Transcript No Slide Title

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-1
Functions, Equations,
and Inequalities
Chapter 2
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2.1
Linear Equations, Functions,
and Models




Solve linear equations.
Solve applied problems using linear models.
Find zeros of linear functions.
Solve a formula for a given variable.
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Linear Equations
An equation is a statement that two expressions are
equal.
One Variable
mx + b = 0, where m and b are real numbers and
m  0.
Examples:
4x + 6 = 18
5(x  2) = 7x + 8
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Slide 2-4
Equivalent Equations
Equations that have the same solution set.
Examples:
4x + 8 = 16 and x = 2 are equivalent because 2 is
the solution of each equation.
5y + 6 = 21 and y = 4 are not equivalent because
y is equal to 3 in the first equation.
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Slide 2-5
Equation-Solving Principles

Addition
If a = b is true, then
a + c = b + c is true.
You can also use the intersect
feature on a graphing calculator
to solve equations.
y1 = 9x  7, y2 = 2
Example:
9x  7  2
9x  7  7  2  7
9x  9
x 1
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Slide 2-6
Equation-Solving Principles


Multiplication
If a = b is true, then ac = bc is
true.
Example:

We graph y1 = 3(5 + 2x) and
y2 = 4 2(x + 3)
3(5  2 x)  4  2( x  3)
15  6 x  4  2 x  6
15  6 x  2  2 x
15  6 x  6 x  2  2 x  6 x
15  2  2  2  8 x
17  8 x
17 8 x

8 8
17
  x or  2.125
8
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Slide 2-7
Applications Using Linear Models

Mathematical techniques to answer questions in realworld situations.

Often modeled by linear equations and functions.
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Slide 2-8
Five Steps for Problem Solving
1. Familiarize yourself with the problem situation.
Make a drawing
Write a list
Assign variables
Organize into a chart or table
Find further information
Guess or estimate the answer
2. Translate to mathematical language or symbolism.
3. Carry out some type of mathematical manipulation.
4. Check to see whether your possible solution actually fits the
problem situation.
5. State the answer clearly using a complete sentence.
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Slide 2-9
The Motion Formula

The distance d traveled by an object moving at rate r in
time t is given by d = rt.

Example: On a bicycle tour, Dave rides his bicycle
10 mph faster than his father, Bill. In the same time that
Dave travels 90 miles, his father travels 60 miles.
Find their speeds.
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Slide 2-10
Example
1. Familiarize. We can organize the information in a table as follows.
Distance
Rate
Time
Dave
90
r + 10
Bill
60
r
t
t
d
2. Translate. Using the formula r  t , we get two expressions for t.
90
t
r  10
and
60
t
r
90
60

Since they occur in the same time, we have the following equation: r  10 r .
90
60
3. Carry out. We solve the equation.

r  10
r
90r  60( r  10)
90r  60r  60r  60r  600
30r  600
r  20
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Slide 2-11
Example
4. Check. If Bill travels at rate of 20 mph for a distance of 60 miles it
will take him 3 hours. If Dave travels 90 miles at a rate of r + 10, or
30 mph, it will also take him 3 hours. Therefore, the answer
checks.
5. State. Dave travels at a rate of 30 mph while his father, Bill, travels
at a rate of 20 mph on their bicycles.
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Slide 2-12
Simple-Interest Formula

I = Prt




I = the simple interest ($)
P = the principal ($)
r = the interest rate (%)
t = time (years)
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Slide 2-13
Example
The Robinson’s have two loans that total $15,000. One loan is at 4%
simple interest and the other is at 6.5%. After 1 year, they owe $700 in
interest. What is the amount of each loan?

