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4.1 Properties of Radicals
4.2 Solving Quadratic Equations by Graphing
4.3 Solving Quadratic Equations Using Square Roots
4.4 Solving Quadratic Equations by Completing the Square
4.5 Solving Quadratic Equations Using the Quadratic Formula
4.6 Complex Numbers
4.7 Solving Quadratic Equations with Complex Solutions
4.8 Solving Nonlinear Systems of Equations
4.9 Quadratic Inequalities
Solving Quadratic
Equations
Parthenon (p. 195)
Electrical Circuits (p. 240)
Half-pipe (p. 223)
Robot-Building Competition (p. 265)
Feeding GannetFe eddiing Gannet (p. 250)((p. 25 )0)
P th ( 195)
SEE the Big Idea
q
Robbot-Builildiding Compe iti ition ((p. 26 5))
Electrical Circuits (p. 240)
Maintaining Mathematical ProficiencyMaintaining Mathematical Proficiency
Factoring Perfect Square Trinomials
Example 1 Factor x^2 + 14 x + 49.
x^2 + 14 x + 49 = x^2 + 2( x )(7) + 72 Write as a^2 + 2 ab + b^2. = ( x + 7)^2 Perfect square trinomial pattern
Factor the trinomial.
1. x^2 + 10 x + 25 2. x^2 − 20 x + 100 3. x^2 + 12 x + 36 4. x^2 − 18 x + 81 5. x^2 + 16 x + 64 6. x^2 − 30 x + 225
Solving Systems of Linear Equations by Graphing
Example 2 Solve the system of linear equations by graphing.
y = 2 x + 1 Equation 1
y = −
— 3 x + 8 Equation 2
Step 1 Graph each equation. Step 2 Estimate the point of intersection. The graphs appear to intersect at (3, 7). Step 3 Check your point from Step 2. Equation 1 Equation 2
y = 2 x + 1 y = −
— 3 x + 8
— 3
7 = 7 ✓ 7 = 7 ✓
The solution is (3, 7).
Solve the system of linear equations by graphing.
7. y = − 5 x + 3 8. y = 3 — 2 x − 2 9. y = 1 — 2 x + 4
y = 2 x − 4 y = − (^1) — 4 x^ +^5 y^ =^ −^3 x^ −^3
10. ABSTRACT REASONING What value of c makes x^2 + bx + c a perfect square trinomial?
2
4
6
− 2 2 4 6 x
y
y = − 13 x + 8
y = 2 x + 1
Dynamic Solutions available at BigIdeasMath.com
Section 4.1 Properties of Radicals 191
Essential QuestionEssential Question How can you multiply and divide square roots?
Operations with Square Roots
Work with a partner. For each operation with square roots, compare the results obtained using the two indicated orders of operations. What can you conclude? a. Square Roots and Addition
Is √
— 36 + √
— 64 equal to √
— 36 + 64?
In general, is √— a + √
— b equal to √
— a + b? Explain your reasoning.
b. Square Roots and Multiplication
Is √
— (^4) ⋅√
— 9 equal to √
— (^4) ⋅ 9?
In general, is √— a (^) ⋅√
— b equal to √
— a (^) ⋅ b? Explain your reasoning.
c. Square Roots and Subtraction
Is √
— 64 − √
— 36 equal to √
— 64 − 36?
In general, is √— a − √
— b equal to √
— a − b? Explain your reasoning.
d. Square Roots and Division
Is
— 100 — √
— 4
equal to (^) √
— —^100 4
In general, is √ — a — √
— b
equal to (^) √
— a — b
? Explain your reasoning.
Writing Counterexamples
Work with a partner. A counterexample is an example that proves that a general statement is not true. For each general statement in Exploration 1 that is not true, write a counterexample different from the example given.
