { 2 x ^ { 2 } } - 5 \sqrt { 4 x ^ { 2 } } \quad\quad\quad\quad\:\:\:\color{Cerulean}{Simplify.} \\ & = 15 x \sqrt { 2 } - 5 \cdot 2 x \\ & = 15 x \sqrt { 2 } - 10 x \end{aligned}\)

\(15 x \sqrt { 2 } - 10 x\)

Example \(\PageIndex{5}\):

Multiply: \(\sqrt [ 3 ] { 6 x ^ { 2 } y } \left( \sqrt [ 3 ] { 9 x ^ { 2 } y ^ { 2 } } - 5 \cdot \sqrt [ 3 ] { 4 x y } \right)\).

Apply the distributive property, and then simplify the result.

\(\begin{aligned} \sqrt [ 3 ] { 6 x ^ { 2 } y } \left( \sqrt [ 3 ] { 9 x ^ { 2 } y ^ { 2 } } - 5 \cdot \sqrt [ 3 ] { 4 x y } \right) & = \color{Cerulean}{\sqrt [ 3 ] { 6 x ^ { 2 } y }}\color{black}{\cdot} \sqrt [ 3 ] { 9 x ^ { 2 } y ^ { 2 } } - \color{Cerulean}{\sqrt [ 3 ] { 6 x ^ { 2 } y }}\color{black}{ \cdot} 5 \sqrt [ 3 ] { 4 x y } \\ & = \sqrt [ 3 ] { 54 x ^ { 4 } y ^ { 3 } } - 5 \sqrt [ 3 ] { 24 x ^ { 3 } y ^ { 2 } } \\ & = \sqrt [ 3 ] { 27 \cdot 2 \cdot x \cdot x ^ { 3 } \cdot y ^ { 3 } } - 5 \sqrt [ 3 ] { 8 \cdot 3 \cdot x ^ { 3 } \cdot y ^ { 2 } } \\ & = 3 x y \sqrt [ 3 ] { 2 x } - 10 x \sqrt [ 3 ] { 3 y ^ { 2 } } \\ & = 3 x y \sqrt [ 3 ] { 2 x } - 10 x \sqrt [ 3 ] { 3 y ^ { 2 } } \end{aligned}\)

\(3 x y \sqrt [ 3 ] { 2 x } - 10 x \sqrt [ 3 ] { 3 y ^ { 2 } }\)

The process for multiplying radical expressions with multiple terms is the same process used when multiplying polynomials. Apply the distributive property, simplify each radical, and then combine like terms.

Example \(\PageIndex{6}\):

Multiply: \(( \sqrt { x } - 5 \sqrt { y } ) ^ { 2 }\).

\(( \sqrt { x } - 5 \sqrt { y } ) ^ { 2 } = ( \sqrt { x } - 5 \sqrt { y } ) ( \sqrt { x } - 5 \sqrt { y } )\)

Begin by applying the distributive property.

6323d469bea405941c7f41b02b259de6.png

\(\begin{array} { l } { = \color{Cerulean}{\sqrt { x }}\color{black}{ \cdot} \sqrt { x } + \color{Cerulean}{\sqrt { x }}\color{black}{ (} - 5 \sqrt { y } ) + ( \color{OliveGreen}{- 5 \sqrt { y }}\color{black}{ )} \sqrt { x } + ( \color{OliveGreen}{- 5 \sqrt { y }}\color{black}{ )} ( - 5 \sqrt { y } ) } \\ { = \sqrt { x ^ { 2 } } - 5 \sqrt { x y } - 5 \sqrt { x y } + 25 \sqrt { y ^ { 2 } } } \\ { = x - 10 \sqrt { x y } + 25 y } \end{array}\)

\(x - 10 \sqrt { x y } + 25 y\)

The binomials \((a + b)\) and \((a − b)\) are called conjugates 18 . When multiplying conjugate binomials the middle terms are opposites and their sum is zero.

Example \(\PageIndex{7}\):

Multiply: \(( \sqrt { 10 } + \sqrt { 3 } ) ( \sqrt { 10 } - \sqrt { 3 } )\).

Apply the distributive property, and then combine like terms.

\(\begin{aligned} ( \sqrt { 10 } + \sqrt { 3 } ) ( \sqrt { 10 } - \sqrt { 3 } ) & = \color{Cerulean}{\sqrt { 10} }\color{black}{ \cdot} \sqrt { 10 } + \color{Cerulean}{\sqrt { 10} }\color{black}{ (} - \sqrt { 3 } ) + \color{OliveGreen}{\sqrt{3}}\color{black}{ (}\sqrt{10}) + \color{OliveGreen}{\sqrt{3}}\color{black}{(}-\sqrt{3}) \\ & = \sqrt { 100 } - \sqrt { 30 } + \sqrt { 30 } - \sqrt { 9 } \\ & = 10 - \color{red}{\sqrt { 30 }}\color{black}{ +}\color{red}{ \sqrt { 30} }\color{black}{ -} 3 \\ & = 10 - 3 \\ & = 7 \\ \end{aligned}\)

It is important to note that when multiplying conjugate radical expressions, we obtain a rational expression. This is true in general

\(\begin{aligned} ( \sqrt { x } + \sqrt { y } ) ( \sqrt { x } - \sqrt { y } ) & = \sqrt { x ^ { 2 } } - \sqrt { x y } + \sqrt {x y } - \sqrt { y ^ { 2 } } \\ & = x - y \end{aligned}\)

Alternatively, using the formula for the difference of squares we have,

\(\begin{aligned} ( a + b ) ( a - b ) & = a ^ { 2 } - b ^ { 2 }\quad\quad\quad\color{Cerulean}{Difference\:of\:squares.} \\ ( \sqrt { x } + \sqrt { y } ) ( \sqrt { x } - \sqrt { y } ) & = ( \sqrt { x } ) ^ { 2 } - ( \sqrt { y } ) ^ { 2 } \\ & = x - y \end{aligned}\)

Exercise \(\PageIndex{1}\)

Multiply: \(( 3 - 2 \sqrt { y } ) ( 3 + 2 \sqrt { y } )\). (Assume \(y\) is positive.)

www.youtube.com/v/HPggvr8g68U

Dividing Radical Expressions

To divide radical expressions with the same index, we use the quotient rule for radicals. Given real numbers \(\sqrt [ n ] { A }\) and \(\sqrt [ n ] { B }\),

\(\frac { \sqrt [ n ] { A } } { \sqrt [ n ] { B } } = \sqrt [n]{ \frac { A } { B } }\)

Example \(\PageIndex{8}\):

Divide: \(\frac { \sqrt [ 3 ] { 96 } } { \sqrt [ 3 ] { 6 } }\).