Solution:
1. Familiarize. We will let x = the amount borrowed at 4% interest. Then
the remainder is $15,000 – x, borrowed at 6.5% interest.
4% Loan
6.5% Loan
Total
Amount
Borrowed
Interest
Rate
Time
x
4% or 0.04
1 yr
15,000 – x
6.5% or
0.065
1 yr
15,000
Amount of
Interest
0.04x
0.065(15,000 – x)
700
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Slide 2-14
Example
2. Translate. The total amount of interest on the two loans is $700. Thus we
write the following equation.
0.04x + 0.065(15,000  x) = 700
3. Carry out. We solve the equation.
0.04x + 0.065(15,000  x) = 700
0.04x + 975  0.065x = 700
 0.025x + 975 = 700
0.025x = 275
x = 11,000
If x = 11,000, then 15,000  11,000 = 4000.
4. Check. The interest on $11,000 at 4% for 1 yr is $11,000(0.04)(1), or $440.
The interest on $4000 at 6.5% for 1 yr is $4000(0.065)(1) or $260.
Since $440 + $260 = $700, the answer checks.
5. State. The Robinson’s borrowed $11,000 at 4% interest and $4000
at
6.5% interest.
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Slide 2-15
Zeros of Linear Functions

An input c of a function f is called a zero of the function,
if the output for c is 0.

f(c) = 0

A linear function f(x) = mx + b, with m  0, has exactly
one zero.
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Slide 2-16
Example

Find the zero of
f(x) = 3x  18.
Algebraic Solution:
3x  18 = 0
3x = 18
x=6

Find the zero of
f(x) = 3x – 18.
Graphic Solution:
The x-intercept of the graph is
(6, 0). Thus, 6 is the zero of the
function.
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Slide 2-17
Formulas


A formula is an equation that can be used to model a
situation.
Example: Solve y = ax + bx  2 for x.
y = ax + bx  2
y + 2 = ax + bx
y + 2 = x(a + b)
y2
x
ab
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Slide 2-18
2.2
The Complex Numbers

Perform computations involving complex numbers.
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The Complex-Number System

The complex-number system is used to find zeros of
functions that are not real numbers.

When looking at a graph of a function, if the graph does
not cross the x-axis, it has no real-number zeros.
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Slide 2-20
The Number i

The number i is defined such that
i  1 and i 2  1.
Examples: Express each number in terms of i.
i is not under
the radical
5  1  5  1  5  i 5 or
5i
 19   1  19   1  19  i 19 or  19i
 81   1  81   1  81  i  9  9i
12  1  12  1  12  i 4  3  i  2 3  2i 3 or 2 3i
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Slide 2-21
Complex Numbers
A complex number is a number of the form a + bi, where
a and b are real numbers. The number a is said to be
the real part of a + bi and the number b is said to be
the imaginary part of a + bi.
Imaginary Number
a + bi, a  0, b  0
Pure Imaginary Number
a + bi, a = 0, b  0
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Slide 2-22
Addition and Subtraction
Complex numbers obey the commutative, associative,
and distributive laws.

Add:

(9 + 5i) + (2 + 4i)
(9 + 2) + (5i + 4i)
11 + (5 + 4)i
11 + 9i
Subtract:
(6 + 7i)  (4  3i)
(6  4) + [7i  (3i)]
2 + 10i
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Slide 2-23
Multiplication
When a and b are real numbers, a b  ab .
Examples: Multiply and simplify.
36 
9  1 
36 
1 
i  6  i  3
 i 2  18
 1  18
9
(7 6i)2 = 72  2  7  6i + (6i)2
= 49  84i + 36i2
= 49  84i  36
= 13  84i
 18
(1 + 5i)(1 + 7i) = 1 + 7i + 5i + 35i2
= 1 + 7i + 5i  35
= 34 + 12i
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Slide 2-24
Simplifying Powers of i

Recall that 1 raised to an even power is 1, and 1
raised to an odd power is 1.

Examples:
 i47 = i46  i = (i2)23  i = (1)23  i = 1  i = i
 i72
= i71  i = (i2)36  i = (1)36  i = 1  i = i
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Slide 2-25
Conjugates
The conjugate of a complex number a + bi is a  bi. The numbers
a + bi and a  bi are complex conjugates.
Examples:
6 + 7i and 6  7i
8  3i and 8 + 3i
14i and 14i

The product of a complex number and its conjugate is a real
number.
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Slide 2-26
Multiplying Conjugates

Example:
(4 + 9i)(4  9i) = 42  (9i)2
= 16  81i2
= 16  81(1)
= 97

Example:
(7i)(7i) =  49i2
= 49(1)
= 49
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Slide 2-27
Dividing Conjugates

