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1
Equations and
Inequalities
Sections 1.1–1.4
Copyright © 2013, 2009, 2005 Pearson Education. Inc.
1-1
1 Equations and Inequalities
1.1 Linear Equations
1.2 Applications and Modeling with
Linear Equations
1.3 Complex Numbers
1.4 Quadratic Equations
Copyright © 2013, 2009, 2005 Pearson Education. Inc.
1-2
1.1 Linear Equations
Basic Terminology of Equations ▪ Solving Linear Equations ▪
Identities, Conditional Equations, and Contradictions ▪ Solving
for a Specified Variable (Literal Equations)
Copyright © 2013, 2009, 2005 Pearson Education. Inc.
1-3
1.1
Example 1 Solving a Linear Equation
Solve
(page 81)
.
Distributive property
Combine terms.
Add 4 to both sides.
Add 12x to both sides.
Combine terms.
Divide both sides by 4.
Solution set: {6}
Copyright © 2013, 2009, 2005 Pearson Education. Inc.
1-4
1.1
Example 2 Solving a Linear Equation with Fractions
(page 81)
3x  6 1
2
33
Solve
 x x
10
2
5
5
3x  6 1
2
33
 x x
10
2
5
5
33  Multiply by 10, the LCD of
 3x  6 1 
2
10 
 x   10  x 

2 
5  all the fractions.
 10
5
Distributive property
3x  6  5x  4x  66
Combine terms.
2x  6  4 x  66
6 x  60
x  10
Add –4x and –6 to both
sides. Combine terms.
Divide both sides by –6.
Solution set: {–10}
Copyright © 2013, 2009, 2005 Pearson Education. Inc.
1-5
1.1
Example 3(a) Identifying Types of Equations (page 82)
Decide whether the equation is an identity, a
conditional equation, or a contradiction. Give the
solution set.
Add –4x and 9 to both
sides. Combine terms.
Divide both sides by 2.
This is a conditional equation.
Solution set: {11}
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1.1
Example 3(b) Identifying Types of Equations (page 82)
Decide whether the equation is an identity, a
conditional equation, or a contradiction. Give the
solution set.
Distributive property
Subtract 14x from both
sides.
This is a contradiction.
Solution set: ø
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1.1
Example 3(c) Identifying Types of Equations (page 82)
Decide whether the equation is an identity, a
conditional equation, or a contradiction. Give the
solution set.
Distributive property
Combine terms.
Add x and –3 to both
sides.
This is an identity.
Solution set: {all real numbers}
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1-8
1.1
Example 4 Solving for a Specified Variable
(page 83)
Solve for the specified variable.
(a) d = rt, for t
Divide both sides by r.
(b)
, for k
Factor out k.
Divide both sides by
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1.1 Example 4 Solving for a Specified Variable (cont.)
Solve for the specified variable.
(c)
, for y
Distributive property
Subtract 8y and 8 from both
sides.
Divide both sides by 3.
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1.1
Example 5 Applying the Simple Interest Formula
(page 84)
Jared North borrowed $2580 to buy new kitchen
appliances for his home. He will pay off the loan in 9
months at an annual simple interest rate of 6.0%.
How much interest will he pay?
Caden will pay $116.10 in interest.
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1.2
Applications and Modeling with Linear
Equations
Solving Applied Problems ▪ Geometry Problems ▪ Motion
Problems ▪ Mixture Problems ▪ Modeling with Linear Equations
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1.2
Example 1 Find the Dimensions of a Square
(page 87)
The length of a rectangle is 2 in. more than the width.
If the length and width are each increased by 3 in.,
the perimeter of the new rectangle will be 4 in. less
than 8 times the width of the original rectangle. Find
the dimensions of the original rectangle.
Assign variables:
Let x = the length of the original rectangle.
Then, x − 2 = the width of the original rectangle.
x + 3 = the length of the new rectangle
(x − 2) + 3 = x + 1 = the width of the new rectangle.
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1.2
Example 1 Find the Dimensions of a Square
(cont.)
The perimeter of the new rectangle is
The perimeter of the new rectangle is 4 in. less than
8 times the width of the original rectangle, so we
have
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1.2
Example 1 Find the Dimensions of a Square
(cont.)
Distributive property.
Combine terms.
Add –4x and 20 to
both sides.
Divide both sides by 4.
The length of the original rectangle is 7 in.
The width of the original rectangle is 7 – 2 = 5 in.
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1.2
Example 2 Solving a Motion Problem
(page 88)
Krissa drove to her grandmother’s house. She
averaged 40 mph driving there. She was able to
average 48 mph returning, and her driving time
was 1 hr less. What is the distance between
Krissa’s house and her grandmother’s house?
Let x = the distance between Krissa’s house and
her grandmother’s house
Use d = rt
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1.2
Example 2 Solving a Motion Problem
(cont.)
The driving time returning was 1 hour less than the
driving time going, so
Multiply by 40·48.