In this case, we can see that \(6\) and \(96\) have common factors. If we apply the quotient rule for radicals and write it as a single cube root, we will be able to reduce the fractional radicand.

\(\begin{aligned} \frac { \sqrt [ 3 ] { 96 } } { \sqrt [ 3 ] { 6 } } & = \sqrt [ 3 ] { \frac { 96 } { 6 } } \quad\color{Cerulean}{Apply\:the\:quotient\:rule\:for\:radicals\:and\:reduce\:the\:radicand.}\\ & = \sqrt [ 3 ] { 16 } \\ & = \sqrt [ 3 ] { 8 \cdot 2 } \color{Cerulean}{Simplify.} \\ & = 2 \sqrt [ 3 ] { 2 } \end{aligned}\)

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

Example \(\PageIndex{9}\):

Divide: \(\frac { \sqrt { 50 x ^ { 6 } y ^ { 4} } } { \sqrt { 8 x ^ { 3 } y } }\).

Write as a single square root and cancel common factors before simplifying.

\(\begin{aligned} \frac { \sqrt { 50 x ^ { 6 } y ^ { 4 } } } { \sqrt { 8 x ^ { 3 } y } } & = \sqrt { \frac { 50 x ^ { 6 } y ^ { 4 } } { 8 x ^ { 3 } y } } \quad\color{Cerulean}{Apply\:the\:quotient\:rule\:for\:radicals\:and\:cancel.}\\ & = \sqrt { \frac { 25 x ^ { 3 } y ^ { 3 } } { 4 } } \quad\color{Cerulean}{Simplify.} \\ & = \frac { \sqrt { 25 x ^ { 3 } y ^ { 3 } } } { \sqrt { 4 } } \\ & = \frac { 5 x y \sqrt { x y } } { 2 } \end{aligned}\)

\(\frac { 5xy \sqrt {x y } } { 2 }\)

Rationalizing the Denominator

When the denominator (divisor) of a radical expression contains a radical, it is a common practice to find an equivalent expression where the denominator is a rational number. Finding such an equivalent expression is called rationalizing the denominator 19 .

\(\begin{array} { c } { \color{Cerulean} { Radical\:expression\quad Rational\: denominator } } \\ { \frac { 1 } { \sqrt { 2 } } \quad\quad\quad=\quad\quad\quad\quad \frac { \sqrt { 2 } } { 2 } } \end{array}\)

To do this, multiply the fraction by a special form of \(1\) so that the radicand in the denominator can be written with a power that matches the index. After doing this, simplify and eliminate the radical in the denominator. For example:

\(\frac { 1 } { \sqrt { 2 } } = \frac { 1 } { \sqrt { 2 } } \cdot \frac { \color{Cerulean}{\sqrt { 2} } } {\color{Cerulean}{ \sqrt { 2} } } \color{black}{=} \frac { \sqrt { 2 } } { \sqrt { 4 } } = \frac { \sqrt { 2 } } { 2 }\)

Remember, to obtain an equivalent expression, you must multiply the numerator and denominator by the exact same nonzero factor.

Example \(\PageIndex{10}\):

Rationalize the denominator: \(\frac { \sqrt { 2 } } { \sqrt { 5 x } }\).

The goal is to find an equivalent expression without a radical in the denominator. The radicand in the denominator determines the factors that you need to use to rationalize it. In this example, multiply by \(1\) in the form \(\frac { \sqrt { 5 x } } { \sqrt { 5 x } }\).

\(\begin{aligned} \frac { \sqrt { 2 } } { \sqrt { 5 x } } & = \frac { \sqrt { 2 } } { \sqrt { 5 x } } \cdot \color{Cerulean}{\frac { \sqrt { 5 x } } { \sqrt { 5 x } } { \:Multiply\:by\: } \frac { \sqrt { 5 x } } { \sqrt { 5 x } } .}\\ & = \frac { \sqrt { 10 x } } { \sqrt { 25 x ^ { 2 } } } \quad\quad\: \color{Cerulean} { Simplify. } \\ & = \frac { \sqrt { 10 x } } { 5 x } \end{aligned}\)

\(\frac { \sqrt { 10 x } } { 5 x }\)

Sometimes, we will find the need to reduce, or cancel, after rationalizing the denominator.

Example \(\PageIndex{11}\):

Rationalize the denominator: \(\frac { 3 a \sqrt { 2 } } { \sqrt { 6 a b } }\).

In this example, we will multiply by \(1\) in the form \(\frac { \sqrt { 6 a b } } { \sqrt { 6 a b } }\).

\(\begin{aligned} \frac { 3 a \sqrt { 2 } } { \sqrt { 6 a b } } & = \frac { 3 a \sqrt { 2 } } { \sqrt { 6 a b } } \cdot \color{Cerulean}{\frac { \sqrt { 6 a b } } { \sqrt { 6 a b } }} \\ & = \frac { 3 a \sqrt { 12 a b } } { \sqrt { 36 a ^ { 2 } b ^ { 2 } } } \quad\quad\color{Cerulean}{Simplify.}\\ & = \frac { 3 a \sqrt { 4 \cdot 3 a b} } { 6 ab } \\ & = \frac { 6 a \sqrt { 3 a b } } { b }\quad\quad\:\:\color{Cerulean}{Cancel.} \\ & = \frac { \sqrt { 3 a b } } { b } \end{aligned}\)

Notice that \(b\) does not cancel in this example. Do not cancel factors inside a radical with those that are outside.

