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. Then a spoonful of this mixture is put into the pitcher of cream. Is there now more cream in the coffee cup or more coffee in the pitcher of cream? 10. A Melting Ice Cube An ice cube is ﬂoating in a cup of water, full to the brim, as shown in the sketch. As the ice melts, what happens? Does the cup overﬂow, or does the water level drop, or does it remain the same? (You need to know Archimedes’ Principle: A ﬂoating object displaces a volume of water whose weight equals the weight of the object.) 11. Wrapping the World A ribbon is tied tightly around the earth at the equator. How much more ribbon would you need if you raised the ribbon 1 ft above the equator everywhere? (You don’t need to know the radius of the earth to solve this problem.) 12. Irrational Powers Prove that it’s possible to raise an irrational number to an irrational General Principles 63 power and get a rational result. If a is rational, you are done. If a is irrational, consider [Hint: The number a 1212 a12.] is either rational or irrational. 13. A Perfect Cube Show that if you multiply three consecutive integers and then add the middle integer to the result, you get a perfect cube. 14. Number Patterns Find the last digit in the number 3459. [Hint: Calculate the ﬁrst few powers of 3, and look for a pattern.] 15. Number Patterns Use the techniques of solving a simpler problem and looking for a pattern to evaluate the number 39999999999992 16. Ending Up Where You Started A woman starts at a point P on the earth’s surface and walks 1 mi south, then 1 mi east, then 1 mi north, and ﬁnds herself back at P, the starting point. Describe all points P for which this is possible (there are inﬁnitely many). 17. Volume of a Truncated Pyramid The ancient Egyptians, as a result of their pyramid building, knew that the volume of a pyramid with height h and square base of side length a is V 1 3ha2 V 1 3h square top and bottom, as shown in the ﬁgure. Prove the truncated pyramid volume formula.. They were able to use this fact to prove that the volume of a truncated pyramid is
a2 ab b2, where h is the height and b and a are the lengths of the sides of the 18. Area of a Ring Find the area of the region between the two concentric circles shown in the ﬁgure. 2 19. Bhaskara’s Proof The Indian mathematician Bhaskara sketched the two ﬁgures shown here and wrote below them, “Behold!” Explain how his sketches prove the Pythagorean Theorem. Bhaskara (born 1114) was an Indian mathematician, astronomer, and astrologer. Among his many accomplishments was an ingenious proof of the Pythagorean Theorem (see Problem 19). His important mathematical book Lilavati [The Beautiful] consists of algebra problems posed in the form of stories to his daughter Lilavati. Many of the problems begin “Oh beautiful maiden, suppose...” The story is told that using astrology, Bhaskara had determined that great misfortune would befall his daughter if she married at any time other than at a certain hour of a certain day. On her wedding day, as she was anxiously watching the water clock, a pearl fell unnoticed from her headdress. It stopped the ﬂow of water in the clock, causing her to miss the opportune moment for marriage. Bhaskara’s Lilavati was written to console her. 64 Focus on Problem Solving 20. Simple Numbers (a) Use a calculator to ﬁnd the value of the expression 13 212 13 212 The number looks very simple. Show that the calculated value is correct. (b) Use a calculator to evaluate 12 16 12 13 Entrance Show that the calculated value is correct. 21. The Impossible Museum Tour A museum is in the shape of a square with six rooms to a side; the entrance and exit are at diagonally opposite corners, as shown in the ﬁgure. Each pair of adjacent rooms is joined by a door. Some very efﬁcient tourists would like to tour the museum by visiting each room exactly once. Can you ﬁnd a path for such a tour? Here are examples of attempts that failed. Exit Oops! Missed this room. Oops! No exit. Here is how you can prove that the museum tour is not possible. Imagine that the rooms are colored black and white like a checkerboard. (a) Show that the room colors alternate between white and black as
the tourists walk through the museum. (b) Use part (a) and the fact that there are an even number of rooms in the museum to con- clude that the tour cannot end at the exit. 1.1 Basic Equations 1.2 Modeling with Equations 1.3 Quadratic Equations 1.4 Complex Numbers 1.5 Other Types of Equations 1.6 Inequalities 1.7 Absolute Value Equations and Inequalities © CHAPTER 1 Equations and Inequalities The Age of the Cell Phone? Cell phones provide us with such an amazing level of communication that some people have labeled our times “the age of the cell phone.” With these small hand-held devices we can contact anyone from anywhere at any time! Of course, mathematics is crucially involved in almost every stage of the design and operation of cell phones—from the digital transmission of sound to the complex routing of calls through the network of cell phone towers. Algebra is also involved in answering much more ordinary questions about cell phones, such as these: Am I paying too much for my cell phone plan? Which plan saves me the most money for my level of calling? In this chapter we will see how equations and inequalities can help us to answer such questions. (See Problem 7 in Focus on Modeling: Making the Best Decisions, page 135.) 6565 65 66 CHAPTER 1 \| Equations and Inequalities 1.1 Basic Equations LEARNING OBJECTIVES After completing this section, you will be able to: ■ Solve linear equations ■ Solve power equations ■ Solve for one variable in terms of others x 3 is a solution of the equation 4x 7 19, because substituting x 3 makes the equation true: x 3 7 19 3 4 1 2 Equations are the basic mathematical tool for solving real-world problems. In this chapter we learn how to solve equations, as well as how to construct equations that model real-life situations. An equation is a statement that two mathematical expressions are equal. For example, 3 5 8 is an equation. Most equations that we study in algebra contain variables, which are symbols (usually letters) that stand for numbers. In the equation 4x 7 19 the letter x is the variable. We think of x as the “unknown” in the equation, and our goal is to ﬁnd the value of x that makes the equation true. The values of the unknown that make the equation true are called the solutions