Distributive property.
Subtract 48x.
Divide by –8.
The distance from Krissa’s home to her
grandmother’s home is 240 mi.
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1.2
Example 3 Solving a Mixture Problem
(page 89)
How many gallons of a 25% anti-freeze solution
should be added to 5 gallons of a 10% solution to
obtain a 15% solution?
Let x = the amount of 25% solution
The number of gallons of pure antifreeze in the
25% solution plus the number of gallons of pure
antifreeze in the 10% solution must equal the
number of gallons of pure antifreeze in the 15%
solution.
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1.2
Example 3 Solving a Mixture Problem
(cont.)
Create a table to show the relationships in the problem.
Write an equation:
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1.2
Example 3 Solving a Mixture Problem
(cont.)
Distributive property.
Subtract .15x and .5.
Divide by .1.
2.5 gallons of the 25% solution should be added.
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1.2
Example 4 Solving an Investment Problem
(page 90)
Last year, Owen earned a total of $784 in interest
from two investments. He invested a total of
$28,000, part at 2.4% and the rest at 3.1%. How
much did he invest at each rate?
Let x = amount invested at 2.4%.
Then 28,000 − x = amount invested at 3.1%.
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1.2
Example 4 Solving an Investment Problem
(cont.)
Create a table to show the relationships in the problem.
Amount in
Account
Interest
Rate
x
28,000 – x
2.4%
3.1%
Interest
0.024x
0.031(28,000 – x)
784
The amount of interest from the 2.4% account
plus the amount of interest from the 3.1%
account must equal the total amount of
interest.
0.024 x  0.031(28,000  x )  784
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1.2
Example 4 Solving an Investment Problem
(cont.)
0.024 x  0.031(28,000  x )  784
0.024 x  868  0.031x  784
868  0.007 x  784
Distributive property
Combine terms.
Subtract 868.
Divide by –.007.
Owen invested $12,000 at 2.4% and
$28,000 − $12,000 = $16,000 at 3.1%.
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1.2
Example 5 Modeling the Prevention of Indoor Pollutants
(page 91)
A range hood removes contaminants at a rate of F
liters of air per second. The percent P of contaminants
that are also removed from the surrounding air can be
modeled by the linear equation
where 10 ≤ F ≤ 75. What flow F must a range hood
have to remove 70% of the contaminants from the air?
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1.2
Example 5 Modeling the Prevention of Indoor Pollutants
(cont.)
Substitute P = 70 into
for F.
and solve
Subtract 7.18.
Divide by 1.06.
The flow rate must be approximately 59.26 L per
second to remove 70% of the contaminants.
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1.2
Example 6(a) Modeling Health Care Costs
(page 91)
The projected per capita health care expenditures
in the United States, where y is in dollars, x years
after 2000, are given by the linear equation
y  343 x  4512
What were the per capita health care expenditures
in 2007?
Let x = 2007 – 2000 = 7, then find the value of y.
y  343 x  4512
y  343(7)  4512  6913
In 2007, per capita health care expenditures were
$6913
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1.2
Example 6(b) Modeling Health Care Costs
(page 91)
When will the per capita expenditures reach $8500?
Let y = 8500, then solve for x.
y  343 x  4512
8500  343 x  4512
3988  343x
11.6  x
Subtract 4512.
Divide by 343.
Per capita health care expenditures are projected
to reach $8500 eleven years after 2000, or in 2011.
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1.3 Complex Numbers
Basic Concepts of Complex Numbers ▪ Operations on Complex
Numbers
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1.3
Example 1 Writing √–a as i√a
(page 98)
Write as the product of a real number and i.
(a)
(b)
(c)
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1.3
Example 2 Finding Products and Quotients Involving
Negative Radicands (page 99)
Multiply or divide. Simplify each answer.
(a)
(b)
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1.3
Example 2 Finding Products and Quotients Involving
Negative Radicands (cont.)
Multiply or divide. Simplify each answer.
(c)
(d)
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1.3
Example 3 Simplifying a Quotient Involving a Negative
Radicand (page 99)
Write
in standard form a + bi.
Factor.
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1.3
Example 4 Adding and Subtracting Complex Numbers
(page 100)
Find each sum or difference.
(a) (4 – 5i) + (–5 + 8i) = [4 + (–5)] + (–5i + 8i)
= –1 + 3i
(b) (–10 + 7i) – (5 – 3i) = (–10 – 5) + (7i + 3i)
= –15 + 10i
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1.3
Example 5 Multiplying Complex Numbers (page 100)
Find each product.
(a) (5 + 3i)(2 – 7i) = 5(2) + (5)(–7i) + (3i)(2) + (3i)(–7i)
= 10 – 35i + 6i – 21i2
= 10 – 29i – 21(–1)
= 31 – 29i
(b) (4 – 5i)2 = 42 + 2(4)(-5i) + (-5i)2
= 16 – 40i + 25i2
= 16 – 40i + 25(–1)
= –9 – 40i
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1.3
Example 5 Multiplying Complex Numbers (cont.)