\(\frac { \sqrt { 3 a b } } { b }\)

Exercise \(\PageIndex{2}\)

Rationalize the denominator: \(\sqrt { \frac { 9 x } { 2 y } }\).

\(\frac { 3 \sqrt { 2xy } } { 2 y }\)

www.youtube.com/v/h-lAW8kI2rA

Up to this point, we have seen that multiplying a numerator and a denominator by a square root with the exact same radicand results in a rational denominator. In general, this is true only when the denominator contains a square root. However, this is not the case for a cube root. For example,

\(\frac { 1 } { \sqrt [ 3 ] { x } } \cdot \color{Cerulean}{\frac { \sqrt [ 3 ] { x } } { \sqrt [ 3 ] { x } }}\color{black}{ =} \frac { \sqrt [ 3 ] { x } } { \sqrt [ 3 ] { x ^ { 2 } } }\)

Note that multiplying by the same factor in the denominator does not rationalize it. In this case, if we multiply by \(1\) in the form of \(\frac { \sqrt [ 3 ] { x ^ { 2 } } } { \sqrt [ 3 ] { x ^ { 2 } } }\), then we can write the radicand in the denominator as a power of \(3\). Simplifying the result then yields a rationalized denominator.

\(\frac { 1 } { \sqrt [ 3 ] { x } } = \frac { 1 } { \sqrt [ 3 ] { x } } \cdot \color{Cerulean}{\frac { \sqrt [ 3 ] { x ^ { 2 } } } { \sqrt [ 3 ] { x ^ { 2 } } }} = \frac { \sqrt [ 3 ] { x ^ { 2 } } } { \sqrt [ 3 ] { x ^ { 3 } } } = \frac { \sqrt [ 3 ] { x ^ { 2 } } } { x }\)

Therefore, to rationalize the denominator of a radical expression with one radical term in the denominator, begin by factoring the radicand of the denominator. The factors of this radicand and the index determine what we should multiply by. Multiply the numerator and denominator by the \(n\)th root of factors that produce nth powers of all the factors in the radicand of the denominator.

Example \(\PageIndex{12}\):

Rationalize the denominator: \(\frac { \sqrt [ 3 ] { 2 } } { \sqrt [ 3 ] { 25 } }\).

The radical in the denominator is equivalent to \(\sqrt [ 3 ] { 5 ^ { 2 } }\). To rationalize the denominator, we need: \(\sqrt [ 3 ] { 5 ^ { 3 } }\). To obtain this, we need one more factor of \(5\). Therefore, multiply by \(1\) in the form of \(\frac { \sqrt [3]{ 5 } } { \sqrt[3] { 5 } }\).

\(\begin{aligned} \frac { \sqrt [ 3 ] { 2 } } { \sqrt [ 3 ] { 25 } } & = \frac { \sqrt [ 3 ] { 2 } } { \sqrt [ 3 ] { 5 ^ { 2 } } } \cdot \color{Cerulean}{\frac { \sqrt [ 3 ] { 5 } } { \sqrt [ 3 ] { 5 } } \:Multiply\:by\:the\:cube\:root\:of\:factors\:that\:result\:in\:powers\:of\:3.} \\ & = \frac { \sqrt [ 3 ] { 10 } } { \sqrt [ 3 ] { 5 ^ { 3 } } } \quad\:\:\:\quad\color{Cerulean}{Simplify.} \\ & = \frac { \sqrt [ 3 ] { 10 } } { 5 } \end{aligned}\)

\(\frac { \sqrt [ 3 ] { 10 } } { 5 }\)

Example \(\PageIndex{13}\):

Rationalize the denominator: \(\sqrt [ 3 ] { \frac { 27 a } { 2 b ^ { 2 } } }\).

In this example, we will multiply by \(1\) in the form \(\frac { \sqrt [ 3 ] { 2 ^ { 2 } b } } { \sqrt [ 3 ] { 2 ^ { 2 } b } }\).

\(\begin{aligned} \sqrt [ 3 ] { \frac { 27 a } { 2 b ^ { 2 } } } & = \frac { \sqrt [ 3 ] { 3 ^ { 3 } a } } { \sqrt [ 3 ] { 2 b ^ { 2 } } } \quad\quad\quad\quad\color{Cerulean}{Apply\:the\:quotient\:rule\:for\:radicals.} \\ & = \frac { 3 \sqrt [ 3 ] { a } } { \sqrt [ 3 ] { 2 b ^ { 2 } } } \cdot \color{Cerulean}{\frac { \sqrt [ 3 ] { 2 ^ { 2 } b } } { \sqrt [ 3 ] { 2 ^ { 2 } b } }\:\:\:Multiply\:by\:the\:cube\:root\:of\:factors\:that\:result\:in\:powers.} \\ & = \frac { 3 \sqrt [ 3 ] { 2 ^ { 2 } ab } } { \sqrt [ 3 ] { 2 ^ { 3 } b ^ { 3 } } } \quad\quad\quad\color{Cerulean}{Simplify.}\\ & = \frac { 3 \sqrt [ 3 ] { 4 a b } } { 2 b } \end{aligned}\)

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

Example \(\PageIndex{14}\):

Rationalize the denominator: \(\frac { 2 x \sqrt [ 5 ] { 5 } } { \sqrt [ 5 ] { 4 x ^ { 3 } y } }\)

In this example, we will multiply by \(1\) in the form \(\frac { \sqrt [ 5 ] { 2 ^ { 3 } x ^ { 2 } y ^ { 4 } } } { \sqrt [ 5 ] { 2 ^ { 3 } x ^ { 2 } y ^ { 4 } } }\)

