Transcript No Slide Title

MTH 209 Week 3
Due for this week…
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Homework 3 (on MyMathLab – via the Materials
Link)  The fifth night after class at 11:59pm.
Read Chapter 6.6, 8.4 and 11.1-11.5
Do the MyMathLab Self-Check for week 3.
Learning team hardest problem assignment.
Complete the Week 3 study plan after submitting
week 3 homework.
Participate in the Chat Discussions in the OLS
Copyright © 2009 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2
Section 7.1
Introduction to
Rational
Expressions
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
•
Basic Concepts
•
Simplifying Rational Expressions
•
Applications
Basic Concepts
Rational expressions can be written as quotients
(fractions) of two polynomials.
Examples include:
5
,
x
x2
,
3x  4
4 x2  6x 1
4 x3  8
Example
If possible, evaluate each expression for the given
value of the variable.
2
4w
1
w
; w  5
a.
b.
c.
;x  3
;w  4
x3
Solution
a. 1 ; x  3
x3
1
1

33 6
w4
3w  4
2
w
b.
;w  4
3w  4
2
(4)
3(4)  4
16

12  4
16

2
8

4w
c.
; w  5
w4
4  (5)

(5)  4

9
 1
9
Try Q 7,11,13,17 pg. 427
Example
Try Q 25,27,31,33 pg. 428
Find all values of the variable for which each
expression is undefined.
2
6
1
w
a. 2
b.
c. 2
x
Solution
a. 12
x
Undefined
when x2 = 0 or
when x = 0.
b.
w4
w 4
w2
w4
6
c. 2
w 4
Undefined
when w – 4 =
0 or when w =
4.
Undefined
when w2 – 4 =
0 or when w =
2.
Example
Try Q 39,43 pg. 428
Simplify each fraction by applying the basic principle
of fractions.
45
20
9
a.
b.
c.135
15
28
Solution
9
a. The GCF of 9 and 15 is 3. 15

33
3

35
5
45 5


28 4  7 7
b. The GCF of 20 and 28 is 4.20
45 1 1


135
45  3 3
c. The GCF of 45 and 135 is 45.45
Example
Simplify each expression.
a. 16 y
b. 3 x  12
4y
4 x  16
2
Solution
a. 16 y
4 y2
4y 4

4y  y
4

y
3 x  12
b.
4 x  16

3( x  4)
4( x  4)
3

4
c.
x 2  25
2 x 2  7 x  15
x 2  25
c. 2
2 x  7 x  15
( x  5)( x  5)

(2 x  3)( x  5)
x5

2x  3
Example
Try Q 51,55,61,79 pg. 428
Simplify each expression.
8 x

y

7

a.
b.
x 8
2 y  14
Solution
a.  y  7  1( y  7)
2 y  14
2( y  7)
1

2
8  x  (8  x)

b.
x 8
x 8
8  x

x 8
x 8

1
x 8
Example
Try Q 105 pg. 429
Suppose that n balls, numbered 1 to n, are placed in
a container and two balls have the winning number.
a. What is the probability of drawing the winning ball
at random?
b. Calculate this probability for n = 100, 1000 and
10,000.
c. What happens to the probability of drawing the
winning ball as the number of balls increases?
Solution
a. There are 2 chances of drawing the winning ball. 2
n
Example (cont)
Try Q 105 pg. 429
b. Calculate this probability for n = 100, 1000 and
10,000.
2
1
2
1
2
1



100 50
1000 500 10,000 5000
c. What happens to the probability of drawing the
winning ball as the number of balls increases?
The probability decreases.
Section 7.2
Multiplication
and Division of
Rational
Expressions
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
•
Review of Multiplication and Division of Fractions
•
Multiplication of Rational Expressions
•
Division of Rational Expressions
Example
Try Q 5,7,9 pg. 435
Multiply and simplify your answers to lowest terms.
4
2 5
4 5
a.
b. 15 
c. 

9 7
5
Solution
a. 4  5  20
9 7 63
4 15 4 60
 12
b. 15    
5 1 5 5
2 5 5 2 5 1 5
  
c.  
7 8 7  8 7 4 28
7 8
Example
Try Q 13,15,17 pg. 435
Divide and simplify your answers to lowest terms.
6
4 11
1 3
a.
b.  18
c. 