Example: Divide 4  3i by 1  8i.
4  3i 4  3i 1  8i


1  8i 1  8i 1  8i
( 4  3i )(1  8i )

(1  8i )(1  8i )
4  29i  24i 2

1  64i 2
4  29i  24

65
28  29i

65
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Slide 2-28
2.3
Quadratic Equations,
Functions, and Models



Find zeros of quadratic functions and solve quadratic
equations by using the principle of zero products,
by using the principle of square roots, by completing
the square, and by using the quadratic formula.
Solve equations that are reducible to quadratic.
Solve applied problems using quadratic equations.
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Quadratic Equations
A quadratic equation is an equation equivalent to
ax2 + bx + c = 0, a  0,
where a, b, and c are real numbers.
A quadratic equation written in this form is said to be in
standard form.
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Slide 2-30
Quadratic Functions
A quadratic function f is a function that can be written in the form
f(x) = ax2 + bx + c, a  0,
where a, b, and c are real numbers.
The zeros of a quadratic function f(x) = ax2 + bx + c are the solutions
of the associated quadratic equation ax2 + bx + c = 0. Quadratic
functions can have real-number or imaginary-number zeros and
quadratic equations can have real-number or imaginary-number
solutions.
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Slide 2-31
Equation-Solving Principles

The Principle of Zero Products: If ab = 0 is true, then a = 0 or
b = 0, and if a = 0 or b = 0, then ab = 0.

Example:
Solve 3x2  5x = 2.

Solution:
3x2  5x = 2
3x2  5x  2 = 0
(3x + 1)(x  2) = 0
3x + 1 = 0 or x  2 = 0
3x = 1 or
x=2
x =
1
3
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Slide 2-32
Checking the Solutions

Check:
2
 1
 1
3    5    2
 3
 3
1 5
3    2
9 3
1 5
 2
3 3
6
2
3
22
3(2)2  5(2) = 2
3(4)  10 = 2
12  10 = 2
2=2
1
The solutions are  and 2.
3
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Slide 2-33
Graphical Solution

The solutions of the equation y = 3x2  5x = 2, or the
equivalent equation 3x2  5x  2 = 0, are the zeros of
the function f(x) = 3x2  5x  2. They are also the first
coordinates of the x-intercepts of the graph of f(x) = 3x2
 5x  2.
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Slide 2-34
Equation-Solving Principles
The Principle of Square Roots:
If x2 = k, then x =
k or x =  k .
Example: Solve 2x2  6 = 0.


Solution: 2x2  6 = 0
2x2 = 6
x2 = 3
x = 3 or x =  3
Check: 2x2  6 = 0
2( 3 )2  6  0
2360
660
0=0
Solutions are
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3 and  3.
Slide 2-35
Completing the Square
To solve a quadratic equation by completing the square:






Isolate the terms with variables on one side of the equation and
arrange them in descending order.
Divide by the coefficient of the squared term if that coefficient is
not 1.
Complete the square by taking half the coefficient of the firstdegree term and adding its square on both sides of the equation.
Express one side of the equation as the square of a binomial.
Use the principle of square roots.
Solve for the variable.
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Slide 2-36
Example
Find the zeros of f(x) = x2 + 6x  16 by completing the
square.
Solution: x2 + 6x  16 = 0
x2 + 6x = 16
6
x2 + 6x + ( 62 ) = 16 + ( 2 )
x2 + 6x + 9 = 25
(x + 3)2 = 25
x + 3 =  25
x+3=5
x = 5  3 or x = 5  3
x=2
or x = 8
2
2
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Slide 2-37
Graphical Solution: x2 + 6x  16 = 0
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Slide 2-38
Quadratic Formula

The solutions of ax2 + bx + c = 0, a  0, are given by
b  b 2  4ac
.
x
2a

This formula can be used to solve any quadratic
equation.
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Slide 2-39
Example
Solve: 4x2 + 3x = 8
Solution:
4x2 + 3x  8 = 0
a = 4, b = 3, c = 8
3  32  4( 4)( 8)
x
2( 4)
The exact solutions are:
3  9  128
x
8
x
3  137
8
and
x
3  137
8
The approximate solutions are 1.088 and 1.838.
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Slide 2-40
Graphical Solution: 4x2 + 3x = 8