or roots of the equation, and the process of ﬁnding the solutions is called solving the equation. Two equations with exactly the same solutions are called equivalent equations. To solve an equation, we try to ﬁnd a simpler, equivalent equation in which the variable stands alone on one side of the “equal” sign. Here are the properties that we use to solve an equation. (In these properties, A, B, and C stand for any algebraic expressions, and the symbol 3 means “is equivalent to.”) PROPERTIES OF EQUALITY Property 1. A B 3 A C B C Adding the same quantity to both sides of an Description 2. A B 3 CA CB C 0) 1 equation gives an equivalent equation. Multiplying both sides of an equation by the same nonzero quantity gives an equivalent equation. These properties require that you perform the same operation on both sides of an equation when solving it. Thus, if we say “add 7” when solving an equation, that is just a short way of saying “add 7 to each side of the equation.” ■ Solving Linear Equations The simplest type of equation is a linear equation, or ﬁrst-degree equation, which is an equation in which each term is either a constant or a nonzero multiple of the variable. LINEAR EQUATIONS A linear equation in one variable is an equation that is equivalent to one of the form ax b 0 where a and b are real numbers and x is the variable. SECTION 1.1 \| Basic Equations 67 Here are some examples that illustrate the difference between linear and nonlinear equations. Linear equations Nonlinear equations 4x 5 3 2 2x 8 x 2x 1 2 x 7 x 6 x 3 1x 6x 0 3 x 2x 1 Not linear; contains the square of the variable Not linear; contains the square root of the variable Not linear; contains the reciprocal of the variable E X AM P L E 1 \| Solving a Linear Equation Solve the equation 7x 4 3x 8. ▼ SO LUTI O N We solve this by changing it to an equivalent equation with all terms that have the variable x on one side and all constant terms on the other. 7x 4 3x 8 Given equation 7x 4 1 2 4 3x 8 2 1 7x 3x 12 3x 12 7x 3x 1 1 4 4x 12 # 4x 1 4 x 3 #
12 4 3x 2 Add 4 Simplify Subtract 3x Simplify Multiply by 1 4 Simplify Because it is important to CHECK YOUR ANSWER, we do this in many of our examples. In these checks, LHS stands for “left-hand side” and RHS stands for “right-hand side” of the original equation. Check Your Answer x 3: LHS RHS ✔ x = 3 LHS 7 3 1 17 2 4 x = 3 RHS 3 3 1 17 2 8 ✎ Practice what you’ve learned: Do Exercise 17. ▲ When a linear equation involves fractions, solving the equation is usually easier if we ﬁrst multiply each side by the lowest common denominator (LCD) of the fractions, as we see in the following examples. E X AM P L E 2 \| Solving an Equation That Involves Fractions Solve the equation x 6 2 3 3 4 x. ▼ SO LUTI O N side of the equation by 12 to clear the denominators. The LCD of the denominators 6, 3, and 4 is 12, so we ﬁrst multiply each 12 # a x 12 # 3 4 x 2 3 b 6 2x 8 9x 8 7x 8 7 x Multiply by LCD Distributive Property Subtract 2x Divide by 7 The solution is ✎ Practice what you’ve learned: Do Exercise 21.. x 8 7 ▲ 68 CHAPTER 1 \| Equations and Inequalities In the next example we solve an equation that doesn’t look like a linear equation, but it simpliﬁes to one when we multiply by the LCD. E X AM P L E 3 \| An Equation Involving Fractional Expressions Solve the equation. x2 x 2. ▼ SO LUTI O N The LCD of the fractional expressions is So as long as x 1 and x 2, we can multiply both sides of the equation by the LCD to get 2x Mulitiply by LCD Expand Simplify Solve The solution is x 4. Check Your Answer x 4: LHS RHS ✔ LHS 1 RHS 4 3 42 4 2 7 10 1 4 2 7 10 4 1 1 1 5 2 ✎ Practice what you’ve learned: Do Exercise 43. ▲ It is always important to check your answer, even if you never make a mistake in your
calculations. This is because you sometimes end up with extraneous solutions, which are potential solutions that do not satisfy the original equation. The next example shows how this can happen. E X AM P L E 4 \| An Equation with No Solution Solve the equation 2 5 x 4 x 1 x 4. First, we multiply each side by the common denominator, which is ▼ SO LUTI O N x 4. Check Your Answer x 4: 2 5 0 LHS 2 5 RHS 4 1 4 4 4 4 5 0 Impossible—can’t divide by 0. LHS and RHS are undeﬁned, so x 4 is not a solution 2x 8 5 x 1 2x 3 x 1 2x x 4 x 4 x 1 x 4 b Multiply by x 4 Expand Distributive Property Simplify Add 3 Subtract x But now if we try to substitute x 4 back into the original equation, we would be dividing by 0, which is impossible. So this equation has no solution. ✎ Practice what you’ve learned: Do Exercise 47. ▲ The ﬁrst step in the preceding solution, multiplying by x 4, had the effect of multiplying by 0. (Do you see why?) Multiplying each side of an equation by an expression that SECTION 1.1 \| Basic Equations 69 contains the variable may introduce extraneous solutions. That is why it is important to check every answer. ■ Solving Power Equations Linear equations have variables only to the ﬁrst power. Now let’s consider some equations that involve squares, cubes, and other powers of the variable. Such equations will be studied more extensively in Sections 1.3 and 1.5. Here we just consider basic equations that can be simpliﬁed into the form Xn a. Equations of this form are called power equations and can be solved by taking radicals of both sides of the equation. “Algebra is a merry science,” Uncle Jakob would say. “We go hunting for a little animal whose name we don’t know, so we call it x. When we bag our game we pounce on it and give it its right name.” ALBERT EINSTEIN SOLVING A POWER EQUATION The power equation X n a has the solution X 1n a X 1n a If n is even and a 0, the equation has no real solution. if n