6
5
7
Solution
5
a. 1 3 1 5
   
6 5
6 3 18
5 15
b.6  18  6  1  1 6  1
7
4 11 4 15 60 12  5 12
c.    


5 15 5 11 55 11 5 11
7 18
7 18
21
Example
Multiply and simplify to lowest terms. Leave your
answers in factored form.
x3 x 4
6x 5

a.
b.

2
2 x  1 3x  9
10 12 x
Solution
a. 6 x  5  6 x  5
2
2
10 12 x
10 12 x
30 x

120 x 2
1

4x
x3 x 4
( x  3)( x  4)
b. 2 x  1  3 x  9 
(2 x  1)(3x  9)

( x  3)( x  4)
3(2 x  1)( x  3)
x4

3(2 x  1)
Example
Try Q 29,31,41,45 pg. 435
Multiply and simplify to lowest terms. Leave your
2
x
 16 x  3
answer in factored form.

x2  9 x  4
Solution
x 2  16 x  3

2
x 9 x 4
( x 2  16)( x  3)
 2
( x  9)( x  4)
( x  4)( x  4)( x  3)

( x  3)( x  3)( x  4)
( x  4)( x  3)( x  4)

( x  3)( x  3)( x  4)
( x  4)

( x  3)
Example
Try Q 49,57,65 pg. 435
Divide and simplify to lowest
terms.
2
a. 3  2 x  1
b. 2x  16  x  4
x
x  2x  8
6x
Solution
a. 3  2 x  1  3  6 x
x
x 2x 1
6x
18 x

x(2 x  1)

18
2x 1
x2
x 2  16
x4

b.2
x  2x  8 x  2
x 2  16 x  2
 2

x  2x  8 x  4
( x  4)( x  4) x  2


( x  2)( x  4) x  4
( x  4)( x  4)( x  2)

1
( x  2)( x  4)( x  4)
Section 7.3
Addition and
Subtraction
with Like
Denominators
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
•
Review of Addition and Subtraction of Fractions
•
Rational Expressions Having Like Denominators
Example
Simplify each expression to lowest terms.
a. 4  1
b.1  5
7
7
Solution
4 1
4 1
5
a.  

7 7
7
7
1 5 1 5
6
2
b.  


9 9
9
9
3
9
9
Example
Try Q 7,9,11,13 pg. 442
Simplify each expression to lowest terms.
a. 13  7
b. 15  11
18 18
30
30
Solution
13 7

a.
18 18
b.
13  7

18
6

18
1

3
15 11
15  11
4
2




30 30
30
30
15
SUMS OF RATIONAL EXPRESSIONS
To add two rational expressions having like
denominators, add their numerators. Keep
the same denominator.
A B A B
 
C C
C
C is nonzero
Example
Try Q 19,25,33,35 pg. 442
Add and simplify to lowest terms.
x
4
x

1
x

2
a.
b. 2

x3
x3
5
 2
x  7 x  10 x  7 x  10
Solution
a. 4 x  1 x  2
5x 1
4x 1 x  2



x3 x3
x3
x3
b.
x5
x
5
x5

 2
 2
2
 x  5 x  2 
x  7 x  10 x  7 x  10
x  7 x  10
1

x2
Example
Try Q 51,53,55 pg. 443
Add and simplify to lowest terms.
7
4
a. 
b.2 w 2 
ab
Solution
a. 7 4
74


ab ab
ab
b.
w y
ab
w
y
 2
2
2
w y
w  y2
y
w2  y 2
11

ab
w y
1
w y


 2
2
( w  y )( w  y )
w y
w y
DIFFERENCES OF RATIONAL EXPRESSIONS
To subtract two rational expressions having
like denominators, subtract their numerators.
Keep the same denominator.
A B A B
 