Communicate Your AnswerCommunicate Your Answer
3. How can you multiply and divide square roots? 4. Give an example of multiplying square roots and an example of dividing square roots that are different from the examples in Exploration 1. 5. Write an algebraic rule for each operation. a. the product of square roots b. the quotient of square roots
REASONING
ABSTRACTLY
To be profi cient in math, you need to recognize and use counterexamples.
Properties of Radicals
192 Chapter 4 Solving Quadratic Equations
4.1 Lesson^ What You Will LearnWhat You Will Learn
Use properties of radicals to simplify expressions. Simplify expressions by rationalizing the denominator. Perform operations with radicals.
Using Properties of Radicals
A radical expression is an expression that contains a radical. An expression involving a radical with index n is in simplest form when these three conditions are met.
- No radicands have perfect n th powers as factors other than 1.
- No radicands contain fractions.
- No radicals appear in the denominator of a fraction. You can use the property below to simplify radical expressions involving square roots.
Using the Product Property of Square Roots
a. √
— 108 = √
— (^36) ⋅ 3 Factor using the greatest perfect square factor.
= √
— (^36) ⋅√
— 3 Product Property of Square Roots
= 6 √
— 3 Simplify.
b. √
— 9 x^3 = √
— (^9) ⋅ x^2 ⋅ x Factor using the greatest perfect square factor.
= √
— (^9) ⋅√
— x^2 ⋅√— x Product Property of Square Roots
= 3 x √— x Simplify.
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Simplify the expression.
1. √
— 24 2. −√
— 80 3. √
— 49 x^3 4. √
— 75 n^5
STUDY TIP
There can be more than one way to factor a radicand. An efficient method is to find the greatest perfect square factor.
STUDY TIP
In this course, whenever a variable appears in the radicand, assume that it has only nonnegative values.
counterexample, p. 191 radical expression, p. 192 simplest form of a radical, p. 192 rationalizing the denominator, p. 194 conjugates, p. 194 like radicals, p. 196 Previous radicand perfect cube
Core VocabularyCore Vocabullarry
CoreCore ConceptConcept
Product Property of Square Roots
Words The square root of a product equals the product of the square roots of the factors. Numbers √
— (^9) ⋅ 5 = √
— (^9) ⋅√
— 5 = 3 √
— 5 Algebra √
— ab = √— a (^) ⋅√
— b , where a , b ≥ 0
CoreCore ConceptConcept
Quotient Property of Square Roots
Words The square root of a quotient equals the quotient of the square roots of the numerator and denominator.
Numbers (^) √
— (^3) — 4
— 3 — √
— 4
— 3 — 2
Algebra (^) √
— a — b
√— a — √
— b
, where a ≥ 0 and b > 0
194 Chapter 4 Solving Quadratic Equations
Rationalizing the Denominator
When a radical is in the denominator of a fraction, you can multiply the fraction by an appropriate form of 1 to eliminate the radical from the denominator. This process is called rationalizing the denominator.
Rationalizing the Denominator
a.
— 5 — √
— 3 n
— 5 — √
— 3 n
⋅
— 3 n — √
— 3 n
Multiply by √
— 3 n — √
— 3 n
— 15 n — √
— 9 n^2
Product Property of Square Roots
— 15 n — √
— (^9) ⋅ √
— n^2
Product Property of Square Roots
— 15 n — 3 n
Simplify.
b.
— 3
— 9
= —^2
3
— 9
⋅
3 √
— 3 — 3 √
— 3
Multiply by
3 √
— 3 — 3 √
— 3
3 √
— 3 — 3 √
— 27
Product Property of Cube Roots
3 √
— 3 — 3
Simplify.
The binomials a √
— b + c √
— d and a √
— b − c √
— d , where a , b , c , and d are rational numbers, are called conjugates. You can use conjugates to simplify radical expressions that contain a sum or difference involving square roots in the denominator.