is odd if n is even and a 0 Here are some examples of solving power equations. The equation x 5 32 has only one real solution: The equation x 4 16 has two real solutions: The equation x 5 32 has only one real solution: The equation x 4 16 has no real solutions because x 25 32 2. x 24 16 2. x 25 32 2. 24 16 does not exist. E X AM P L E 5 \| Solving Power Equations Solve each equation. x 2 5 0 (a) x 4 2 5 (b) 1 2 x ▼ SO LUTI O N 2 5 0 (a) 2 5 x x 15 The solutions are Add 5 Take the square root x 15. and x 15 © Euclid (circa 300 B.C.) taught in Alexandria, Egypt. His Elements is the most widely inﬂuential scientiﬁc book in history. For 2000 years it was the standard introduction to geometry in the schools, and for many generations it was considered the best way to develop logical reasoning. Abraham Lincoln, for instance, studied the Elements as a way to sharpen his mind. The ing by this that mathematics does not respect wealth or social status. Euclid was revered in his own time and was referred to by the title “The Geometer” or “The Writer of the Elements.” The greatness of the Elements stems from its precise, logical, and systematic treatment of geometry. For dealing with equality, Euclid lists the following rules, which he calls “common notions”: 1. Things that are equal to the same thing are equal to each other. 2. If equals are added to equals, the sums are equal. 3. If equals are subtracted from equals, the remainders are equal. story is told that King Ptolemy once asked Euclid whether there was a faster way to learn geometry than through the Elements. Euclid replied that there is “no royal road to geometry”—mean- 4. Things that coincide with one another are equal. 5. The whole is greater than the part. 70 CHAPTER 1 \| Equations and Inequalities (b) We can take the square root of each side of this equation as well. 1 x 4 2 5 x 4 15 2 Take the square root x 4 15 Add 4 The solutions are x 4 15 and x 4 15. Be sure to check that each answer satisﬁes the original equation
. ✎ Practice what you’ve learned: Do Exercises 53 and 59. We will revisit equations like the ones in Example 5 in Section 1.3. E X AM P L E 6 \| Solving Power Equations Find all real solutions for each equation. (a) x 3 8 16x4 81 (b) ▼ SO LUTI O N (a) Since every real number has exactly one real cube root, we can solve this equation by taking the cube root of each side. 1/3 3 x 1 2 1/3 8 1 x 2 2 (b) Here we must remember that if n is even, then every positive real number has two real nth roots, a positive one and a negative one. If n is even, the equation c 0 n c x x c1/n and x c1/n. 2 1 has two solutions, x 4 81 16 1/4 Divide by 16 4 x 81 16B A x 3 2 ✎ Practice what you’ve learned: Do Exercises 61 and 63. Take the fourth root 1/4 1 2 ▲ ▲ The next example shows how to solve an equation that involves a fractional power of the variable. E X AM P L E 7 \| Solving an Equation with a Fractional Power Solve the equation 5x 2/3 2 43. ▼ SO LUTI O N raise both sides of the equation to the reciprocal of that exponent. The idea is to ﬁrst isolate the term with the fractional exponent, then If n is even, the equation x n/m c has two solutions, x c m/n and x c m/n. 5x 5x 2/3 2 43 2/3 45 2/3 9 x 93/2 x 27 x Add 2 Divide by 5 Raise both sides to power 3 2 Simplify The solutions are x 27 and x 27. SECTION 1.1 \| Basic Equations 71 Check Your Answers x 27: x 27: 1 2 27 2/3 2 2 2 LHS 5 5 9 1 43 RHS 43 LHS RHS ✔ 2/3 2 2 1 27 2 2 LHS 5 5 9 1 43 RHS 43 LHS RHS ✔ ✎ Practice what you’ve learned: Do Exercise 73. ▲ ■ Solving for One Variable in Terms of Others Many formulas in the sciences involve several variables, and it is often necessary to express one of the variables in
terms of the others. In the next example we solve for a variable in Newton’s Law of Gravity. E X AM P L E 8 \| Solving for One Variable in Terms of Others Solve for the variable M in the equation This is Newton’s Law of Gravity. It gives the gravitational force F between two masses m and M that are a distance r apart. The constant G is the universal gravitational constant. F G mM 2 r ▼ SO LUTI O N Although this equation involves more than one variable, we solve it as usual by isolating M on one side and treating the other variables as we would numbers. F Gm r 2 b a M Factor M from RHS 2 r Gm b a F 2 r Gm b a Gm r 2 b a M Multiply by reciprocal of Gm 2 r 2F r Gm M Simplify The solution is M r 2F Gm. ✎ Practice what you’ve learned: Do Exercise 83. ▲ E X AM P L E 9 \| Solving for One Variable in Terms of Others The surface area A of the closed rectangular box shown in Figure 1 can be calculated from the length l, the width „, and the height h according to the formula A 2l„ 2„h 2lh Solve for „ in terms of the other variables in this equation. ▼ SO LUTI O N Although this equation involves more than one variable, we solve it as usual by isolating „ on one side, treating the other variables as we would numbers. h l „ FIGURE 1 A closed rectangular box A 2l„ 2„h 1 A 2lh 2l„ 2„h 2lh 2 Collect terms involving w Subtract 2Ih 72 CHAPTER 1 \| Equations and Inequalities 2l 2h „ 2 Factor w from RHS Divide by 2I + 2h 1 „ A 2lh A 2lh 2l 2h „ A 2lh 2l 2h. The solution is ✎ Practice what you’ve learned: Do Exercise 85. ▲ 1. ▼ CONCE PTS 1. Which of the following equations are linear? (a) x 2 2x 10 (b) 2 x 2x 1 (c) x 7 5 3x 2. Explain why each of the following equations is not linear. (a) x x 1 6 (b
) 1x 2 x (c) 3x2 2x 1 0 2 1 3. True or false? (a) Adding the same number to each side of an equation always gives an equivalent equation. (b) Multiplying each side of an equation by the same number always gives an equivalent equation. (c) Squaring each side of an equation always gives an equiva- lent equation. 4. To solve the equation x3 125, we take the root of each side. So the solution is x =. ▼ SKI LLS 5–12 ■ Determine whether the given value is a solution of the equation. 5. 4x 7 9x 3 (a) x 2 6. 2 5x 8 x (a) x 1 2 1 3 (a) x 2 (b) x 1 6 x 2 (b) x 4 (b) x 2 4x 3 x 7. 2 4 1 1 1 8. 1 1 x 4 x (a) x 2 9. 2x 1/3 3 1 (a) x 1 10. 11. 12. x 8 3/2 x x 6 (aa) x 0 x2 bx 1 4 x b 2 (a) b 0 1 2 b2 0 (b) x 4 (b) x 8 (b) x 8 (b) x b (b) x 1 b 13–50 ■ The given equation is either linear or equivalent to a linear equation. Solve the equation. 13. 2x 7 31 2 x 8 1 15. 17. x 3 2x 6 19. 7„ 15 2„ 14. 5x 3 4 3 x 5 3 1 16. 18. 4x 7 9x 13 20. 5t 13 12 5t 1 ✎ ✎ 21. 1 2 y 2 1 3 y 22. z 5 3 10 z 7 23. 24 2x 25. 4 y 1 2B A y 6 5 y 1 2 27. 29. 31. 33. 34. 35. 37. 39. 41 3x 1 2 t 6 3 t 1 61 1 3 2 2r 13 x 112 x 5 13 2 2 2 4 2 32 3x 3 2x 7 2x 4 2 3 ✎ 43. 1 z 1 2z 1 5z 10 z 1 45. u u u 1 2 4 46. 1 3 t 4 3 t 16 9 t 2 0 26. 28. 30. 32. 36. 38. 40. 42. 44.