\(\begin{aligned} \frac{2x\sqrt[5]{5}}{\sqrt[5]{4x^{3}y}} & = \frac{2x\sqrt[5]{5}}{\sqrt[5]{2^{2}x^{3}y}}\cdot\color{Cerulean}{\frac{\sqrt[5]{2^{3}x^{2}y^{4}}}{\sqrt[5]{2^{3}x^{2}y^{4}}} \:\:Multiply\:by\:the\:fifth\:root\:of\:factors\:that\:result\:in\:pairs.} \\ & = \frac { 2 x \sqrt [ 5 ] { 5 \cdot 2 ^ { 3 } x ^ { 2 } y ^ { 4 } } } { \sqrt [ 5 ] { 2 ^ { 5 } x ^ { 5 } y ^ { 5 } } } \quad\quad\:\:\color{Cerulean}{Simplify.} \\ & = \frac { 2 x \sqrt [ 5 ] { 40 x ^ { 2 } y ^ { 4 } } } { 2 x y } \\ & = \frac { \sqrt [ 5 ] { 40 x ^ { 2 } y ^ { 4 } } } { y } \end{aligned}\)

\(\frac { \sqrt [ 5 ] { 40 x ^ { 2 } y ^ { 4 } } } { y }\)

When two terms involving square roots appear in the denominator, we can rationalize it using a very special technique. This technique involves multiplying the numerator and the denominator of the fraction by the conjugate of the denominator. Recall that multiplying a radical expression by its conjugate produces a rational number.

Example \(\PageIndex{15}\):

Rationalize the denominator: \(\frac { 1 } { \sqrt { 5 } - \sqrt { 3 } }\).

In this example, the conjugate of the denominator is \(\sqrt { 5 } + \sqrt { 3 }\). Therefore, multiply by \(1\) in the form \(\frac { ( \sqrt { 5 } + \sqrt { 3 } ) } { ( \sqrt {5 } + \sqrt { 3 } ) }\).

\(\begin{aligned} \frac { 1 } { \sqrt { 5 } - \sqrt { 3 } } & = \frac { 1 } { ( \sqrt { 5 } - \sqrt { 3 } ) } \color{Cerulean}{\frac { ( \sqrt { 5 } + \sqrt { 3 } ) } { ( \sqrt { 5 } + \sqrt { 3 } ) } \:\:Multiply \:numerator\:and\:denominator\:by\:the\:conjugate\:of\:the\:denominator.} \\ & = \frac { \sqrt { 5 } + \sqrt { 3 } } { \sqrt { 25 } + \sqrt { 15 } - \sqrt{15}-\sqrt{9} } \:\color{Cerulean}{Simplify.} \\ & = \frac { \sqrt { 5 } + \sqrt { 3 } } { 5-3 } \\ & = \frac { \sqrt { 5 } + \sqrt { 3 } } { 2 } \end{aligned}\)

\( \frac { \sqrt { 5 } + \sqrt { 3 } } { 2 } \)

Notice that the terms involving the square root in the denominator are eliminated by multiplying by the conjugate. We can use the property \(( \sqrt { a } + \sqrt { b } ) ( \sqrt { a } - \sqrt { b } ) = a - b\) to expedite the process of multiplying the expressions in the denominator.

Example \(\PageIndex{16}\):

Rationalize the denominator: \(\frac { \sqrt { 10 } } { \sqrt { 2 } + \sqrt { 6 } }\).

Multiply by \(1\) in the form \(\frac { \sqrt { 2 } - \sqrt { 6 } } { \sqrt { 2 } - \sqrt { 6 } }\).

\(\begin{aligned} \frac{\sqrt{10}}{\sqrt{2}+\sqrt{6} }&= \frac{(\sqrt{10})}{(\sqrt{2}+\sqrt{6})} \color{Cerulean}{\frac{(\sqrt{2}-\sqrt{6})}{(\sqrt{2}-\sqrt{6})}\quad\quad Multiple\:by\:the\:conjugate.} \\ &= \frac { \sqrt { 20 } - \sqrt { 60 } } { 2 - 6 } \quad\quad\quad\quad\quad\quad\:\:\:\color{Cerulean}{Simplify.} \\ &= \frac { \sqrt { 4 \cdot 5 } - \sqrt { 4 \cdot 15 } } { - 4 } \\ &= \frac { 2 \sqrt { 5 } - 2 \sqrt { 15 } } { - 4 } \\ &=\frac{2(\sqrt{5}-\sqrt{15})}{-4} \\ &= \frac { \sqrt { 5 } - \sqrt { 15 } } { - 2 } = - \frac { \sqrt { 5 } - \sqrt { 15 } } { 2 } = \frac { - \sqrt { 5 } + \sqrt { 15 } } { 2 } \end{aligned}\)

\(\frac { \sqrt { 15 } - \sqrt { 5 } } { 2 }\)

Example \(\PageIndex{17}\):

Rationalize the denominator: \(\frac { \sqrt { x } - \sqrt { y } } { \sqrt { x } + \sqrt { y } }\).

In this example, we will multiply by \(1\) in the form \(\frac { \sqrt { x } - \sqrt { y } } { \sqrt { x } - \sqrt { y } }\).

\(\begin{aligned} \frac { \sqrt { x } - \sqrt { y } } { \sqrt { x } + \sqrt { y } } & = \frac { ( \sqrt { x } - \sqrt { y } ) } { ( \sqrt { x } + \sqrt { y } ) } \color{Cerulean}{\frac { ( \sqrt { x } - \sqrt { y } ) } { ( \sqrt { x } - \sqrt { y } ) } \quad \quad Multiply\:by\:the\:conjugate\:of\:the\:denominator.} \\ & = \frac { \sqrt { x ^ { 2 } } - \sqrt { x y } - \sqrt { x y } + \sqrt { y ^ { 2 } } } { x - y } \:\:\color{Cerulean}{Simplify.} \\ & = \frac { x - 2 \sqrt { x y } + y } { x - y } \end{aligned}\)

\(\frac { x - 2 \sqrt { x y } + y } { x - y }\)

Exercise \(\PageIndex{3}\)