Rationalizing the Denominator Using Conjugates
Simplify
— 2 − √
— 3
SOLUTION
— 2 − √
— 3
— 2 − √
— 3
⋅
— 3 — 2 + √
— 3
The conjugate of 2 − √
— 3 is 2 + √
—
— 3 ) —
— 3 )
2 Sum and difference pattern
= 14 +^7 √
— 3 — 1
Simplify.
= 14 + 7 √
— 3 Simplify.
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Simplify the expression.
13. —^1
— 5
— 10 — √
— 3
15. —^7
— 2 x
16. (^) √
— 2 y^2 — 3
17. —^5
3 √
— 32
18. —^8
— 3
— 13 — √
— 5 − 2
20. —^12
— 2 + √
— 7
LOOKING FOR
STRUCTURE
Notice that the product of two conjugates a √
— b + c √
— d and a √
— b − c √
— d does not contain a radical and is a rational number.
( a √
— b + c √
—
d )(^ a √
— b − c √
—
d )
= (^ a √
—
b )
2
− (^ c √
—
d )
2
= a^2 b − c^2 d
STUDY TIP
Rationalizing the denominator works because you multiply the numerator and denominator by the same nonzero number a , which is the same as multiplying by a — a
, or 1.
Section 4.1 Properties of Radicals 195
Solving a Real-Life Problem
The distance d (in miles) that you can see to the horizon with your eye level h feet above the water is given by d = (^) √
— —^3 h 2
. How far can you see when your eye level is 5 feet above the water?
SOLUTION
d = (^) √
— —^ 3(5) 2
Substitute 5 for h.
— 15 — √
— 2
Quotient Property of Square Roots
— 15 — √
— 2 ⋅
— 2 — √
— 2
Multiply by
— —^2 √
— 2
— —^30 2
Simplify.
You can see √^
— —^30 2
, or about 2.74 miles.
Modeling with Mathematics
The ratio of the length to the width of a golden rectangle is (^1 + √
— 5 ) : 2. The dimensions of the face of the Parthenon in Greece form a golden rectangle. What is the height h of the Parthenon?
SOLUTION
1. Understand the Problem Think of the length and height of the Parthenon as the length and width of a golden rectangle. The length of the rectangular face is 31 meters. You know the ratio of the length to the height. Find the height h. 2. Make a Plan Use the ratio (^1 + √
— 5 ) : 2 to write a proportion and solve for h.
3. Solve the Problem^1 +^ √
— —^5 2
h
Write a proportion.
h (^1 + √
— 5 )^ = 62 Cross Products Property
h =
— 1 + √
— 5
Divide each side by 1 + √
—
h =
— 1 + √
— 5
⋅
— 5 — 1 − √
— 5
Multiply the numerator and denominator by the conjugate.
h = 62 −^62 √
— —^5 − 4
Simplify.
h ≈ 19.16 Use a calculator.
The height is about 19 meters.
4. Look Back
— 5 — 2
≈ 1.62 and —^31
≈ 1.62. So, your answer is reasonable.
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21. WHAT IF? In Example 6, how far can you see when your eye level is 35 feet above the water? 22. The dimensions of a dance floor form a golden rectangle. The shorter side of the dance fl oor is 50 feet. What is the length of the longer side of the dance floor?
5 ft
h
31 m
Section 4.1 Properties of Radicals 197
4.1 Exercises Dynamic Solutions available at BigIdeasMath.com
In Exercises 5–12, determine whether the expression is in simplest form. If the expression is not in simplest form, explain why.
5. √
— 19 6. (^) √
— (^1) — 7
7. √
— 48 8. √
— 34
— √
— 2
10.^3 √
— 10 — 4
11.
— 2 +
3 √
— 2
3
— 54
In Exercises 13–20, simplify the expression. (See Example 1.)
13. √
— 20 14. √
— 32
15. √
— 128 16. −√
— 72
17. √
— 125 b 18. √
— 4 x^2
19. −√
— 81 m^3 20. √
— 48 n^5
In Exercises 21–28, simplify the expression. (See Example 2.)