y 1 4 6x 2x x 4 2 2x 1 x 2 4 5 1 t 1 t 3t 35 x 2 1 x 1 2 5 x 12x 5 6x 3 1 1 3 2 „ 60 ✎ 47. x 2x 4 2 1 49. 3 x 4 1 x x 2 6x 12 2 4x x 48. 50 2x 1 1 2 x 2x ✎ ✎ ✎ ✎ ✎ ✎ ✎ 51–74 ■ The given equation involves a power of the variable. Find all real solutions of the equation. 51. x 2 49 53. x 2 24 0 55. 8x 2 64 0 57. x 2 16 0 2 4 x 2 59. 2 1 61. x 3 27 63. x 4 16 0 65. x 4 64 0 x 2 2 x 3 52. x 2 18 54. x 2 7 0 56. 5x 2 125 0 58. 6x 2 100 0 2 15 x 5 60. 62. x 5 32 0 64. 64x 6 27 16 0 5 1 4 81 0 3 375 68. 67. 66. 70. 69 23 x 5 71. 73. 2x 5/3 64 0 1 2 72. x 4/3 16 0 74. 6x 2/3 216 0 75–82 ■ Find the solution of the equation correct to two decimals. 75. 3.02x 1.48 10.92 77. 2.15x 4.63 x 1.19 x 4.63 x 4.06 2.14 0.26x 1.94 3.03 2.44x x 7.24 4.19 1 2.27 0.11x 76. 8.36 0.95x 9.97 78. 3.95 x 2.32x 2.00 1.76 3.16 81. 79. 80. 2 2 2 1 1 82. 1.73x 2.12 x 1.51 83–96 ■ Solve the equation for the indicated variable. 83. PV nRT; for R 84. F G 86. 1 R for x x; for x 85. P 2l 2„; for „ 87. 88. 89. 90. ax b cx d a 2 2x a a 1 b 3 2 ; for x 6 ; 91. V 1 3 pr 2 h ; for r 93. a 2 b 2 c 2; 1 i A P 94. for b a 100 b 2 ; for
i for a 92. F G mM 2 r ; for r 95. V 4 3 pr 3 ; for r 96. x 4 y 4 z 4 100; for x SECTION 1.1 \| Basic Equations 73 ▼ APPLICATIONS 97. Shrinkage in Concrete Beams As concrete dries, it shrinks; the higher the water content, the greater the shrinkage. If a concrete beam has a water content of „ kg/m3, then it will shrink by a factor S 0.032„ 2.5 10,000 where S is the fraction of the original beam length that disappears owing to shrinkage. (a) A beam 12.025 m long is cast in concrete that contains 250 kg/m3 water. What is the shrinkage factor S? How long will the beam be when it has dried? (b) A beam is 10.014 m long when wet. We want it to shrink to 10.009 m, so the shrinkage factor should be S 0.00050. What water content will provide this amount of shrinkage? 98. Manufacturing Cost A toy maker ﬁnds that it costs C 450 3.75x dollars to manufacture x toy trucks. If the budget allows $3600 in costs, how many trucks can be made? 99. Power Produced by a Windmill When the wind blows with speed √ km/h, a windmill with blade length 150 cm generates P watts (W) of power according to the formula P 15.6 √ (a) How fast would the wind have to blow to generate. 3 10,000 W of power? (b) How fast would the wind have to blow to generate 50,000 W of power? ; mM 2 r 1 R 2 for m ; for R1 100. Food Consumption The average daily food consumption F of a herbivorous mammal with body weight x, where both F and x are measured in pounds, is given approximately by the equation F 0.3x 3/4. Find the weight x of an elephant that consumes 300 lb of food per day. 1 R1 ▼ DISCOVE RY • DISCUSSION • WRITI NG 101. A Family of Equations The equation 3x k 5 kx k 1 is really a family of equations, because for each value of k, 74 CHAPTER 1 \| Equations and Inequalities we get a different equation with the unknown x. The letter k is called a parameter
for this family. What value should we pick for k to make the given value of x a solution of the resulting equation? (a) x 0 (b) x 1 (c) x 2 102. Proof That 0 1? The following steps appear to give equivalent equations, which seem to prove that 1 0. Find the error. x 1 x 2 x x2 Given Multiply by x Subtract x Factor Divide by x 1 Given x = 1 1.2 Modeling with Equations LEARNING OBJECTIVES After completing this section, you will be able to: ■ Make equations that model real-world situations ■ Solve problems about interest ■ Solve problems about areas and lengths ■ Solve problems about mixtures and concentrations ■ Solve problems about the time needed to do a job ■ Solve problems about distance, speed, and time Many problems in the sciences, economics, ﬁnance, medicine, and numerous other ﬁelds can be translated into algebra problems; this is one reason that algebra is so useful. In this section we use equations as mathematical models to solve real-life problems. ■ Making and Using Models We will use the following guidelines to help us set up equations that model situations described in words. To show how the guidelines can help you to set up equations, we note them as we work each example in this section. GUIDELINES FOR MODELING WITH EQUATIONS 1. Identify the Variable. Identify the quantity that the problem asks you to ﬁnd. This quantity can usually be determined by a careful reading of the question that is posed at the end of the problem. Then introduce notation for the variable (call it x or some other letter). 2. Translate from Words to Algebra. Read each sentence in the problem again, and express all the quantities mentioned in the problem in terms of the variable you deﬁned in Step 1. To organize this information, it is sometimes helpful to draw a diagram or make a table. 3. Set Up the Model. Find the crucial fact in the problem that gives a relationship between the expressions you listed in Step 2. Set up an equation (or model) that expresses this relationship. 4. Solve the Equation and Check Your Answer. Solve the equation, check your answer, and express it as a sentence that answers the question posed in the problem. The following example illustrates how these guidelines are used to translate a “word problem” into the language of algebra
. SE CTI O N 1.2 \| Modeling with Equations 75 E X AM P L E 1 \| Renting a Car A car rental company charges $30 a day and 15¢ a mile for renting a car. Helen rents a car for two days, and her bill comes to $108. How many miles did she drive? ▼ SO LUTI O N Identify the variable. We are asked to ﬁnd the number of miles Helen has driven. So we let x number of miles driven Translate from words to algebra. Now we translate all the information given in the problem into the language of algebra. Check Your Answer total cost mileage cost daily cost 0.15 108 1 320 2 2 30 1 2 ✔ In Words In Algebra Number of miles driven Mileage cost (at $0.15 per mile) Daily cost (at $30 per day) x 0.15x 30 2 1 2 Set up the model. Now we set up the model. mileage cost daily cost total cost 0.15x 2 108 30 1 2 Solve. Now we solve for x. 0.15x 48 Subtract 60 x 48 0.15 x 320 Divide by 0.15 Calculator Helen drove her rental car 320 miles. ✎ Practice what you’ve learned: Do Exercise 19. ▲ In the examples and exercises that follow, we construct equations that model problems in many different real-life situations. ■ Problems About Interest When you borrow money from a bank or when a bank “borrows” your money by keeping it for you in a savings account, the borrower in each case must pay for the privilege of using the money. The fee that is paid is called interest. The most basic type of interest is simple interest, which is just an annual percentage of the total amount borrowed or deposited. The amount of a loan or deposit is called the principal P. The annual percentage paid for the use of this money is the interest rate r. We will use the variable t to stand for the number of years that the money is on deposit and the variable I to stand for the total interest earned. The following simple interest formula gives the amount of interest I earned when a principal P is deposited for t years at an interest rate r. I Prt When using this formula, remember to convert r from a percentage to a decimal. For example, in decimal form, 5% is 0.05. So at an interest rate of 5%, the interest paid on a $1000 deposit