Graph y1 = 4x2 + 3x and y2 = 8
The approximate solutions are 1.088 and 1.838.
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Slide 2-41
Discriminant

When you apply the quadratic formula to any quadratic
equation, you find the value of b2  4ac, which can be
positive, negative, or zero.
This expression is called the discriminant.
For ax2 + bx + c = 0:
b2  4ac = 0 One real-number solution;
b2  4ac > 0 Two different real-number solutions;
b2  4ac < 0 Two different imaginary-number
solutions, complex conjugates.
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Slide 2-42
Equations Reducible to Quadratic

Some equations can be treated as quadratic, provided
that we make a suitable substitution.

Example: x4  10x2 + 9 = 0
Knowing that x4 = (x2)2, we can substitute u for x2 and
the resulting equation is then u2  10u + 9 = 0. This
equation can then be solved for u by factoring or using
the quadratic formula. Then the substitution can be
reversed by replacing u with x2, and solving for x.
Equations like this are said to be reducible to quadratic,
or quadratic in form.
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Slide 2-43
Solving an Equation Reducible to
Quadratic
Solve:
x4  10x2 + 9 = 0
u2  10u + 9 = 0
(substituting u for x2)
(u  9)(u  1) = 0
u  9 = 0 or u  1 = 0
u = 9 or
u=1
x2 = 9 or
x2 = 1 (substitute x2 for u and solve for x)
x = ±3 or
x = ±1
The solutions are 3, 3, 1, and 1.
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Slide 2-44
Application
Some applied problems can be translated to quadratic
equations.
Example: Free Fall. An acorn falls from the top of a tree
that is 32 feet tall. How long will it take to reach the
ground?
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Slide 2-45
Solving an Application
1. Familiarize. The formula s = 16t2 is used to approximate the distance s, in
feet, that an object falls freely from rest in t seconds.
2. Translate. Substitute 32 for s in the formula: 32 = 16t2.
3. Carry out. Use the principle of square roots.
32 = 16t2
32 2
t
16
32
t
16
2  t  1.414
4. Check. In 1.414 seconds, an acorn would travel a distance of 16(1.414)2, or
about 32 ft. The answer checks.
5. State. It would take about 1.414 sec for an acorn to reach the ground from
the top of a 32 ft tall tree.
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Slide 2-46
Graphical Solution: 32 = 16t2

Use the Intersect method, we
replace t with x, graph
y1 = 32 and y2 = 16x2, and find
the first coordinates of the
points of intersection. Time
cannot be negative in this
application, so we need to find
only the point of intersection
with a positive first coordinate.
The window [-5, 5, -10, 40],
with Xscl = 1 and Yscl = 5.
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Slide 2-47
2.4
Analyzing Graphs of
Quadratic Functions



Find the vertex, the axis of symmetry, and the maximum
or minimum value of a quadratic function using the
method of completing the square.
Graph quadratic functions.
Solve applied problems involving maximum and
minimum function values.
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Graphing Quadratic Functions of the
Type f(x) = a(x  h)2 + k
The graph of a quadratic function is called a parabola.
The point (h, k) at which the graph turns is called the vertex.
The maximum or minimum value of f(x) occurs at the vertex.
Each graph has a line x = h that is called the axis of symmetry.
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Slide 2-49
Example
Find the vertex, the axis of symmetry, and the maximum or minimum
value of f(x) = x2  8x + 12.
Solution: Complete the square.
f ( x)  x 2  8 x  12
 x 2  8 x  16  16  12
 ( x 2  8 x  16)  16  12
 ( x  4) 2  4
Vertex: (4, 4)
Axis of symmetry: x = 4
Minimum value of the function: 4
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Slide 2-50
Graph
f(x) = x2  8x + 12
x
y
2
0
3
3
4
4
5
3
6
0
vertex
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Slide 2-51
Example
Find the vertex, the axis of symmetry, and the maximum
or minimum value of f(x) = 3x2 + 6x  1.
Solution: Complete the square.
f ( x)  3 x 2  6 x  1
 3( x 2  2 x)  1
 3( x 2  2 x  1  1)  1
 3( x 2  2 x  1)  (3)(1)  1
 3( x 2  2 x  1)  3  1
 3( x  1) 2  2
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Slide 2-52
Example continued
Graph: f(x) = 3x2 + 6x  1
Vertex: (1, 2)
Axis of symmetry: x = 1
Maximum value of the function: 2
x
y
0
1
1
2
2
1
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Slide 2-53
Vertex of a Parabola
The vertex of the graph of f(x) = ax2 + bx + c is
 b
  2a ,