21. (^) √
— —^4 49
22. −√
— —^7 81
23. −√
— (^23) — 64
24. (^) √
— —^65 121
25. (^) √
— —^ a^3 49
26. (^) √
— (^144) — k^2
27. (^) √
— (^100) — 4 x^2
28. (^) √
— (^25) — v^2 36
In Exercises 29–36, simplify the expression. (See Example 3.)
29. 3
— 16 30. 3 √
— − 108
3 √
— − 64 x^5 32. − 3 √
— 343 n^2
33. (^) √^3
— —^6 c − 125
34. (^) √^3
— —^8 h^4 27
35. − (^) √^3
— —^81 y^2 1000 x^3
36. (^) √^3
— —^21 − 64 a^3 b^6
ERROR ANALYSIS In Exercises 37 and 38, describe and correct the error in simplifying the expression.
37.
—
— (^4) ⋅ 18 = √
— (^4) ⋅√
— 18
— 18
✗
3 √
— 128 y^3 — 125 =
3
— 128 y^3 — 125
3
— (^64) ⋅ (^2) ⋅ y^3 —— 125
3 √
— (^64) ⋅ 3 √
— (^2) ⋅ 3
— y^3 —— 125
=^4 y^
3 √
— 2 — 125
✗
Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with Mathematics
1. COMPLETE THE SENTENCE The process of eliminating a radical from the denominator of a radical expression is called _______________. 2. VOCABULARY What is the conjugate of the binomial √
— 6 + 4?
3. WRITING Are the expressions
— 3
— 2 x and (^) √
— 2 x — 9
equivalent? Explain your reasoning.
4. WHICH ONE DOESN’T BELONG? Which expression does not belong with the other three? Explain your reasoning.
— − 1 — 3 √ 3
— 6 1 — 6 √
— 6 √ 3
— 3
Vocabulary and Core Concept CheckVocabulary and Core Concept Check
198 Chapter 4 Solving Quadratic Equations
In Exercises 39– 44, write a factor that you can use to rationalize the denominator of the expression.
39.
— √
— 6
—
— 13 z
41.
— 3 √
— x^2
3 m — 3 √
— 4
43.
— 2 — √
— 5 − 8
— √
— 3 + √
— 7
In Exercises 45–54, simplify the expression. (See Example 4.)
45.
— √
— 2
— √
— 3
47.
— 5 — √
— 48
48. (^) √
— 4 — 52
— √— a
— √
— 2 x
51. √
— 3 d^2 — 5
— 8 — √
— 3 n^3
53.
— 3
— 25
54. (^) √^3
— 1 — 108 y^2
In Exercises 55– 60, simplify the expression. (See Example 5.)
55.
— √
— 7 + 1
— 5 − √
— 3
57.
— 10 — 7 − √
— 2
— 5 — 6 + √
— 5
59.
— √
— 5 − √
— 2
— 3 — √
— 7 + √
— 3
61. MODELING WITH MATHEMATICS The time t (in seconds) it takes an object to hit the ground is given by t = (^) √
— h — 16
, where h is the height (in feet) from which the object was dropped. (See Example 6.)
a. How long does it take an earring to hit the ground when it falls from the roof of the building? b. How much sooner does the earring hit the ground when it is dropped from two stories (22 feet) below the roof?
62. MODELING WITH MATHEMATICS The orbital period of a planet is the time it takes the planet to travel around the Sun. You can find the orbital period P (in Earth years) using the formula P = √
— d^3 , where d is the average distance (in astronomical units, abbreviated AU) of the planet from the Sun.
d = 5.2 AU
Sun
Jupiter
a. Simplify the formula. b. What is Jupiter’s orbital period?
63. MODELING WITH MATHEMATICS The electric current I (in amperes) an appliance uses is given by the formula I = (^) √
— P — R
, where P is the power (in watts) and R is the resistance (in ohms). Find the current an appliance uses when the power is 147 watts and the resistance is 5 ohms.