over a 3-year period is I Prt 1000 $150. 0.05 3 1 2 1 2 76 CHAPTER 1 \| Equations and Inequalities E X AM P L E 2 \| Interest on an Investment Mary inherits $100,000 and invests it in two certiﬁcates of deposit. One certiﬁcate pays 6% 41 and the other pays % simple interest annually. If Mary’s total interest is $5025 per year, 2 how much money is invested at each rate? ▼ SO LUTI O N Identify the variable. The problem asks for the amount she has invested at each rate. So we let x the amount invested at 6% Translate from words to algebra. Since Mary’s total inheritance is $100,000, it follows that she invested 100,000 x at %. We translate all the information given into the language of algebra. 4 1 2 In Words In Algebra Amount invested at 6% 1 Amount invested at % 4 2 Interest earned at 6% 1 4 Interest earned at % 2 x 100,000 x 0.06x 0.045 1 100,000 x 2 Set up the model. We use the fact that Mary’s total interest is $5025 to set up the model. interest at 6% interest at % 4 1 2 total interest 0.06x 0.045 100,000 x 1 Solve. Now we solve for x. 5025 2 0.06x 4500 0.045x 5025 Multiply 0.015x 4500 5025 0.015x 525 Combine the x-terms Subtract 4500 x 525 0.015 35,000 Divide by 0.015 So Mary has invested $35,000 at 6% and the remaining $65,000 at %. 4 1 2 Check Your Answer total interest 6% of $35,000 41 2% of $65,000 $2100 $2925 $5025 ✔ ✎ Practice what you’ve learned: Do Exercise 21. ▲ ■ Problems About Area or Length When we use algebra to model a physical situation, we must sometimes use basic formulas from geometry. For example, we may need a formula for an area or a perimeter, or the formula that relates the sides of similar triangles, or the Pythagorean Theorem. Most of these formulas are listed in the front endpapers of this book. The next two examples use these geometric formulas to solve some real
-world problems. x FIGURE 1 SE CTI O N 1.2 \| Modeling with Equations 77 E X AM P L E 3 \| Dimensions of a Garden A square garden has a walkway 3 ft wide around its outer edge, as shown in Figure 1. If the area of the entire garden, including the walkway, is 18,000 ft2, what are the dimensions of the planted area? 3 ft ▼ SO LUTI O N Identify the variable. We are asked to ﬁnd the length and width of the planted area. So we let x the length of the planted area 3 ft Translate from words to algebra. Next, translate the information from Figure 1 into the language of algebra. In Words In Algebra Length of planted area Length of entire garden Area of entire garden x x 6 x 6 1 2 2 Set up the model. We now set up the model. area of entire garden 18,000 ft2 x 6 1 2 2 18,000 Solve. Now we solve for x. x 6 118,000 Take square roots x 118,000 6 x 128 Subtract 6 The planted area of the garden is about 128 ft by 128 ft. ✎ Practice what you’ve learned: Do Exercise 41. ▲ E X AM P L E 4 \| Determining the Height of a Building Using Similar Triangles A man 6 ft tall wishes to ﬁnd the height of a certain four-story building. He measures its shadow and ﬁnds it to be 28 ft long, while his own shadow is ft long. How tall is the building? 31 2 ▼ SO LUTI O N Identify the variable. The problem asks for the height of the building. So let h the height of the building Translate from words to algebra. We use the fact that the triangles in Figure 2 on the next page are similar. Recall that for any pair of similar triangles the ratios of corresponding sides are equal. Now we translate these observations into the language of algebra. In Words Height of building Ratio of height to base in large triangle Ratio of height to base in small triangle In Algebra h h 28 6 3.5 78 CHAPTER 1 \| Equations and Inequalities FIGURE 2 h 6 ft 28 ft 1 3 ft 2 Set up the model. Since the large and small triangles are similar, we get the equation ratio of height to base in large triangle ratio of height to base in small triangle Solve. Now we
solve for h. h 28 6 3.5 h 6 # 28 3.5 Multiply by 28 48 Multiply by 28 So the building is 48 ft tall. ✎ Practice what you’ve learned: Do Exercise 43. ▲ ■ Problems About Mixtures Many real-world problems involve mixing different types of substances. For example, construction workers may mix cement, gravel, and sand; fruit juice from concentrate may involve mixing different types of juices. Problems involving mixtures and concentrations make use of the fact that if an amount x of a substance is dissolved in a solution with volume V, then the concentration C of the substance is given by C x V So if 10 g of sugar is dissolved in 5 L of water, then the sugar concentration is C 10/5 2 g/L. Solving a mixture problem usually requires us to analyze the amount x of the substance that is in the solution. When we solve for x in this equation, we see that x CV. Note that in many mixture problems the concentration C is expressed as a percentage, as in the next example. E X AM P L E 5 \| Mixtures and Concentration A manufacturer of soft drinks advertises their orange soda as “naturally ﬂavored,” although it contains only 5% orange juice. A new federal regulation stipulates that to be called “natural,” a drink must contain at least 10% fruit juice. How much pure orange juice must this manufacturer add to 900 gal of orange soda to conform to the new regulation? ▼ SO LUTI O N Identify the variable. The problem asks for the amount of pure orange juice to be added. So let x the amount in gallons 1 2 of pure orange juice to be added SE CTI O N 1.2 \| Modeling with Equations 79 In any problem of this type—in which two different Translate from words to algebra. substances are to be mixed—drawing a diagram helps us to organize the given information (see Figure 3). 5% juice 100% juice 10% juice Volume Amount of orange juice 900 gallons x gallons 900+x gallons 5% of 900 gallons =45 gallons 100% of x gallons =x gallons 10% of 900+x gallons =0.1(900+x) gallons FIGURE 3 We now translate the information in the ﬁgure into the language of algebra. In Words In Algebra Amount of orange juice to be added Amount of the mixture Amount of orange juice in the