We calculate the
x-coordinate.
 b 
f    .
 2a  
We substitute to find
the y-coordinate.
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Slide 2-54
Example
For the function f(x) = 3x2 + 18x  21:
a) Find the vertex.
b) Determine whether there is a maximum or minimum
value and find that value.
c) Find the range.
d) On what intervals is the function increasing?
decreasing?
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-55
Solution
a) f(x) = 3x2 + 18x  21
b
18
The x-coordinate of the vertex is: 

3
2a
2(3)
Since f(3) = 3(3)2 + 18(3)  21 = 6,
the vertex is (3, 6).
b) Since a is negative, the graph opens down, so the
second coordinate of the vertex, 6, is the maximum
value of the function.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-56
Solution continued
c) The range is (, 6].
d) Since the graph opens down, function values increase
to the left of the vertex and decrease to the right of the
vertex. Thus the function is increasing on the interval
(, 3) and decreasing on (3, )
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-57
Application
A farmer has enough fence to enclose a rectangular pasture with
240 ft of fencing. If the barn forms one side of the rectangle, what
is the maximum area that the farmer can enclose? What should
the dimensions of the fence be in order to yield this area?
1. Familiarize. Make a drawing of the situation, using w to
represent the width of the fencing.
2. Translate. Since the area of a rectangle is
given by length times width, we have
A(w) = (240  2w)w
= 240w  2w2.
w
w
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
240  2w
Slide 2-58
Application continued
3. Carry out. We need to find the maximum value of A(w) and find the
dimensions for which that maximum occurs. The maximum will
occur at the vertex of the parabola.
b
240
240
w


 60 ft.
2a
2(2)
4
Thus, if w = 60, then the length l = 240  2(60) = 120 ft and the
area is (60)(120) = 7200 ft2.
4. Check. (60 + 60 + 120) = 240 feet of fencing.
5. State. The maximum possible area is 7200 ft2 when the pasture is
60 feet wide and 120 feet long.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-59
2.5
More Equation Solving

Solve rational and radical equations and equations with
absolute value.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Rational Equations
Equations containing rational expressions are called
rational equations.
Solving such equations requires multiplying both sides
by the least common denominator (LCD) to clear the
equation of fractions.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-61
Example
Solve:
x 1 x  3

3
2
3
 x 1 x  3 
6

  3 6
3 
 2
x 1
x3
6
 6
 18
2
3
3( x  1)  2( x  3)  18
3 x  3  2 x  6  18
x  9  18
x9
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-62
Example continued


The possible solution is 9.
Check:
x 1 x  3

3
2
3
9 1 9  3

? 3
2
3
10 6
 ? 3
2 3
52 ? 3
33

The solution is 9.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-63
Example
x2
16

x4 x4
Solve:
The LCD is x  4.
x2
16

x4 x4
x2
16
( x  4) 

 ( x  4)
x4 x4
x 2  16
x  4
or
x4
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-64
Example continued


The possible solutions are 4
and 4.
Check x = 4:
42
16

44 44
16 16

0
0
Division by 0 is not defined, so
4 is not a solution.