64. MODELING WITH MATHEMATICS You can find the average annual interest rate r (in decimal form) of a savings account using the formula r = (^) √
— V 2 — V 0
where V 0 is the initial investment and V 2 is the balance of the account after 2 years. Use the formula to compare the savings accounts. In which account would you invest money? Explain.
Account Initial investment
Balance after 2 years 1 $275 $ 2 $361 $ 3 $199 $ 4 $254 $ 5 $386 $
55 ft
200 Chapter 4 Solving Quadratic Equations
REASONING In Exercises 99 and 100, use the table shown.
—
— 3 π
2 1
— 3
— 3
π
99. Copy and complete the table by (a) fi nding each sum ( 2 + 2, 2 + 1 — 4 , etc. (^) ) and (b) fi nding each product ( (^2) ⋅ 2, 2^ ⋅ (^1) — 4 , etc.^ ). 100. Use your answers in Exercise 99 to determine whether each statement is always , sometimes , or never true. Justify your answer. a. The sum of a rational number and a rational number is rational. b. The sum of a rational number and an irrational number is irrational. c. The sum of an irrational number and an irrational number is irrational. d. The product of a rational number and a rational number is rational. e. The product of a nonzero rational number and an irrational number is irrational. f. The product of an irrational number and an irrational number is irrational. 101. REASONING Let m be a positive integer. For what values of m will the simplifi ed form of the expression √
— 2 m^ contain a radical? For what values will it not contain a radical? Explain.
102. HOW DO YOU SEE IT? The edge length s of a cube is an irrational number, the surface area is an irrational number, and the volume is a rational number. Give a possible value of s.
s
s
s
103. REASONING Let a and b be positive numbers. Explain why √
— ab lies between a and b on a number line. ( Hint: Let a < b and multiply each side of a < b by a. Then let a < b and multiply each side by b .)
104. MAKING AN ARGUMENT Your friend says that you can rationalize the denominator of the expression
— 4 + 3
— 5
by multiplying the numerator
and denominator by 4 − 3
—
- Is your friend correct? Explain. 105. PROBLEM SOLVING The ratio of consecutive terms an — an − 1
in the Fibonacci sequence gets closer and
closer to the golden ratio
— 5 — 2
as n increases. Find the term that precedes 610 in the sequence.
106. THOUGHT PROVOKING Use the golden ratio
— 5 — 2 and the golden ratio conjugate
— 5 — 2
for each of the following. a. Show that the golden ratio and golden ratio conjugate are both solutions of x^2 − x − 1 = 0. b. Construct a geometric diagram that has the golden ratio as the length of a part of the diagram.
107. CRITICAL THINKING Use the special product pattern ( a + b )( a^2 − ab + b^2 ) = a^3 + b^3 to simplify the expression
— 3 √ — x + 1. Explain your reasoning.
Maintaining Mathematical ProficiencyMaintaining Mathematical Proficiency
Graph the linear equation. Identify the x -intercept. (Skills Review Handbook)
108. y = x − 4 109. y = − 2 x + 6 110. y = − —^13 x − 1 111. y = (^3) — 2 x + 6
Find the product. (Section 2.3)
112. ( z + 3)^2 113. (3 a − 5 b )^2 114. ( x + 8)( x − 8) 115. (4 y + 2)(4 y − 2)
Reviewing what you learned in previous grades and lessons
Section 4.2 Solving Quadratic Equations by Graphing 201
Essential QuestionEssential Question How can you use a graph to solve a quadratic
equation in one variable?
Based on the definition of an x -intercept of a graph, it follows that the x -intercept of the graph of the linear equation y = ax + b 2 variables is the same value as the solution of ax + b = 0. 1 variable You can use similar reasoning to solve quadratic equations.