ﬁrst vat Amount of orange juice in the second vat Amount of orange juice in the mixture x 900 x 900 0.05 1 # x x 1 0.10 1 2 900 x 2 45 Set up the model. To set up the model, we use the fact that the total amount of orange juice in the mixture is equal to the orange juice in the ﬁrst two vats. amount of orange juice in ﬁrst vat amount of orange juice in second vat amount of orange juice in mixture 45 x 0.1 900 x 1 2 From Figure 3 Solve. Now we solve for x. 45 x 90 0.1x Distributive Property Subtract 0.1x and 45 0.9x 45 x 45 0.9 50 Divide by 0.9 The manufacturer should add 50 gal of pure orange juice to the soda. Check Your Answer amount of juice before mixing 5% of 900 gal 50 gal pure juice 45 gal 50 gal 95 gal amount of juice after mixing 10% of 950 gal 95 gal Amounts are equal. ✔ ✎ Practice what you’ve learned: Do Exercise 45. ▲ 80 CHAPTER 1 \| Equations and Inequalities ■ Problems About the Time Needed to Do a Job When solving a problem that involves determining how long it takes several workers to complete a job, we use the fact that if a person or machine takes H time units to complete the task, then in one time unit the fraction of the task that has been completed is 1/H. For example, if a worker takes 5 hours to mow a lawn, then in 1 hour the worker will mow 1/5 of the lawn. E X AM P L E 6 \| Time Needed to Do a Job Because of an anticipated heavy rainstorm, the water level in a reservoir must be lowered by 1 ft. Opening spillway A lowers the level by this amount in 4 hours, whereas opening the smaller spillway B does the job in 6 hours. How long will it take to lower the water level by 1 ft if both spillways are opened? ▼ SO LUTI O N Identify the variable. We are asked to ﬁnd the time needed to lower the level by 1 ft if both spillways are open. So let x the time (in hours) it takes to lower the water level by 1 ft if both spillways are open Translate from words to algebra. Finding an equation relating x to the other quantities
in this problem is not easy. Certainly x is not simply 4 6, because that would mean that together the two spillways require longer to lower the water level than either spillway alone. Instead, we look at the fraction of the job that can be done in 1 hour by each spillway. In Words In Algebra B A Time it takes to lower level 1 ft with A and B together Distance A lowers level in 1 h Distance B lowers level in 1 h Distance A and B together lower levels in 1 h Set up the model. Now we set up the model. ft x h 1 4 1 6 1 x ft ft fraction done by A fraction done by B fraction done by both Solve. Now we solve for x. 1 4 1 6 1 x 3x 2x 12 5x 12 x 12 5 Multiply by the LCD, 12x Add Divide by 5 hours, or 2 h 24 min, to lower the water level by 1 ft if both spillways are open. 2 5 2 It will take ✎ Practice what you’ve learned: Do Exercise 53. ▲ ■ Problems About Distance, Rate, and Time The next example deals with distance, rate (speed), and time. The formula to keep in mind here is distance rate time SE CTI O N 1.2 \| Modeling with Equations 81 where the rate is either the constant speed or average speed of a moving object. For example, driving at 60 mi/h for 4 hours takes you a distance of 60 # 4 240 mi. E X AM P L E 7 \| Distance, Speed, and Time Bill left his house at 2:00 P.M. and rode his bicycle down Main Street at a speed of 12 mi/h. When his friend Mary arrived at his house at 2:10 P.M., Bill’s mother told her the direction in which Bill had gone, and Mary cycled after him at a speed of 16 mi/h. At what time did Mary catch up with Bill? ▼ SO LUTI O N Identify the variable. We are asked to ﬁnd the time that it took Mary to catch up with Bill. Let t the time in hours 1 2 it took Mary to catch up with Bill In problems involving motion, it is often helpful to orTranslate from words to algebra. ganize the information in a table, using the formula distance rate time. First we ﬁll in the “Speed” column in the table, since
we are told the speeds at which Mary and Bill cycled. Then we ﬁll in the “Time” column. (Because Bill had a 10-minute, or -hour head start, he cycled for hours.) Finally, we multiply these columns to calculate the entries in the “Distance” column. t 1 6 1 6 Distance (mi) Speed (mi/h) Time (h) Mary Bill 16t t 1 6B A 12 16 12 t t 1 6 Set up the model. At the instant when Mary caught up with Bill, they had both cycled the same distance. We use this fact to set up the model for this problem. distance traveled by Mary distance traveled by Bill Solve. Now we solve for t. 16t 12 t 1 6B A From table 16t 12t 2 Distributive Property 4t 2 t 1 2 Subtract 12t Divide by 4 Mary caught up with Bill after cycling for half an hour, that is, at 2:40 P.M. Check Your Answer Bill traveled for 1 2 1 6 2 3 h, so distance Bill traveled 12 mi/h 2 distance Mary traveled 16 mi/h 1 ✔ 3 h 8 mi 2 h 8 mi Distances are equal. ✎ Practice what you’ve learned: Do Exercise 57. ▲ 82 CHAPTER 1 \| Equations and Inequalities 1. ▼ CONCE PTS 1. Explain in your own words what it means for an equation to model a real-world situation, and give an example. 2. In the formula I = Prt for simple interest, P stands for, r for, and t for. 3. Give a formula for the area of the geometric ﬁgure. (a) A square of side x: A =. (b) A rectangle of length l and width „: A =. (c) A circle of radius r: A =. 4. Balsamic vinegar contains 5% acetic acid, so a 32-oz bottle of balsamic vinegar contains ounces of acetic acid. ✎ ▼ APPLICATIONS 19. Renting a Truck A rental company charges $65 a day and 20 cents a mile for renting a truck. Michael rents a truck for 3 days, and his bill comes to $275. How many miles did he drive? 20. Cell Phone Costs A cell phone company charges a monthly fee of $10 for the ﬁrst 1000 text messages and 10 cents