C C
C
C is nonzero
Example
Try Q 21,27,67 pg. 442-3
Subtract and simplify to lowest terms.
6 x6
2x  3 x  4
a. 2  2
b. 2
 2
x
x 1
x
x 1
Solution
a. 6  x  6  6   x  6   6  x  6   x   1
2
2
2
2
2
x
x
x
x
x
x
x 1
2x  3 x  4 2x  3  x  4
x 1 
 2

 2
b. 2
2
 x  1 x  1
x 1 x 1
x 1
x 1
1

x 1
Example
Try Q 31,59 pg. 442-3
Subtract and simplify to lowest terms.
Solution
7a
a2
7 a  (a  2)
6a  2



a2 a2
a2
a2
7a
a2

a2 a2
Section 7.4
Addition and
Subtraction
with Unlike
Denominators
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
•
Finding Least Common Multiples
•
Review of Fractions Having Unlike Denominators
•
Rational Expressions Having Unlike Denominators
FINDING THE LEAST COMMON MULTIPLE
The least common multiple (LCM) of two or
more polynomials can be found as follows.
Step 1: Factor each polynomial completely.
Step 2: List each factor the greatest number
of
times that it occurs in either
factorization.
Step 3: Find the product of this list of factors.
The
result is the LCM.
Example
Find the least common multiple of each pair of
expressions.
a. 6x, 9x4
b. x2 + 7x + 12, x2 + 8x + 16
Solution
Step 1: Factor each polynomial completely.
6x = 3 ∙ 2 ∙ x
9x4 = 3 ∙ 3 ∙ x ∙ x ∙ x ∙ x
Step 2: List each factor the greatest number of
times.
3∙3∙2∙x∙x∙x∙x
Step 3: The LCM is 18x4.
Example (cont)Try Q 15,19,27,29
pg. 451
b. x2 + 7x + 12, x2 + 8x + 16
Step 1: Factor each polynomial completely.
x2 + 7x + 12 = (x + 3)(x + 4)
x2 + 8x + 16 = (x+ 4)(x + 4)
Step 2: List each factor the greatest number of times.
(x + 3), (x + 4), and (x + 4)
Step 3: The LCM is (x + 3)(x + 4)2.
Example
Try Q 45,47 pg. 452
Simplify each expression.
a. 4  1
b. 5  11
7
6
12
30
Solution
a. The LCD is the LCM, 42.
24 7
4 1 4 6 1 7
31

     

42 42 42
7 6 7 6 6 7
b. The LCD is 60.
1
5 11 5 5 11 2
25 22 3


    


20
12 30 12 5 30 2
60 60 60
Example
Try Q 53,65,71 pg. 452
Find each sum and leave your answer in factored
form.
4
3
2 5
a.
b.

 2
x
x
x 1 1 x
Solution
a. The LCD is x2.
2 5
2 x 5
2x 5 2x  5
 2    2  2 2 
x x
x x x
x
x
x2
4
3
4
3 1
4
3
1
b. x  1  1  x  x  1  1  x  1  x  1  x  1  x  1
Example
Simply the expression. Write your answer in lowest
terms and leave it in factored form. x  3
5

Solution
x
x7
The LCD is x(x + 7).
x 3
5
x 3 x 7
5 x  x  3 x  7 
5x




 

x
x7
x x7 x7 x
x  x  7
x  x  7
x  3 x  7   5 x


x  x  7
x 2  4 x  21  5 x
x 2  x  21


x  x  7
x  x  7
Example
Try Q 63,77,79,81 pg. 452
Simplify the expression. Write your answer in lowest
5
terms and leave it in factored form. 2 6
 2
x  6x  9 x  9
Solution
6
5
6
5



x 2  6 x  9 x 2  9  x  3 x  3  x  3 x  3
x  3
x  3


5




 x  3 x  3  x  3  x  3 x  3  x  3
6
6  x  3
5  x  3


 x  3 x  3 x  3  x  3 x  3 x  3

6 x  18  5 x  15
 x  3 x  3 x  3

x  33
 x  3 x  3 x  3
Example
Add
1 1
 ,
R S
Try Q 101 pg. 453
and then find the reciprocal of the result.
Solution
The LCD is RS.
1 1 1 S 1 R
    
