Section 1.4 b

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Transcript Section 1.4 b

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Hawkes Learning Systems: College Algebra

Section 1.4b: Properties of Radicals

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Objectives o Combining radical expressions.

o Rational number exponents.

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Combining Radical Expressions Often, a sum of two or more radical expressions can be combined into one. This can be done if the radical expressions are

like radicals

, meaning that they have the same index and the same radicand. It is frequently necessary to simplify the radical expressions before it can be determined if they are like or not.

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Like and Unlike Radicals

Like

The expression below can be combined because both the index and radicand are the same, therefore making the radical expressions like radicals.

2  6 3

x

3   (2  4 )

x

4 3 3

x x

Unlike

The expression below cannot be combined because the radicand is not the same, therefore making this radical expression an unlike radical.

2 4

x

 4 5

x

Likewise, the below expression also cannot be combined because the index is not the same, therefore making this radical expression an unlike radical.

x

 4 4

x

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Example 1: Combining Radical Expressions Combine the radical expressions, if possible.

a.

 5 12

x

5  27

x

  5 3 2   10

x

2 2

x x

3

x

 3 3

x

 10

x

2  3  3

x

2  

x

We begin by simplifying the radicals. Note: the two radicals have the same index and the same radicand and can be combined.

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Example 1: Combining Radical Expressions (cont.) Combine the radical expressions, if possible.

b.

3 432

x

3  32

x

2  3 3     6

x

3 2  4

x x

3  2 4 2

x

2 We begin by simplifying the radicals.

Note: the two radicals have the same radicand but not the same index. Therefore, they cannot be combined.

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Example 2: Combining Radical Expressions Combine the radical expression, if possible.

1 20  16 45     2 1 2 1  2 5 5       4 3 5 3 5 2        4 3 5      2 5 2 5     3 5  8 5   5 5   30 5 6 Here, we rationalize the denominator.

Note: the two radicals have the same index and the same radicand and can be combined.

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Rational Number Exponents

1

a

real number, then

n a

1

n

is a natural number and if is a 

n a

.

a

  1 5  5

x

and   1 3  3

x

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Rational Number Exponents

m a

m

and

a n m

n n a m

are natural numbers , then   

m

.

n a m

 

m

they are equal.

m a n

, Note:

a

m n

is defined to be 1

m

.

a n

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Example 3: Rational Number Exponents Simplify the following expression, writing your answer using the same notation as the original expression.

a.

32  3 5   32 5 1    3 2  3 To simplify the expression, we begin by noting that 32  3 5   32 1    3 , then simplify the expression.

 1 2 3  1 8

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Example 3: Rational Number Exponents (cont.) Simplify the following expression, writing your answer using the same notation as the original expression.

b.

10 16

x

2  10 2 4

x

2 4 2  2 10

x

10 2  2 5

x

5 1  5 4

x

2 4  16 Simplify.

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Example 3: Rational Number Exponents (cont.) Simplify the following expression, writing your answer using the same notation as the original expression.

c.

 7

x

2  6 9   7

x

2  6   4 5 

7

x

2  6

9 5 4     

7

x

2   7

x

2  6

5    6  Recall, (

a n

)(

a m

) 

a

.

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Example 4: Rational Number Exponents Simplify each of the following expressions.

a.

7 3

x

4  (7)(3)

x

4  21

x

4 Since

m n a

mn a

Because 4 and 21 have no common factors, the radical is in simplest form.

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Example 4: Rational Number Exponents (cont.) Simplify each of the following expressions.

b.

 7

x

7

x

  2

y

2

y

  1 5   7

x

 2

y

 1   5    7

x

 2

y

 6 5

c.

6

x

3   

 

1 6 1

x

2

x

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Example 5: Rational Number Exponents Write the expression as a single radical.

5    2  4 3 1 1 2 5  3 4 4 5 2 20  3 20 2 4 1 20  3 5 1 20 First make the exponents equal. We do so by finding the least common denominator of and and writing 5 4 both fractions with this common denominator.

 

m

 16 1  243 1 20 

3888

20 1  20 3888

n

 

n

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Heron’s Formula Copyright © 2010 Hawkes Learning Systems. All rights reserved.

Heron’s formula applied to an equilateral triangle of length

d

is

A

  3

d

2    3

d

2 

d

 3 . Expressing this in terms of

d

, each triangle has area

A

 3

d d

3 4 or . 16

d

Simplifying this radical, we obtain

A

d

2 4 3 .

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Example 6: Rational Number Exponents

Find the area of a regular hexagon (pictured on the right).

Since the hexagon is made up of six of these triangles, the total area

A

of the 2 hexagon is . 2

d