Rationalize the denominator: \(\frac { 2 \sqrt { 3 } } { 5 - \sqrt { 3 } }\)

\(\frac { 5 \sqrt { 3 } + 3 } { 11 }\)

www.youtube.com/v/gYNvQ3NVxCc

Key Takeaways

  • To multiply two single-term radical expressions, multiply the coefficients and multiply the radicands. If possible, simplify the result.
  • Apply the distributive property when multiplying a radical expression with multiple terms. Then simplify and combine all like radicals.
  • Multiplying a two-term radical expression involving square roots by its conjugate results in a rational expression.
  • It is common practice to write radical expressions without radicals in the denominator. The process of finding such an equivalent expression is called rationalizing the denominator.
  • If an expression has one term in the denominator involving a radical, then rationalize it by multiplying the numerator and denominator by the \(n\)th root of factors of the radicand so that their powers equal the index.
  • If a radical expression has two terms in the denominator involving square roots, then rationalize it by multiplying the numerator and denominator by the conjugate of the denominator.

Exercise \(\PageIndex{4}\)

Multiply. (Assume all variables represent non-negative real numbers.)

  • \(\sqrt { 3 } \cdot \sqrt { 7 }\)
  • \(\sqrt { 2 } \cdot \sqrt { 5 }\)
  • \(\sqrt { 6 } \cdot \sqrt { 12 }\)
  • \(\sqrt { 10 } \cdot \sqrt { 15 }\)
  • \(\sqrt { 2 } \cdot \sqrt { 6 }\)
  • \(\sqrt { 5 } \cdot \sqrt { 15 }\)
  • \(\sqrt { 7 } \cdot \sqrt { 7 }\)
  • \(\sqrt { 12 } \cdot \sqrt { 12 }\)
  • \(2 \sqrt { 5 } \cdot 7 \sqrt { 10 }\)
  • \(3 \sqrt { 15 } \cdot 2 \sqrt { 6 }\)
  • \(( 2 \sqrt { 5 } ) ^ { 2 }\)
  • \(( 6 \sqrt { 2 } ) ^ { 2 }\)
  • \(\sqrt { 2 x } \cdot \sqrt { 2 x }\)
  • \(\sqrt { 5 y } \cdot \sqrt { 5 y }\)
  • \(\sqrt { 3 a } \cdot \sqrt { 12 }\)
  • \(\sqrt { 3 a } \cdot \sqrt { 2 a }\)
  • \(4 \sqrt { 2 x } \cdot 3 \sqrt { 6 x }\)
  • \(5 \sqrt { 10 y } \cdot 2 \sqrt { 2 y }\)
  • \(\sqrt [ 3 ] { 3 } \cdot \sqrt [ 3 ] { 9 }\)
  • \(\sqrt [ 3 ] { 4 } \cdot \sqrt [ 3 ] { 16 }\)
  • \(\sqrt [ 3 ] { 15 } \cdot \sqrt [ 3 ] { 25 }\)
  • \(\sqrt [ 3 ] { 100 } \cdot \sqrt [ 3 ] { 50 }\)
  • \(\sqrt [ 3 ] { 4 } \cdot \sqrt [ 3 ] { 10 }\)
  • \(\sqrt [ 3 ] { 18 } \cdot \sqrt [ 3 ] { 6 }\)
  • \(( 5 \sqrt [ 3 ] { 9 } ) ( 2 \sqrt [ 3 ] { 6 } )\)
  • \(( 2 \sqrt [ 3 ] { 4 } ) ( 3 \sqrt [ 3 ] { 4 } )\)
  • \(( 2 \sqrt [ 3 ] { 2 } ) ^ { 3 }\)
  • \(( 3 \sqrt [ 3 ] { 4 } ) ^ { 3 }\)
  • \(\sqrt [ 3 ] { 3 a ^ { 2 } } \cdot \sqrt [ 3 ] { 9 a }\)
  • \(\sqrt [ 3 ] { 7 b } \cdot \sqrt [ 3 ] { 49 b ^ { 2 } }\)
  • \(\sqrt [ 3 ] { 6 x ^ { 2 } } \cdot \sqrt [ 3 ] { 4 x ^ { 2 } }\)
  • \(\sqrt [ 3 ] { 12 y } \cdot \sqrt [ 3 ] { 9 y ^ { 2 } }\)
  • \(\sqrt [ 3 ] { 20 x ^ { 2 } y } \cdot \sqrt [ 3 ] { 10 x ^ { 2 } y ^ { 2 } }\)
  • \(\sqrt [ 3 ] { 63 x y } \cdot \sqrt [ 3 ] { 12 x ^ { 4 } y ^ { 2 } }\)
  • \(\sqrt { 5 } ( 3 - \sqrt { 5 } )\)
  • \(\sqrt { 2 } ( \sqrt { 3 } - \sqrt { 2 } )\)
  • \(3 \sqrt { 7 } ( 2 \sqrt { 7 } - \sqrt { 3 } )\)
  • \(2 \sqrt { 5 } ( 6 - 3 \sqrt { 10 } )\)
  • \(\sqrt { 6 } ( \sqrt { 3 } - \sqrt { 2 } )\)
  • \(\sqrt { 15 } ( \sqrt { 5 } + \sqrt { 3 } )\)
  • \(\sqrt { x } ( \sqrt { x } + \sqrt { x y } )\)
  • \(\sqrt { y } ( \sqrt { x y } + \sqrt { y } )\)
  • \(\sqrt { 2 a b } ( \sqrt { 14 a } - 2 \sqrt { 10 b } )\)
  • \(\sqrt { 6 a b } ( 5 \sqrt { 2 a } - \sqrt { 3 b } )\)
  • \(\sqrt [ 3 ] { 6 } ( \sqrt [ 3 ] { 9 } - \sqrt [ 3 ] { 20 } )\)
  • \(\sqrt [ 3 ] { 12 } ( \sqrt [ 3 ] { 36 } + \sqrt [ 3 ] { 14 } )\)
  • \(( \sqrt { 2 } - \sqrt { 5 } ) ( \sqrt { 3 } + \sqrt { 7 } )\)
  • \(( \sqrt { 3 } + \sqrt { 2 } ) ( \sqrt { 5 } - \sqrt { 7 } )\)
  • \(( 2 \sqrt { 3 } - 4 ) ( 3 \sqrt { 6 } + 1 )\)
  • \(( 5 - 2 \sqrt { 6 } ) ( 7 - 2 \sqrt { 3 } )\)
  • \(( \sqrt { 5 } - \sqrt { 3 } ) ^ { 2 }\)
  • \(( \sqrt { 7 } - \sqrt { 2 } ) ^ { 2 }\)
  • \(( 2 \sqrt { 3 } + \sqrt { 2 } ) ( 2 \sqrt { 3 } - \sqrt { 2 } )\)
  • \(( \sqrt { 2 } + 3 \sqrt { 7 } ) ( \sqrt { 2 } - 3 \sqrt { 7 } )\)
  • \(( \sqrt { a } - \sqrt { 2 b } ) ^ { 2 }\)
  • \(( \sqrt { a b } + 1 ) ^ { 2 }\)
  • What is the perimeter and area of a rectangle with length measuring \(5\sqrt{3}\) centimeters and width measuring \(3\sqrt{2}\) centimeters?
  • What is the perimeter and area of a rectangle with length measuring \(2\sqrt{6}\) centimeters and width measuring \(\sqrt{3}\) centimeters?
  • If the base of a triangle measures \(6\sqrt{2}\) meters and the height measures \(3\sqrt{2}\) meters, then calculate the area.
  • If the base of a triangle measures \(6\sqrt{3}\) meters and the height measures \(3\sqrt{6}\) meters, then calculate the area.