Solving a Quadratic Equation by Graphing
Work with a partner. a. Sketch the graph of y = x^2 − 2 x. b. What is the definition of an x -intercept of a graph? How many x -intercepts does this graph have? What are they? c. What is the definition of a solution of an equation in x? How many solutions does the equation x^2 − 2 x = 0 have? What are they? d. Explain how you can verify the solutions you found in part (c).
Solving Quadratic Equations by Graphing
Work with a partner. Solve each equation by graphing. a. x^2 − 4 = 0 b. x^2 + 3 x = 0 c. − x^2 + 2 x = 0 d. x^2 − 2 x + 1 = 0 e. x^2 − 3 x + 5 = 0 f. − x^2 + 3 x − 6 = 0
Communicate Your AnswerCommunicate Your Answer
3. How can you use a graph to solve a quadratic equation in one variable? 4. After you fi nd a solution graphically, how can you check your result algebraically? Check your solutions for parts (a)−(d) in Exploration 2 algebraically. 5. How can you determine graphically that a quadratic equation has no solution?
MAKING SENSE
OF PROBLEMS
To be profi cient in math, you need to check your answers to problems using a different method and continually ask yourself, “Does this make sense?”
Solving Quadratic Equations
by Graphing
− 4
− 2
2
4
6
− 6 − 4 2 4 6 x
y
(−2, 0)
22
44
66
22 44 66
The x -intercept of the graph of y = x + 2 is −2. (^) The solution of the equation x + 2 = 0 is x = −2.
2
4
6
8
10
− 6 − 4 − 2 2 4 6 x − 2
− 4
y
Section 4.2 Solving Quadratic Equations by Graphing 203
Solving a Quadratic Equation: One Real Solution
Solve x^2 − 8 x = −16 by graphing.
SOLUTION
Step 1 Write the equation in standard form. x^2 − 8 x = − 16 Write original equation. x^2 − 8 x + 16 = 0 Add 16 to each side. Step 2 Graph the related function y = x^2 − 8 x + 16. Step 3 Find the x -intercept. The only x -intercept is at the vertex, (4, 0).
So, the solution is x = 4.
Solving a Quadratic Equation: No Real Solutions
Solve − x^2 = 2 x + 4 by graphing.
SOLUTION
Method 1 Write the equation in standard form, x^2 + 2 x + 4 = 0. Then graph the related function y = x^2 + 2 x + 4, as shown at the left.
There are no x -intercepts. So, − x^2 = 2 x + 4 has no real solutions.
Method 2 Graph each side of the equation. y = − x^2 Left side y = 2 x + 4 Right side
The graphs do not intersect. So, − x^2 = 2 x + 4 has no real solutions.
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Solve the equation by graphing.
4. x^2 + 36 = 12 x 5. x^2 + 4 x = 0 6. x^2 + 10 x = − 25 7. x^2 = 3 x − 3 8. x^2 + 7 x = − 6 9. 2 x + 5 = − x^2
ANOTHER WAY
You can also solve the equation in Example 2 by factoring. x^2 − 8 x + 16 = 0 ( x − 4)( x − 4) = 0 So, x = 4.
Number of Solutions of a Quadratic Equation
A quadratic equation has:
- two real solutions when the graph of its related function has two x -intercepts.
- one real solution when the graph of its related function has one x -intercept.
- no real solutions when the graph of its related function has no x -intercepts.
Concept SummaryConcept Summary
x
y
2 4 6
2
4
6
y = x^2 − 8 x + 16
− 2
2
4
2 x
y
22 x
y = − x^2
22
44 y = 2 x + 4 x
y
− 4 − 2 2
2
4
6
22 y = x^2 + 2 x + 4
204 Chapter 4 Solving Quadratic Equations
Finding Zeros of Functions
Recall that a zero of a function is an x -intercept of the graph of the function.
Finding the Zeros of a Function
The graph of f ( x ) = − x^2 + 2 x + 3 is shown. Find the zeros of f.
SOLUTION
The x -intercepts are −1 and 3. So, the zeros of f are −1 and 3.
