Section 3Chapter 8. 1 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Objectives 2 6 5 3 4...

Section 3Chapter 8

1

Copyright © 2012, 2008, 2004 Pearson Education, Inc.

Objectives

2

6

5

3

4

Simplifying Radical Expressions

Use the product rule for radicals.

Use the quotient rule for radicals.

Simplify radicals.

Simplify products and quotients of radicals with different indexes.

Use the Pythagorean theorem.

Use the distance formula.

8.3



Objective 1

Slide 8.3- 3


Product Rule for Radicals

If and are real numbers and n is a natural number, then

That is, the product of two nth roots is the nth root of the product.

n a n b

.n n na b ab

Slide 8.3- 4


Use the product rule only when the radicals have the same index.


Multiply. Assume that all variables represent positive real numbers.

5 13 5 13

7 xy 7xy

65

Slide 8.3- 5

CLASSROOM EXAMPLE 1

Using the Product Rule

Solution:


Multiply. Assume that all variables represent positive real numbers.

3 32 7 3 2 7

66 2 38 2r r 6 516r

2 25 59 8y x xy 4 25 72y x

37 5

3 14

This expression cannot be simplified by using the product rule.

Slide 8.3- 6

CLASSROOM EXAMPLE 2

Using the Product Rule

Solution:



Objective 2

Slide 8.3- 7


Quotient Rule for Radicals

If and are real numbers, b ≠ 0, and n is a natural number,

then

That is, the nth root of a quotient is the quotient of the nth roots.

n a n b

.n

nn

a a

b b

Slide 8.3- 8



Simplify. Assume that all variables represent positive real numbers.

100

81

10

9

11

2511

5

8

16

y3

18

125

3

3

18

125

2

312

x

r

3 2

3 12

x

r

3 18

5

8

16

y

4

4

y

3 2

4

x

r

Slide 8.3- 9

CLASSROOM EXAMPLE 3

Using the Quotient Rule

Solution:


Simplify radicals.

Objective 3

Slide 8.3- 10


Simplify.

32 216 216 4 2

300 3100 3100 10 3

35 Cannot be simplified further.

Slide 8.3- 11

CLASSROOM EXAMPLE 4

Simplifying Roots of Numbers

Solution:

3 54 3 227 33 27 2 33 2

4 243 44 3 3 43 3


Conditions for a Simplified Radical

1. The radicand has no factor raised to a power greater than or equal to the index.

2. The radicand has no fractions.

3. No denominator contains a radical.

4. Exponents in the radicand and the index of the radical have greatest common factor 1.

Slide 8.3- 12

Simplify radicals.

Be careful with which factors belong outside the radical sign and which belong inside.



725p 3 225 ( )p p 35p p

372y x 2236 y y x 6 2y yx

7 5 63 27y x z3 6 3 2 63 3 y y x x z 2 2 233y xz yx

5 74 32a b4 4 4 34 2 2 a a b b 342 2ab ab

Slide 8.3- 13

CLASSROOM EXAMPLE 5

Simplifying Radicals Involving Variables

Solution:



12 32 1/1232 1/ 42

6 2t 1/ 62t 3 t

4 2

1/3t

Slide 8.3- 14

CLASSROOM EXAMPLE 6

Simplifying Radicals by Using Smaller Indexes

Solution:


If m is an integer, n and k are natural numbers, and all indicated roots exist, then

.kn nkm ma a

Slide 8.3- 15

Simplify radicals.

kn kma


Simplify products and quotients of radicals with different indexes.

Objective 4

Slide 8.3- 16


Simplify.

35 4

1/ 2 3/ 6 6 3 6 125 5 5 5

1/3 2/ 6 6 2 6 164 4 4

35 4 6 6125 16 6 2000

The indexes, 2 and 3, have a least common index of 6, use rational exponents to write each radical as a sixth root.

Slide 8.3- 17

CLASSROOM EXAMPLE 7

Multiplying Radicals with Different Indexes

Solution:



Objective 5

Slide 8.3- 18


Pythagorean Theorem

If c is the length of the longest side of a right triangle and a and b are lengths of the shorter sides, then

c2 = a2 + b2.

The longest side is the hypotenuse, and the two shorter sides are the legs, of the triangle. The hypotenuse is the side opposite the right angle.

Slide 8.3- 19



Find the length of the unknown side of the triangle.

c2 = a2 + b2

c2 = 142 + 82

c2 = 196 + 64

c2 = 260

260c

4 65c

4 65c

2 65c The length of the hypotenuse is 2 65.

Slide 8.3- 20

CLASSROOM EXAMPLE 8

Using the Pythagorean Theorem

Solution:


Find the length of the unknown side of the triangle.

c2 = a2 + b2

62 = 42 + b2

36 = 16 + b2

20 = b2

20 b

4 5 b

4 5 b

2 5 bThe length of the leg is 2 5.

Slide 8.3- 21

CLASSROOM EXAMPLE 8

Using the Pythagorean Theorem (cont’d)

Solution:



Objective 6

Slide 8.3- 22


Distance Formula

The distance between points (x1, y1) and (x2, y2) is

2 22 1 2 1( ) ( )d x x y y

Slide 8.3- 23



Find the distance between each pair of points.

(2, –1) and (5, 3)

Designate which points are (x1, y1) and (x2, y2).

(x1, y1) = (2, –1) and (x2, y2) = (5, 3)

2 22 1 2 1( ) ( )d x x y y

2 2( ) (5 32 ( ))1d

2 2(3) (4)d

9 16d

5 52d

Start with the x-value and y-value of the same point.

Slide 8.3- 24

CLASSROOM EXAMPLE 9

Using the Distance Formula

Solution:


Find the distance between each pairs of points.

(–3, 2) and (0, 4)

Designate which points are (x1, y1) and (x2, y2).

(x1, y1) = (–3, 2) and (x2, y2) = (0, 4)

2 22 1 2 2( ) ( )d x x y y

2 2( ( )) (3 )0 4 2d

2 2(3) ( 6)d

9 36d

3 545d Slide 8.3- 25

CLASSROOM EXAMPLE 9

Using the Distance Formula (cont’d)

Solution:

Section 3Chapter 8. 1 Copyright © 2012, 2008, 2004 Pearson Education, Inc. Objectives 2 6 5 3 4...

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