1. \(\sqrt{21}\)

3. \(6\sqrt{2}\)

5. \(2\sqrt{3}\)

9. \(70\sqrt{2}\)

15. \(6\sqrt{a}\)

17. \(24x\sqrt{3}\)

21. \(5 \sqrt [ 3 ] { 3 }\)

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

25. \(30 \sqrt [ 3 ] { 2 }\)

31. \(2 x \sqrt [ 3 ] { 3 x }\)

33. \(2 x y \sqrt [ 3 ] { 25 x }\)

35. \(3\sqrt{5}-5\)

37. \(42 - 3 \sqrt { 21 }\)

39. \(3 \sqrt { 2 } - 2 \sqrt { 3 }\)

41. \(x + x \sqrt { y }\)

43. \(2 a \sqrt { 7 b } - 4 b \sqrt { 5 a }\)

45. \(3 \sqrt [ 3 ] { 2 } - 2 \sqrt [ 3 ] { 15 }\)

47. \(\sqrt { 6 } + \sqrt { 14 } - \sqrt { 15 } - \sqrt { 35 }\)

49. \(18 \sqrt { 2 } + 2 \sqrt { 3 } - 12 \sqrt { 6 } - 4\)

51. \(8 - 2 \sqrt { 15 }\)

55. \(a - 2 \sqrt { 2 a b } + 2 b\)

57. Perimeter: \(( 10 \sqrt { 3 } + 6 \sqrt { 2 } )\) centimeters; area \(15\sqrt{6}\) square centimeters

59. \(18\) square meters

Exercise \(\PageIndex{5}\)

Divide. (Assume all variables represent positive real numbers.)

  • \(\frac { \sqrt { 75 } } { \sqrt { 3 } }\)
  • \(\frac { \sqrt { 360 } } { \sqrt { 10 } }\)
  • \(\frac { \sqrt { 72 } } { \sqrt { 75 } }\)
  • \(\frac { \sqrt { 90 } } { \sqrt { 98 } }\)
  • \(\frac { \sqrt { 90 x ^ { 5 } } } { \sqrt { 2 x } }\)
  • \(\frac { \sqrt { 96 y ^ { 3 } } } { \sqrt { 3 y } }\)
  • \(\frac { \sqrt { 162 x ^ { 7 } y ^ { 5 } } } { \sqrt { 2 x y } }\)
  • \(\frac { \sqrt { 363 x ^ { 4 } y ^ { 9 } } } { \sqrt { 3 x y } }\)
  • \(\frac { \sqrt [ 3 ] { 16 a ^ { 5 } b ^ { 2 } } } { \sqrt [ 3 ] { 2 a ^ { 2 } b ^ { 2 } } }\)
  • \(\frac { \sqrt [ 3 ] { 192 a ^ { 2 } b ^ { 7 } } } { \sqrt [ 3 ] { 2 a ^ { 2 } b ^ { 2 } } }\)

3. \(\frac { 2 \sqrt { 6 } } { 5 }\)

5. \(3 x ^ { 2 } \sqrt { 5 }\)

7. \(9 x ^ { 3 } y ^ { 2 }\)

Exercise \(\PageIndex{6}\)

Rationalize the denominator. (Assume all variables represent positive real numbers.)