for each additional text message. Miriam’s bill for text messages for the month of June is $38.50. How many text messages did she send that month? ✎ 21. Investments Phyllis invested $12,000, a portion earning a 1 2 4 simple interest rate of % per year and the rest earning a rate of 4% per year. After 1 year the total interest earned on these investments was $525. How much money did she invest at each rate? 5. A painter paints a wall in x hours, so the fraction of the wall 22. Investments If Ben invests $4000 at 4% interest per year, that she paints in 1 hour is. 6. The formula d rt models the distance d traveled by an object moving at the constant rate r in time t. Find formulas for the following quantities. r t. ▼ SKI LLS 7–18 ■ Express the given quantity in terms of the indicated variable. n ﬁrst integer of the 7. The sum of three consecutive integers; three 8. The sum of three consecutive integers; n middle integer of the three 9. The average of three test scores if the ﬁrst two scores are 78 and 82; s third test score 10. The average of four quiz scores if each of the ﬁrst three scores is 8; q fourth quiz score 11. The interest obtained after one year on an investment at 21 2% x number of dollars invested simple interest per year; 12. The total rent paid for an apartment if the rent is $795 a month; n number of months 13. The area (in ft2) of a rectangle that is three times as long as it is wide; „ width of the rectangle (in ft) 14. The perimeter (in cm) of a rectangle that is 5 cm longer than it is wide; „ width of the rectangle (in cm) 15. The distance (in mi) that a car travels in 45 min; s speed of the car (in mi/h) 16. The time (in hours) it takes to travel a given distance at 55 mi/h; d given distance (in mi) 17. The concentration (in oz/gal) of salt in a mixture of 3 gal of brine containing 25 oz of salt to which some pure water has been added; x volume of pure water added (in gal) 18. The value (in cents) of the change in a purse that contains twice as
many nickels as pennies, four more dimes than nickels, and as many quarters as dimes and nickels combined; p number of pennies how much additional money must he invest at % annual interest to ensure that the interest he receives each year is % of the total amount invested? 4 5 1 2 1 2 23. Investments What annual rate of interest would you have to earn on an investment of $3500 to ensure receiving $262.50 interest after 1 year? 24. Investments Jack invests $1000 at a certain annual interest rate, and he invests another $2000 at an annual rate that is onehalf percent higher. If he receives a total of $190 interest in 1 year, at what rate is the $1000 invested? 25. Salaries An executive in an engineering ﬁrm earns a monthly salary plus a Christmas bonus of $8500. If she earns a total of $97,300 per year, what is her monthly salary? 26. Salaries A woman earns 15% more than her husband. Together they make $69,875 per year. What is the husband’s annual salary? 27. Inheritance Craig is saving to buy a vacation home. He inherits some money from a wealthy uncle, then combines this with the $22,000 he has already saved and doubles the total in a lucky investment. He ends up with $134,000—just enough to buy a cabin on the lake. How much did he inherit? 28. Overtime Pay Helen earns $7.50 an hour at her job, but if times she works more than 35 hours in a week, she is paid her regular salary for the overtime hours worked. One week her gross pay was $352.50. How many overtime hours did she work that week? 1 1 2 29. Labor Costs A plumber and his assistant work together to replace the pipes in an old house. The plumber charges $45 an hour for his own labor and $25 an hour for his assistant’s labor. The plumber works twice as long as his assistant on this job, and the labor charge on the ﬁnal bill is $4025. How long did the plumber and his assistant work on this job? 30. A Riddle A father is four times as old as his daughter. In 6 years, he will be three times as old as she is. How old is the daughter now? 31. A Riddle A movie star, unwilling to give his
age, posed the following riddle to a gossip columnist: “Seven years ago, I was eleven times as old as my daughter. Now I am four times as old as she is.” How old is the movie star? 32. Career Home Runs During his major league career, Hank Aaron hit 41 more home runs than Babe Ruth hit during his career. Together they hit 1469 home runs. How many home runs did Babe Ruth hit? 33. Value of Coins A change purse contains an equal number of pennies, nickels, and dimes. The total value of the coins is $1.44. How many coins of each type does the purse contain? 34. Value of Coins Mary has $3.00 in nickels, dimes, and quarters. If she has twice as many dimes as quarters and ﬁve more nickels than dimes, how many coins of each type does she have? 35. Length of a Garden A rectangular garden is 25 ft wide. If its area is 1125 ft 2, what is the length of the garden? 25 ft x ft 36. Width of a Pasture A pasture is twice as long as it is wide. Its area is 115,200 ft 2. How wide is the pasture? 37. Dimensions of a Lot A square plot of land has a building 60 ft long and 40 ft wide at one corner. The rest of the land outside the building forms a parking lot. If the parking lot has area 12,000 ft 2, what are the dimensions of the entire plot of land? 38. Dimensions of a Lot A half-acre building lot is ﬁve times as long as it is wide. What are its dimensions? [Note: 1 acre 43,560 ft 2.] 39. Geometry Find the length y in the ﬁgure if the shaded area is 120 in2. y SECTION 1.2 \| Modeling with Equations 83 ✎ 41. Framing a Painting Ali paints with watercolors on a sheet of paper 20 in. wide by 15 in. high. He then places this sheet on a mat so that a uniformly wide strip of the mat shows all around the picture. The perimeter of the mat is 102 in. How wide is the strip of the mat showing around the picture? x. 15 in. 20 in 42. A poster has a rectangular printed area 100 cm by 140 cm and a blank strip of uniform width around the edges. The perimeter of the
poster is 1 times the perimeter of the printed area. What is the width of the blank strip? 1 2 100 cm 140 cm x x ✎ 43. Length of a Shadow A man is walking away from a lamppost with a light source 6 m above the ground. The man is 2 m tall. How long is the man’s shadow when he is 10 m from the lamppost? [Hint: Use similar triangles.] y y 40. Geometry Find the length x in the ﬁgure if the shaded area 6 m is 144 cm2. x 2 m 10 m x 10 cm 6 cm x 44. Height of a Tree A woodcutter determines the height of a tall tree by ﬁrst measuring a smaller one 125 ft away, then moving so that his eyes are in the line of sight along the tops of the trees and measuring how far he is standing from the small tree (see the ﬁgure). Suppose the small tree is 20 ft tall, the 84 CHAPTER 1 \| Equations and Inequalities man is 25 ft from the small tree, and his eye level is 5 ft above the ground. How tall is the taller tree? 5 ft 20 ft 25 ft 125 ft ✎ 45. Mixture Problem What quantity of a 60% acid solution must be mixed with a 30% solution to produce 300 mL of a 50% solution? ✎ 46. Mixture Problem What quantity of pure acid must be added to 300 mL of a 50% acid solution to produce a 60% acid solution? 47. Mixture Problem A jeweler has ﬁve rings, each weighing 18 g, made of an alloy of 10% silver and 90% gold. She decides to melt down the rings and add enough silver to reduce the gold content to 75%. How much silver should she add? 48. Mixture Problem A pot contains 6 L of brine at a concen- tration of 120 g/L. How much of the water should be boiled off to increase the concentration to 200 g/L? 49. Mixture Problem The radiator in a car is ﬁlled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests that for summer driving, optimal cooling of the engine is obtained with only 50% antifreeze. If the capacity of the radiator is 3.6 L, how much coolant should be drained and replaced with water to reduce the