R S R S S R
S
R


RS RS
SR

RS
The reciprocal is
RS
.
SR
Section 7.6
Rational
Equations and
Formulas
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
•
Solving Rational Equations
•
Rational Expressions and Equations
•
Graphical and Numerical Solutions
•
Solving a Formula for a Variable
•
Applications
Rational Equations
If an equation contains one or more rational
expressions, it is called a rational equation.
Example
Try Q 9,15,31,41 pg. 473
Solve each equation.
a. 9  4
b. 6  x
7
2x 1
x
Solution
a. 9  4
7
6
b.  x
2x 1
6
x

2x 1 1
x  2 x  1  6(1)
x
9x  28
28
x
9
The solution
2x2  x  6
28
is9 .
2x  x  6  0
2
 2x  3 x  2  0
2x  3  0
x20
3
x
2
x  2
The solutions are
2 and
3
.
2
Example
Determine whether you are given an expression or
an equation. If it is an expression, simplify it and then
evaluate it for x = 4. If it is an equation, solve it.
a. 2 x  4  x
b. x 2  3x  10
x3
x3
x2
x2
Solution
a. There is an equal sign, so it is an equation.
 x  3
2x
x
 4  x  3   x  3
x3
x3
2x  4x 12  x
3x  12
x  4
The answer checks.
The solution is −4.
x 2  3x 10

x2
x2
Example (cont)
b. There is no equals sign, so it is an expression.
The common denominator is x – 2, so we can add
the numerators.
x 2  3x 10

x2
x2
x 2  3x  10

x2
x  2  x  5 


x2
 x5
When x = 4, the expression evaluates 4 + 5 = 9.
Try Q 49,55 pg. 473
Example
Solve
2
x
x 1
Solution
2
Graph y1 
x 1
x
−3 −2 −1
graphically and numerically.
andy2  x.
0
−1 −2 −− 2
1
1
2
3
2
3
1
2
The solutions are −2 and 1.
(1, 1)
(−2, −2)
Example (cont)
Solve
2
x
x 1
Try Q 75 pg. 474
graphically and numerically.
Solution
Numerical Solution
−3 −2 −1
2 −1 −2 −−
y1 
0
2
1
1
2
3
2
3
1
2
−3 −2 −1
0
1
2
3
x
x 1
y2  x
The solutions are −2 and 1.
Example
Solve the equation for the specified variable.
C  2 r for r
Solution
C  2 r for r
C
2 r

2
2
C
r
2
Example
Solve the equation for the specified variable.
2A
h
for A
Bb
Solution
2A
h
Bb
h( B  b)  2 A
h( B  b)
A
2
Example
Try Q 89,91,97 pg. 474
Solve the equation for the specified variable.
S  B  2sl for s
Solution
S  B  2sl for s
S  B  B  B  2sl
S  B  2sl
S  B 2sl

2l
2l
S B
s
2l
Example
Try Q 102 pg. 474
A pump can fill a swimming pool ¾ full in 6 hours,
another can fill the pool ¾ full in 9 hours. How long
would it take the pumps to fill the pool ¾ full, working
together?
t t 3
 
Solution
6 9 4
t t
3
36     36  
6 9
4
6t  4t  27
7
t2
10
7
The two pumps can fill the pool ¾ full in 2
10
hours.
Section 7.7
Proportions
and Variation
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
•
Proportions
•
Direct Variation
•
Inverse Variation
•
Analyzing Data
•
Joint Variation
Proportions
A proportion is a statement (equation) that two
ratios (fractions) are equal.
The following property is a convenient way to solve
proportions:
a c

b d
is equivalent to ad  bc,
provided b ≠ 0 and d ≠ 0.
Example
Try Q 65 pg. 488
On an elliptical machine, Francis can burn 370
calories in 25 minutes. If he increases his work time
to 30 minutes, how many calories will he burn?
Solution
Let x be the equivalent amount of calories.
25 30

370 x
Minutes Minutes
=
Calories Calories
25 x  11,100
x  444
Thus, in 30 minutes,
Francis will burn 444
calories.
Example
Try Q 56 pg. 488
A 6-foot tall person casts a shadow that is 8-foot
long. If a nearby tree casts a 32-foot long shadow,
estimate the height of the tree.
Solution
6 ft
h
The triangles are similar because
8 ft
32 ft
the measures of its corresponding
angles are equal. Therefore corresponding sides are
proportional. h 32 Height Shadow length