Check x = 4:
(4) 2
16
?
4  4
4  4
16
16
?
8
8
2  2
The number 4 checks, so it
is a solution.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-65
Radical Equations

A radical equation is an equation in which variables
appear in one or more radicands. For example:
3x  4  2 x  6  3

The Principle of Powers
For any positive integer n:
If a = b is true, then an = bn is true.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-66
Example

Solve 1  x  5  x
First we isolate the power.
1  x  5  x
x 1  x  5
( x  1) 
2

x5

2
x2  2x  1  x  5
x 2  3x  4  0
( x  4)( x  1)  0
x4
or x  1
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-67
Example continued

1  x  5  x
Check for 4
1  4  5  4

Since 4 checks but 1 does
not, the only solution is 4.
1  9  4
1  3  4

Check for 1
1   1
1  x  5  x
1  ?
1  5  (1)
1  ? 4  1
1  ? 2  1
1  3
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-68
Example

Solve:
2y  5  y  3 1
2y  5 1 y  3

2y  5
 
2
 1 y  3

2
2y  5 1 2 y  3  y  3
2y  5  y  2  2 y  3
y 3 2 y 3

( y  3)  2 y  3
2

2
y 2  6 y  9  4( y  3)
y 2  6 y  9  4 y  12
y 2  10 y  21  0
( y  3)( y  7)  0
y3
or
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
y7
Slide 2-69
Equations with Absolute Value
For a > 0 and an algebraic expression X:
|X| = a is equivalent to X = a or X = a.
x6 3
Solve:
x  6  3 or
x  6  3
x9
x3
or
The solutions are 9 and 3.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-70
2.6
Solving Linear Inequalities




Solve linear inequalities.
Solve compound inequalities.
Solve inequalities with absolute value.
Solve applied problems using inequalities.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Inequalities

An inequality is a sentence with <, >, , or  as its verb.
Examples: 5x  7 < 3 + 4x
3(x + 6)  4(x  3)
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-72
Principles for Solving Inequalities
For any real numbers a, b, and c:
The Addition Principle for Inequalities: If a < b is true,
then a + c < b + c is true.
The Multiplication Principle for Inequalities: If a < b and c > 0 are true,
then ac < bc is true. If a < b and c < 0, then ac > bc is true.
Similar statements hold for a  b.
When both sides of an inequality are multiplied or divided by a
negative number, we must reverse the inequality sign.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-73
Examples
Solve:
Solve:
4 x  6  2 x  10
4x  2x  4
6( x  3)  7( x  2)
6 x  18  7 x  14
2 x  4
x  4
x  2
x  4
{x|x < 2} or (, 2)
[
)
–5
{x|x  4} or (4, )
0
5
–5
0
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
5
Slide 2-74
Compound Inequalities
When two inequalities are joined by the word and or the
word or, a compound inequality is formed.
Conjunction contains the word and.
Example: 7 < 3x + 5 and 3x + 9  6
Disjunction contains the word or.
Example: 3x + 5  6 or 3x + 6 > 12
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-75
Examples
Solve:
4  3x  8  11
4  3x  8  11
12  3x  3
4  x  1
4 x  5  3 or
4x  2
0
4x  5  3
4x  8
1
x
2
x2
]
(
–5
Solve:
4x  5  3 or 4x  5 > 3
] (
5
–5
0
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
5
Slide 2-76
Inequalities with Absolute Value
Inequalities sometimes contain absolute-value notation.
The following properties are used to solve them.
For a > 0 and an algebraic expression X:
|X| < a is equivalent to a < X < a.
|X| > a is equivalent to X < a or X > a.
Similar statements hold for |X|  a and |X|  a.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-77
Example

Solve:
4x  1  3
3  4 x  1  3
4  4 x  2
1
1  x 
2
( )
–5
0
5
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-78
Application
Johnson Catering charges $100 plus $30 per hour
to cater an event. Catherine’s Catering charges
$50 per hour. For what lengths of time does it cost
less to hire Catherine’s Catering?
1. Familiarize. Read the problem.
2. Translate. Catherine’s
50x
is less than Johnson
<
100 + 30x
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-79
Application continued
3. Carry out.
4. Check.
50 x  100  30 x
20 x  100
x5
50(5)  ? 100  30(5)
250  ? 100  150
250  250
5. State. For values of x < 5 hr, Catherine’s Catering
will cost less.
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 2-80