Find the product.
(c) (9 – 8i)(9 + 8i) = 92 – (8i)2
= 81 – 64i2
= 81 – 64(–1)
= 81 + 64
= 145 or 145 + 0i
This screen shows how the
TI–83/84 Plus displays the
results in this example.
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1.3
Example 6(a) Dividing Complex Numbers (page 101)
Write in standard form a + bi.
Multiply the numerator and
denominator by the complex
conjugate of the denominator.
Multiply.
i2 = –1
Combine terms.
Lowest terms; standard form
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1.3
Example 6(b) Dividing Complex Numbers (page 101)
Write in standard form a + bi.
Multiply the numerator and
denominator by the complex
conjugate of the denominator.
Multiply.
–i2 = 1
Lowest terms; standard form
This screen shows how the
TI–83/84 Plus displays the
results in this example.
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1.3
Example 7 Simplifying Powers of i (page 102)
Simplify each power of i.
(a)
(b)
Write the given power as a product involving
or
.
(a)
(b)
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1.4 Quadratic Equations
Solving a Quadratic Equation ▪ Completing the Square ▪
The Quadratic Formula ▪ Solving for a Specified Variable ▪
The Discriminant
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1.4
Example 1 Using the Zero-Factor Property (page 105)
Solve
.
Factor.
or
or
Set each factor
equal to 0 and then
solve for x.
or
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1.4
Example 1 Using the Zero-Factor Property (cont.)
Now check.
Solution set:
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1.4
Example 2 Using the Square Root Property (page 106)
Solve each quadratic equation.
(a)
(b)
(c)
Generalized square root property
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1.4
Example 3 Using the Method of Completing the Square,
a = 1 (page 107)
Solve
by completing the square.
Rewrite the equation so that the constant is alone on one side
of the equation.
Square half the coefficient of x, and add this square to both
sides of the equation.
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1.4
Example 3 Using the Method of Completing the Square,
a = 1 (cont.)
Factor the resulting trinomial as a perfect square and combine
terms on the other side.
Use the square root property to complete the solution.
Solution set:
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1.4
Example 4 Using the Method of Completing the Square,
a ≠ 1 (page 107)
Solve
by completing the square.
Divide both sides of the equation by a, 4.
Rewrite the equation so that the constant is alone on one side
of the equation.
Square half the coefficient of x, and add this square to both
sides of the equation.
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1.4
Example 4 Using the Method of Completing the Square,
a ≠ 1 (cont.)
Factor the resulting trinomial as a perfect square and combine
terms on the other side.
Use the square root property to complete the solution.
Solution set:
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1.4
Example 5 Using the Quadratic Formula (Real Solutions)
(page 108)
Solve
.
Write the equation in standard form.
a = 1, b = 6, c = –3
Quadratic formula
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1.4
Example 5 Using the Quadratic Formula (Real Solutions)
(cont.)
Solution set:
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1.4
Example 6 Using the Quadratic Formula
(Nonreal Complex Solutions)
Solve
(page 109)
.
Write the equation in standard form.
a = 4, b = –3, c = 5
Quadratic formula
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1.4
Example 6 Using the Quadratic Formula
(Nonreal Complex Solutions)
(page 109)
Solution set:
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1.4
Example 7 Solving a Cubic Equation (page 109)
Solve
.
Factor as a difference
of cubes.
Zero-factor property
or
Quadratic formula
with a = 1, b = 5,
c = 25
Solution set:
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1.4
Example 8(a) Solving for a Quadratic Variable in a
Formula (page 110)
Solve
roots.
for r. Use ± when taking square
Goal: Isolate r.
Multiply by 3.
Divide by πh.
Square root property
Rationalize the
denominator.
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1-52
1.4
Example 8(b) Solving for a Quadratic Variable in a
Formula (page 110)
Solve
taking square roots.
for y. Use ± when
Write in standard form.
Use the quadratic
formula with a = 2m,
b = –n, c = –3p.
Simplify.
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1.4
Example 9(a) Using the Discriminant (page 111)
Determine the number of distinct solutions, and tell
whether they are rational, irrational, or nonreal
complex numbers.
a = 4, b = –12, c = 9
There is one distinct rational solution.
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1.4
Example 9(b) Using the Discriminant (page 111)
Determine the number of distinct solutions, and tell
whether they are rational, irrational, or nonreal
complex numbers.
Write in standard form.
a = 3, b = 1, c = 5
There are two distinct nonreal complex solutions.
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1.4
Example 9(c) Using the Discriminant (page 111)
Determine the number of distinct solutions, and tell
whether they are rational, irrational, or nonreal
complex numbers.
Write in standard form.
a = 2, b = –6, c = –7
There are two distinct irrational solutions.
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