6 8
Height
8h  192
h  24
=
Shadow length
The tree is 24 feet tall.
Example
Try Q 33 pg. 487
Let y be directly proportional to x, or vary directly with
x. Suppose y = 9 when x = 6. Find y when x = 13.
Solution
y  kx
Step 1 The general equation is y = kx. 9  6k
Step 2 Substitute 9 for y and 6 for x in 9
k
y = kx. Solve for k.
6
Step 3 Replace k with 9/6 in the equation y = 9x/6.
Step 4 To find y, let x = 13. y  9 x
6
9
y  (13)
6
y  19.5
Example
The table lists the amount of pay for various hours
worked.
Pay
$138
$253
$345
$529
$713
800
31, 713
700
600
Pay (dollars)
Hours
6
11
15
23
31
23, 529
500
400
15, 345
300
11, 253
200
6, 138
100
0
0
10
20
Hours
a. Find the constant of proportionality.
b. Predict the pay for 19 hours of work.
30
40
Example (cont)
Try Q 73 pg. 488
The slope of the line equals the proportionality, k. If
we use the first and last data points (6, 138) and (31,
713), the slope is
713  138
k
 23
31  6
The amount of pay per hour is $23. The graph of the
line y = 23x, models the given graph.
To find the pay for 19 hours, substitute 19 for x.
y = 23x,
y = 23(19) 19 hours of work would
pay $437.00
y = 437
Example
Try Q 39 pg. 487
Let y be inversely proportional to x, or vary inversely
with x. Suppose y = 6 when x = 4. Find y when x =k 8.
y
Solution
x
Step 1 The general equation is y = k/x.
k
6
Step 2 Substitute 6 for y and 4 for x in
4
Solve for k.
24  k
Step 3 Replace k with 24 in the equation y = k/x.
k
y
Step 4 To find y, let x = 8.
x
24
y
8
y3
Example
Try Q 51a,53a,55a pg. 487-488
Determine whether the data in each table represent
direct variation, inverse variation, or neither. For
direct and inverse variation, find the equation.
Neither the product xy nor the ratio y/x
a. x
3
7
9 12
b.
c.
y
12
28
32
48
x
y
5
12
10
6
12
5
15
4
x
y
8
48
11
66
14
84
21
126
are constant in the data in the table.
Therefore there is neither direct
variation nor indirect variation in this
table.
As x increases, y decreases. Because
xy = 60 for each data point, the
equation
y = 60/x models the data. This
represents an inverse variation.
The equation y = 6x models the
data. The data represents direct
variation.
f  x   a , a  0 and a  1,
JOINT VARIATION
x
Let x, y, and z denote three quantities.
Then z varies jointly with x and y if
there is a nonzero number k such that
z  kxy.
Example
Try Q 83 pg. 488
The strength S of a rectangular beam varies jointly
as its width w and the square of its thickness t. If a
beam 5 inches wide and 2 inches thick supports 280
pounds, how much can a similar beam 4 inches wide
and 3 inches thick support?
Solution
2
280

k

5

2
The strength of the beam is
modeled by S = kwt2.
280
k
5 4
k  14
Example (cont)
Try Q 83 pg. 488
Thus S = 14wt2 models the strength of this type of
beam. When w = 4 and t = 3, the beam can support
S = 14 ∙ 4 ∙ 32= 504 pounds
Section 10.1
Radical
Expressions
and Functions
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
•
Radical Notation
•
The Square Root Function
•
The Cube Root Function
Radical Notation
Every positive number a has two square roots, one
positive and one negative. Recall that the positive
square root is called the principal square root.
The symbol
is called the radical sign.
The expression under the radical sign is called the
radicand, and an expression containing a radical
sign is called a radical expression.
Examples of radical expressions:
7,
6  x  2, and
5x
3x  4
Example
Try Q 15,17,19,21 pg. 641
Evaluate each square root.
a.
36  6
b.
0.64  0.8
c.
4
16