  • \(\frac { 1 } { \sqrt { 5 } }\)
  • \(\frac { 1 } { \sqrt { 6 } }\)
  • \(\frac { \sqrt { 2 } } { \sqrt { 3 } }\)
  • \(\frac { \sqrt { 3 } } { \sqrt { 7 } }\)
  • \(\frac { 5 } { 2 \sqrt { 10 } }\)
  • \(\frac { 3 } { 5 \sqrt { 6 } }\)
  • \(\frac { \sqrt { 3 } - \sqrt { 5 } } { \sqrt { 3 } }\)
  • \(\frac { \sqrt { 6 } - \sqrt { 2 } } { \sqrt { 2 } }\)
  • \(\frac { 1 } { \sqrt { 7 x } }\)
  • \(\frac { 1 } { \sqrt { 3 y } }\)
  • \(\frac { a } { 5 \sqrt { a b } }\)
  • \(\frac { 3 b ^ { 2 } } { 2 \sqrt { 3 a b } }\)
  • \(\frac { 2 } { \sqrt [ 3 ] { 36 } }\)
  • \(\frac { 14 } { \sqrt [ 3 ] { 7 } }\)
  • \(\frac { 1 } { \sqrt [ 3 ] { 4 x } }\)
  • \(\frac { 1 } { \sqrt [ 3 ] { 3 y ^ { 2 } } }\)
  • \(\frac { 9 x \sqrt[3] { 2 } } { \sqrt [ 3 ] { 9 x y ^ { 2 } } }\)
  • \(\frac { 5 y ^ { 2 } \sqrt [ 3 ] { x } } { \sqrt [ 3 ] { 5 x ^ { 2 } y } }\)
  • \(\frac { 3 a } { 2 \sqrt [ 3 ] { 3 a ^ { 2 } b ^ { 2 } } }\)
  • \(\frac { 25 n } { 3 \sqrt [ 3 ] { 25 m ^ { 2 } n } }\)
  • \(\frac { 3 } { \sqrt [ 5 ] { 27 x ^ { 2 } y } }\)
  • \(\frac { 2 } { \sqrt [ 5 ] { 16 x y ^ { 2 } } }\)
  • \(\frac { a b } { \sqrt [ 5 ] { 9 a ^ { 3 } b } }\)
  • \(\frac { a b c } { \sqrt [ 5 ] { a b ^ { 2 } c ^ { 3 } } }\)
  • \(\sqrt [ 5 ] { \frac { 3 x } { 8 y ^ { 2 } z } }\)
  • \(\sqrt [ 5 ] { \frac { 4 x y ^ { 2 } } { 9 x ^ { 3 } y z ^ { 4 } } }\)
  • \(\frac { 3 } { \sqrt { 10 } - 3 }\)
  • \(\frac { 2 } { \sqrt { 6 } - 2 }\)
  • \(\frac { 1 } { \sqrt { 5 } + \sqrt { 3 } }\)
  • \(\frac { 1 } { \sqrt { 7 } - \sqrt { 2 } }\)
  • \(\frac { \sqrt { 3 } } { \sqrt { 3 } + \sqrt { 6 } }\)
  • \(\frac { \sqrt { 5 } } { \sqrt { 5 } + \sqrt { 15 } }\)
  • \(\frac { 10 } { 5 - 3 \sqrt { 5 } }\)
  • \(\frac { - 2 \sqrt { 2 } } { 4 - 3 \sqrt { 2 } }\)
  • \(\frac { \sqrt { 3 } + \sqrt { 5 } } { \sqrt { 3 } - \sqrt { 5 } }\)
  • \(\frac { \sqrt { 10 } - \sqrt { 2 } } { \sqrt { 10 } + \sqrt { 2 } }\)
  • \(\frac { 2 \sqrt { 3 } - 3 \sqrt { 2 } } { 4 \sqrt { 3 } + \sqrt { 2 } }\)
  • \(\frac { 6 \sqrt { 5 } + 2 } { 2 \sqrt { 5 } - \sqrt { 2 } }\)
  • \(\frac { x - y } { \sqrt { x } + \sqrt { y } }\)
  • \(\frac { x - y } { \sqrt { x } - \sqrt { y } }\)
  • \(\frac { x + \sqrt { y } } { x - \sqrt { y } }\)
  • \(\frac { x - \sqrt { y } } { x + \sqrt { y } }\)
  • \(\frac { \sqrt { a } - \sqrt { b } } { \sqrt { a } + \sqrt { b } }\)
  • \(\frac { \sqrt { a b } + \sqrt { 2 } } { \sqrt { a b } - \sqrt { 2 } }\)
  • \(\frac { \sqrt { x } } { 5 - 2 \sqrt { x } }\)
  • \(\frac { 1 } { \sqrt { x } - y }\)
  • \(\frac { \sqrt { x } + \sqrt { 2 y } } { \sqrt { 2 x } - \sqrt { y } }\)
  • \(\frac { \sqrt { 3 x } - \sqrt { y } } { \sqrt { x } + \sqrt { 3 y } }\)
  • \(\frac { \sqrt { 2 x + 1 } } { \sqrt { 2 x + 1 } - 1 }\)
  • \(\frac { \sqrt { x + 1 } } { 1 - \sqrt { x + 1 } }\)
  • \(\frac { \sqrt { x + 1 } + \sqrt { x - 1 } } { \sqrt { x + 1 } - \sqrt { x - 1 } }\)
  • \(\frac { \sqrt { 2 x + 3 } - \sqrt { 2 x - 3 } } { \sqrt { 2 x + 3 } + \sqrt { 2 x - 3 } }\)
  • The radius of the base of a right circular cone is given by \(r = \sqrt { \frac { 3 V } { \pi h } }\) where \(V\) represents the volume of the cone and \(h\) represents its height. Find the radius of a right circular cone with volume \(50\) cubic centimeters and height \(4\) centimeters. Give the exact answer and the approximate answer rounded to the nearest hundredth.
  • The radius of a sphere is given by \(r = \sqrt [ 3 ] { \frac { 3 V } { 4 \pi } }\) where \(V\) represents the volume of the sphere. Find the radius of a sphere with volume \(135\) square centimeters. Give the exact answer and the approximate answer rounded to the nearest hundredth.