The zeros of a function are not necessarily integers. To approximate zeros, analyze the signs of function values. When two function values have different signs, a zero lies between the x -values that correspond to the function values.
Approximating the Zeros of a Function
The graph of f ( x ) = x^2 + 4 x + 1 is shown. Approximate the zeros of f to the nearest tenth.
SOLUTION
There are two x -intercepts: one between −4 and −3, and another between −1 and 0. Make tables using x -values between −4 and −3, and between −1 and 0. Use an increment of 0.1. Look for a change in the signs of the function values.
x −3.9 −3.8 −3.7 −3.6 −3.5 −3.4 −3.3 −3.2 −3.
f ( x ) 0.61 0.24 −0.11 −0.44 −0.75 −1.04 −1.31 −1.56 −1.
change in signs
x −0.9 −0.8 −0.7 −0.6 −0.5 −0.4 −0.3 −0.2 −0.
f ( x ) −1.79 −1.56 −1.31 −1.04 −0.75 −0.44 −0.11 0.24 0.
The function values that are closest to 0 correspond change in signs to x -values that best approximate the zeros of the function.
In each table, the function value closest to 0 is −0.11. So, the zeros of f are about −3.7 and −0.3.
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10. Graph f ( x ) = x^2 + x − 6. Find the zeros of f. 11. Graph f ( x ) = − x^2 + 2 x + 2. Approximate the zeros of f to the nearest tenth.
Check f (−1) = −(−1) 2 + 2(−1) + 3 = 0 ✓ f (3) = − 3 2 + 2(3) + 3 = 0 ✓
ANOTHER WAY
You could approximate one zero using a table and then use the axis of symmetry to find the other zero.
x
y 4
− 4
− 2
− 2 2 4
− 44
f ( x ) = − x^2 + 2 x + 3
− 3
2
− 4 − 2 1 x
y
− 33 f ( x ) = x^2 + 4 x + 1
206 Chapter 4 Solving Quadratic Equations
4.2 Exercises Dynamic Solutions available at BigIdeasMath.com
1. VOCABULARY What is a quadratic equation? 2. WHICH ONE DOESN’T BELONG? Which equation does not belong with the other three? Explain your reasoning.
x^2 + 5 x = 20 x^2 + x − 4 = 0 x^2 − 6 = 4 x 7 x + 12 = x^2
3. WRITING How can you use a graph to find the number of solutions of a quadratic equation? 4. WRITING How are solutions, roots, x -intercepts, and zeros related?
Vocabulary and Core Concept CheckVocabulary and Core Concept Check
In Exercises 5–8, use the graph to solve the equation.
5. − x^2 + 2 x + 3 = 0 6. x^2 − 6 x + 8 = 0
y = − x^2 + 2 x + 3
2 4 x
y
− 2
4
y = x^2 − 6 x + 8
1 3 5 x
y 3
1
7. x^2 + 8 x + 16 = 0 8. − x^2 − 4 x − 6 = 0
y = x^2 + 8 x + 16
− 6 − 4 − 2 x
y 4
2
y = − x^2 − 4 x − 6
− 2
− 6
− 4 − 2 x
y
In Exercises 9–12, write the equation in standard form.
9. 4 x^2 = 12 10. − x^2 = 15 11. 2 x − x^2 = 1 12. 5 + x = 3 x^2
In Exercises 13–24, solve the equation by graphing. (See Examples 1, 2, and 3.)
13. x^2 − 5 x = 0 14. x^2 − 4 x + 4 = 0 15. x^2 − 2 x + 5 = 0 16. x^2 − 6 x − 7 = 0 17. x^2 = 6 x − 9 18. − x^2 = 8 x + 20 19. x^2 = − 1 − 2 x 20. x^2 = − x − 3 21. 4 x − 12 = − x^2 22. 5 x − 6 = x^2 23. x^2 − 2 = − x 24. 16 + x^2 = − 8 x 25. ERROR ANALYSIS Describe and correct the error in solving x^2 + 3 x = 18 by graphing.