25 5
Example
Approximate
Try Q 39 pg. 641
38 to the nearest thousandth.
Solution
 6.164
Example
Try Q 23,25,27,41 pg. 641
Evaluate the cube root.
a. 64  4
3
b. 3 125  5
c.
3
1
1

2
8
Example
Try Q 33,35,37 pg. 641
Find each root, if possible.
a.
4
256
b.5 243
c.1296
4
Solution
a.
b.
c.
4
5
4
256  4 because 4  4  4  4  256.
243
1296
 3 because (  3)5  243.
An even root of a negative number
is not a real number.
Example
Try Q 45,49,51 pg. 641
Write each expression in terms of an absolute value.
a.
( 5) 2
2
(
x

3)
b.
c. w2  6w  9
Solution
a. (5) 2  5  5
b. ( x  3) 2  x  3
c.
w2  6w  9  ( w  3) 2
 w3
Example
Try Q 61,63 pg. 641
If possible, evaluate f(1) and f(2) for each f(x).
2
f
(
x
)

x
4
f
(
x
)

5
x

1
a.
b.
Solution
a. f ( x )  5 x  1
f (1)  5(1)  1
 6
f ( 2)  5( 2)  1
 9  undefined
2
f
(
x
)

x
4
b.
f (1)  12  4
 5
f ( 2)  ( 2)2  4
 8
Example
Try Q 75,89 pg. 642
Calculate the hang time for a ball that is kicked 75
feet into the air. Does the hang time double when a
ball is kicked twice as high? Use the formula
1
x
2
Solution
1
The hang time isT (75)  75  4.3 sec
2
T (x) 
The hang time isT (150) 
1
150  6.1 sec
2
The hang times is less than double.
Example
Try Q 75,89 pg. 641
Find the domain of each function. Write your answer
in interval notation.
a. f ( x )  3  4 x
b.
f (x)  x2  4
Solution
Solve 3 – 4x  0.
3  4x  0
4 x  3
3
x
4
3

The domain is , 4  .
b. Regardless of the value
of x; the expression is
always positive. The
function is defined for all
real numbers, and it
,   . is
domain
Section 10.2
Rational
Exponents
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
•
Basic Concepts
•
Properties of Rational Exponents
Example
Try Q 37,54,59,63 pg. 650
Write each expression in radical notation. Then
evaluate the expression and round to the nearest
hundredth when appropriate.
a. 49
1/2
1/2
b. 26
Solution
a. 491/2  49  7
c. (6 x)1/2  6x
c.
 6x 
1/5
b.261/5
Example
Write each expression in radical notation. Evaluate
the expression by hand when possible.
2/3
103/4
a.  8 
b.
Solution
2
2/3
3
a.  8    8    2 2  4
b. 103/4  4 103  4 1000
Example
Try Q 47,51 pg. 650
Write each expression in radical notation. Evaluate
the expression by hand when possible.
a. 813/4
b.144/5
Solution
a. 813/4
Take the fourth root of
4/5 Take the fifth root of 14
14
b.
3/ 4 81 and then cube it.
and then fourth it.
 (81)

 81
4
3
3
 27
3
 144/5

 14 
5
4
Cannot be evaluated
by hand.
Example
Try Q 53,55 pg. 650
Write each expression in radical notation and then
evaluate.
a. 811/4
b.64 2/3
Solution
1
a. 811/4  1/4
81
1
 4
81
1

3
b.64 2/3  12/3
64


1
3
64

2
1
1
 2 
4
16
Example
Try Q 53,55 pg. 650
Use rational exponents to write each radical
expression.
a. 7 x3  x 3/7
1
b.
c.
d.
b
3
 b 3/2
 ( x  1)2/5
5
( x  1) 2
4
2
2 1/4
a 2  b2  ( a  b )
Example
Write each expression using rational exponents and
simplify. Write the answer with a positive exponent.
Assume that all variables are positive numbers.
a.
x4 x
 x x
1/2
1/4
b.4 256x3  (256 x3 )1/4
 x1/21/4
 2561/4 ( x3 )1/4
 x 3/4
 4x 3/4
Example (cont)
Try Q 77,83,91,97 pg. 650
Write each expression using rational exponents and
simplify. Write the answer with a positive exponent.
Assume that all variables are positive numbers.
5
c.
32x
4
x
(32 x)1/5