1. \(\frac { \sqrt { 5 } } { 5 }\)

3. \(\frac { \sqrt { 6 } } { 3 }\)

5. \(\frac { \sqrt { 10 } } { 4 }\)

7. \(\frac { 3 - \sqrt { 15 } } { 3 }\)

9. \(\frac { \sqrt { 7 x } } { 7 x }\)

11. \(\frac { \sqrt { a b } } { 5 b }\)

13. \(\frac { \sqrt [ 3 ] { 6 } } { 3 }\)

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

17. \(\frac { 3 \sqrt [ 3 ] { 6 x ^ { 2 } y } } { y }\)

19. \(\frac { \sqrt [ 3 ] { 9 a b } } { 2 b }\)

21. \(\frac { \sqrt [ 5 ] { 9 x ^ { 3 } y ^ { 4 } } } { x y }\)

23. \(\frac { \sqrt [ 5 ] { 27 a ^ { 2 } b ^ { 4 } } } { 3 }\)

25. \(\frac { \sqrt [ 5 ] { 12 x y ^ { 3 } z ^ { 4 } } } { 2 y z }\)

27. \(3\sqrt { 10 } + 9\)

29. \(\frac { \sqrt { 5 } - \sqrt { 3 } } { 2 }\)

31. \(- 1 + \sqrt { 2 }\

33. \(\frac { - 5 - 3 \sqrt { 5 } } { 2 }\)

35. \(- 4 - \sqrt { 15 }\)

37. \(\frac { 15 - 7 \sqrt { 6 } } { 23 }\)

39. \(\sqrt { x } - \sqrt { y }\)

41. \(\frac { x ^ { 2 } + 2 x \sqrt { y } + y } { x ^ { 2 } - y }\)

43. \(\frac { a - 2 \sqrt { a b + b } } { a - b }\)

45. \(\frac { 5 \sqrt { x } + 2 x } { 25 - 4 x }\)

47. \(\frac { x \sqrt { 2 } + 3 \sqrt { x y } + y \sqrt { 2 } } { 2 x - y }\)

49. \(\frac { 2 x + 1 + \sqrt { 2 x + 1 } } { 2 x }\)

51. \(x + \sqrt { x ^ { 2 } - 1 }\)

53. \(\frac { 5 \sqrt { 6 \pi } } { 2 \pi }\) centimeters; \(3.45\) centimeters

Exercise \(\PageIndex{7}\)

  • Research and discuss some of the reasons why it is a common practice to rationalize the denominator.
  • Explain in your own words how to rationalize the denominator.

1. Answer may vary

18 The factors \((a+b)\) and \((a-b)\) are conjugates.

19 The process of determining an equivalent radical expression with a rational denominator.

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Course: Algebra 2   >   Unit 9

  • Graphing square and cube root functions

Radical functions & their graphs

  • Graphs of square and cube root functions

Introduction

In this article, we will practice a couple of problems where we should match the appropriate graph to a given radical function., practice question 1: square-root function.

A square root function graph on an x y coordinate plane. It has an endpoint (zero, zero) and passes through (four, two).
  • (Choice A)   A
  • (Choice B)   B
  • (Choice C)   C
  • (Choice D)   D

Practice question 2: Cube-root function

A cube root function graph on an x y coordinate plane. Its middle point is at (zero, zero). It passes through (negative eight, negative two) and (eight, two).

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Common Monomial Factor

32.2k plays, 7th -  8th  , 48.5k plays, comparing numbers, trigonometric ratios of special angles.

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Laws of Radicals

Mathematics.

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  36 a 2 \sqrt{36a^2} 3 6 a 2 ​    

  6 a 2 6a^2 6 a 2    

  9 5 9\sqrt{5} 9 5 ​    

  3 5 3\sqrt{5} 3 5 ​    

  2 5 2\sqrt{5} 2 5 ​    

  3 15 3\sqrt{15} 3 1 5 ​    

  S i m p l i f y   20 a 2 b 4 Simplify\ \sqrt{20a^2b^4} S i m p l i f y   2 0 a 2 b 4 ​    

  2 a b 2 5 2ab^2\sqrt{5} 2 a b 2 5 ​    

  4 a b 2 5 4ab^2\sqrt{5} 4 a b 2 5 ​    

  5 a b 2 5 5ab^2\sqrt{5} 5 a b 2 5 ​    

  2 a 2 b 2 5 2a^2b^2\sqrt{5} 2 a 2 b 2 5 ​    

  49 81 \sqrt{\frac{49}{81}} 8 1 4 9 ​ ​    

  7 9 \frac{7}{9} 9 7 ​    

  7 6 \frac{7}{6} 6 7 ​    

  6 7 \frac{6}{7} 7 6 ​    

  9 7 \frac{9}{7} 7 9 ​    

  W h i c h   o f   t h e   f o l l o w i n g   i s   e q u a l   t o   72 x 2 ? Which\ of\ the\ following\ is\ equal\ to\ \sqrt{72x^2}? W h i c h   o f   t h e   f o l l o w i n g   i s   e q u a l   t o   7 2 x 2 ​ ?    

  x 72 x\sqrt{72} x 7 2 ​    

  6 x 3 6x\sqrt{3} 6 x 3 ​    

  6 x 2 6x\sqrt{2} 6 x 2 ​    

  3 x 8 3x\sqrt{8} 3 x 8 ​    

3 \sqrt{3} 3 ​

  W h i c h   i s   t h e   s a m e   a s   5 y ? Which\ is\ the\ same\ as\ \sqrt{5y}? W h i c h   i s   t h e   s a m e   a s   5 y ​ ?    

  5    +   y \sqrt{5\ \ }+\ \sqrt{y} 5     ​ +   y ​    

  5 − y \sqrt{5}-\sqrt{y} 5 ​ − y ​    

  5 y \frac{\sqrt{5}}{\sqrt{y}} y ​ 5 ​ ​    

  5   ⋅   y \sqrt{5}\ \cdot\ \sqrt{y} 5 ​   ⋅   y ​    

  W h i c h   i s   e q u i v a l e n t   t o   5 m ? Which\ is\ equivalent\ to\ \sqrt{\frac{5}{m}}? W h i c h   i s   e q u i v a l e n t   t o   m 5 ​ ​ ?    

  5   + m \sqrt{5}\ +\sqrt{m} 5 ​   + m ​    

  5   −   m \sqrt{5\ }-\ \sqrt{m} 5   ​ −   m ​    

  5   ⋅   m \sqrt{5\ }\cdot\ \sqrt{m} 5   ​ ⋅   m ​    

  5 m \frac{\sqrt{5}}{\sqrt{m}} m ​ 5 ​ ​    

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