− 6 − 4 2 x
y
− 4
2
4
y = x − (^2) + 3 x
The solutions of the equation x^2 + 3 x = 18 are x = − 3 and x = 0.
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26. ERROR ANALYSIS Describe and correct the error in solving x^2 + 6 x + 9 = 0 by graphing.
12
18
− 6 6 12 x
y
y = x^2 + 6 x + 9
The solution of the equation x^2 + 6 x + 9 = 0 is x = 9.
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Monitoring Progress and Modeling with MathematicsMonitoring Progress and Modeling with Mathematics
Section 4.2 Solving Quadratic Equations by Graphing 207
27. MODELING WITH MATHEMATICS The height y (in yards) of a flop shot in golf can be modeled by y = − x^2 + 5 x , where x is the horizontal distance (in yards).
a. Interpret the x -intercepts of the graph of the equation. b. How far away does the golf ball land?
28. MODELING WITH MATHEMATICS The height h (in feet) of an underhand volleyball serve can be modeled by h = − 16 t^2 + 30 t + 4, where t is the time (in seconds). a. Do both t -intercepts of the graph of the function have meaning in this situation? Explain. b. No one receives the serve. After how many seconds does the volleyball hit the ground?
In Exercises 29–36, solve the equation by using Method 2 from Example 3.
29. x^2 = 10 − 3 x 30. 2 x − 3 = x^2 31. 5 x − 7 = x^2 32. x^2 = 6 x − 5 33. x^2 + 12 x = − 20 34. x^2 + 8 x = 9 35. − x^2 − 5 = − 2 x 36. − x^2 − 4 = − 4 x
In Exercises 37– 42, find the zero(s) of f****. (See Example 4.)
37. 38.
In Exercises 43–46, approximate the zeros of f to the nearest tenth. (See Example 5.)
43.
x
y
− 2
2
2 5
f ( x ) = x^2 − 5 x + 3
x
y
− 1
− 4
2
− 4 − 2 1
− 44
f ( x ) = x^2 + 3 x − 1
x
y 2
− 2 2 4
f ( x ) = − x^2 + 2 x + 1
x
y
2
4
6
222 444 x f ( x ) = − x^2 + 6 x − 2
In Exercises 47–52, graph the function. Approximate the zeros of the function to the nearest tenth, if necessary.
47. f ( x ) = x^2 + 6 x + 1 48. f ( x ) = x^2 − 3 x + 2 49. y = − x^2 + 4 x − 2 50. y = − x^2 + 9 x − 6 51. f ( x ) = 1 — 2 x^2 + 2 x − 5 52. f ( x ) = − 3 x^2 + 4 x + 3 53. MODELING WITH MATHEMATICS At a Civil War reenactment, a cannonball is fired into the air with an initial vertical velocity of 128 feet per second. The release point is 6 feet above the ground. The function h = − 16 t^2 + 128 t + 6 represents the height h (in feet) of the cannonball after t seconds. (See Example 6.) a. Find the height of the cannonball each second after it is fired. b. Use the results of part (a) to estimate when the height of the cannonball is 150 feet. c. Using a graph, after how many seconds is the cannonball 150 feet above the ground?
x
y 4
− 2
− 2 4 − 22
f ( x ) = x^2 − 2 x
x
y 4
− 8
− 2
− 88
f ( x ) = x^2 + 3 x − 4
x
y 4
2
−− 444 −− 222 222 x
f ( x ) = x^2 + 2 x + 1
x
y
− 5
− 1
1
− 3
1 3
− 55
f ( x ) = − x^2 + 6 x − 9
x
y 2
− 2
− 1 − 22
f ( x ) = 2 x^2 − x − 1
x
y 2
− 1 1 2
f ( x ) = − 2 x^2 + 3 x