x1/4
321/5 x1/5

x1/4
 2x1/51/4
 2x 1/20
2
 1/20
x
x 
d. 27 
3
1/3
1/3
 27 
 3 
x 
271/3
 3 1/3
(x )
3

x
Example
Try Q 85,89,95 pg. 650
Write each expression with positive rational
exponents and simplify, if1/4
possible.
y
4
a. x  2
b. 1/5
x
Solution
a. 4 x  2  ( x  2)1/2 1/4


 ( x  2)1/8
y 1/4
b. 1/5
x
x1/5
 1/4
y
Section 10.3
Simplifying
Radical
Expressions
Copyright © 2013, 2009, and 2005 Pearson Education, Inc.
Objectives
•
Product Rule for Radical Expressions
•
Quotient Rule for Radical Expressions
Product Rule for Radicals
Consider the following example:
4  25  2  5  10
4  25  100  10
Note: the product rule only works when the radicals
have the same index.
Example
Try Q 13,15,21 pg. 659
Multiply each radical expression.
a. 36  4
b.
3
8  3 27
 36  4  144  12
 3 8  27  3 216  6
c. 4 1  4 1  4 1  4 1  1  1  4 1  1
4 16 4
4 16 4
256
4
Example
Try Q 23,51,57,61 pg. 659-60
Multiply each radical expression.
x 2  x 4  x 2  x 4  x 6  x3
a.
b. 5a  10a
c.
3
3
4
3x 4 7 y

y
x
2
 3 5a 10a2  3 50a3  a 3 50
3x 7 y
21xy 4
4

4
 21
y x
xy
Example
Try Q 73,75,77,79 pg. 660
Simplify each expression.
a. 500  100  5  10 5
b.
3
40  3 8  3 5  2 3 5
c. 72  36  2  6 2
Example
Try Q 45,85,89,91 pg. 660
Simplify each expression. Assume that all variables
are positive.
5
4
75
y
4
4
2

25y
   3y
a. 49x  49  x  7 x
b.
 25y 4  3 y
c. 3 3a  3 9a2w  3 3a  9a2w
 3  27a3  w
 3  27a3   3 w
 3a 3 w
 5 y 2 3y
Example
Try Q 101,103,107 pg. 660
Simplify each expression.
a.
7  7  7 7
3
1/2
1/3
b.
3
a  5 a  a1/3  a1/5
7
 a1/31/5
 7 5/6
 a 8/15
1/21/3
Quotient Rule
Consider the following examples of dividing radical
expressions: 4
2 2 2
9


3 3

4
4 2


9
9 3
3
Example
Try Q 25,27,29 pg. 659
Simplify each radical expression. Assume that all
variables are positive.
a.
3
3
7
7
3
27
27
3
7

3
5
x
x
b.5
5
32
32
5
x

2
Example
Try Q 33,39,41 pg. 659
Simplify each radical expression. Assume that all
variables are positive.
a.
90
90

10
10
 9
3
b.
x4 y
y

x4 y
y
 x4
 x2
Example
Try Q 95,97 pg. 660
Simplify the radical expression. Assume that all
variables are positive.
5
32x 4
y5

5
32x 4
5

5
y5
32  5 x 4
5
y5
2 5 x4

y
Example
Simplify the expression. x  1  x  1
Solution
x  1  x  1  ( x  1)( x  1)
 x2 1
Example
Try Q 63,67 pg. 660
Simplify the expression.
3
x2  5x  6
3
x2
Solution
3
x2  5x  6
( x  3)( x  2)
3

3
x2
( x  2)
 3 x3
End of week 3





You again have the answers to those problems not
assigned
Practice is SOOO important in this course.
Work as much as you can with MyMathLab, the
materials in the text, and on my Webpage.
Do everything you can scrape time up for, first the
hardest topics then the easiest.
You are building a skill like typing, skiing, playing a
game, solving puzzles.