antifreeze concentration to the recommended level? 50. Mixture Problem A health clinic uses a solution of bleach to sterilize petri dishes in which cultures are grown. The sterilization tank contains 100 gal of a solution of 2% ordinary household bleach mixed with pure distilled water. New research indicates that the concentration of bleach should be 5% for complete sterilization. How much of the solution should be drained and replaced with bleach to increase the bleach content to the recommended level? 51. Mixture Problem A bottle contains 750 mL of fruit punch with a concentration of 50% pure fruit juice. Jill drinks 100 mL of the punch and then reﬁlls the bottle with an equal amount of a cheaper brand of punch. If the concentration of juice in the bottle is now reduced to 48%, what was the concentration in the punch that Jill added? 52. Mixture Problem A merchant blends tea that sells for $3.00 a pound with tea that sells for $2.75 a pound to produce 80 lb of a mixture that sells for $2.90 a pound. How many pounds of each type of tea does the merchant use in the blend? ✎ 53. Sharing a Job Candy and Tim share a paper route. It takes Candy 70 min to deliver all the papers, and it takes Tim 80 min. How long does it take the two when they work together? 54. Sharing a Job Stan and Hilda can mow the lawn in 40 min if they work together. If Hilda works twice as fast as Stan, how long does it take Stan to mow the lawn alone? 55. Sharing a Job Betty and Karen have been hired to paint the houses in a new development. Working together, the women can paint a house in two-thirds the time that it takes Karen working alone. Betty takes 6 h to paint a house alone. How long does it take Karen to paint a house working alone? 56. Sharing a Job Next-door neighbors Bob and Jim use hoses from both houses to ﬁll Bob’s swimming pool. They know that it takes 18 h using both hoses. They also know that Bob’s hose, used alone, takes 20% less time than Jim’s hose alone. How much time is required to ﬁll the pool by each hose alone? 57. Distance, Speed, and Time Wendy took a trip from Davenport to Omaha, a distance of 300 mi. She traveled part of the way by bus, which arrived at
the train station just in time for Wendy to complete her journey by train. The bus averaged 40 mi/h, and the train averaged 60 mi/h. The entire trip took 5 h. How long did Wendy spend on the train? 1 2 58. Distance, Speed, and Time Two cyclists, 90 mi apart, start riding toward each other at the same time. One cycles twice as fast as the other. If they meet 2 h later, at what average speed is each cyclist traveling? 59. Distance, Speed, and Time A pilot ﬂew a jet from Montreal to Los Angeles, a distance of 2500 mi. On the return trip, the average speed was 20% faster than the outbound speed. The round-trip took 9 h 10 min. What was the speed from Montreal to Los Angeles? 60. Distance, Speed, and Time A woman driving a car 14 ft long is passing a truck 30 ft long. The truck is traveling at 50 mi/h. How fast must the woman drive her car so that she can pass the truck completely in 6 s, from the position shown in ﬁgure (a) to the position shown in ﬁgure (b)? and seconds instead of miles and hours.] [Hint: Use feet 50 mi /h 50 mi /h (a) (b) 61. Law of the Lever The ﬁgure shows a lever system, similar to a seesaw that you might ﬁnd in a children’s playground. For the system to balance, the product of the weight and its distance from the fulcrum must be the same on each side; that is, „1x1 „2 x 2 This equation is called the law of the lever and was ﬁrst discovered by Archimedes (see page 557). A woman and her son are playing on a seesaw. The boy is at one end, 8 ft from the fulcrum. If the son weighs 100 lb and the mother weighs 125 lb, where should the woman sit so that the seesaw is balanced? „ ¤ „⁄ x⁄ x ¤ 62. Law of the Lever A plank 30 ft long rests on top of a ﬂat- roofed building, with 5 ft of the plank projecting over the edge, as shown in the ﬁgure. A worker weighing 240 lb sits on one end of the plank. What is the largest weight that can
be hung on the projecting end of the plank if it is to remain in balance? (Use the law of the lever stated in Exercise 61.) 5 ft SE CTI O N 1.2 \| Modeling with Equations 85 65. Dimensions of a Structure A storage bin for corn consists of a cylindrical section made of wire mesh, surmounted by a conical tin roof, as shown in the ﬁgure. The height of the roof is one-third the height of the entire structure. If the total volume of the structure is 1400p ft 3 and its radius is 10 ft, what is its height? side back cover of this book.] [Hint: Use the volume formulas listed on the in- 1 3 h h 10 ft 66. An Ancient Chinese Problem This problem is taken from a Chinese mathematics textbook called Chui-chang suan-shu, or Nine Chapters on the Mathematical Art, which was written about 250 B.C. A 10-ft-long stem of bamboo is broken in such a way that its tip touches the ground 3 ft from the base of the stem, as shown in the ﬁgure. What is the height of the break? [Hint: Use the Pythagorean Theorem.] 63. Dimensions of a Lot A rectangular parcel of land is 50 ft wide. The length of a diagonal between opposite corners is 10 ft more than the length of the parcel. What is the length of the parcel? 64. Dimensions of a Track A running track has the shape shown in the ﬁgure, with straight sides and semicircular ends. If the length of the track is 440 yd and the two straight parts are each 110 yd long, what is the radius of the semicircular parts (to the nearest yard)? 110 yd r 3 ft ▼ DISCOVE RY • DISCUSSION • WRITI NG 67. Historical Research Read the biographical notes on Pythagoras (page 284), Euclid (page 69), and Archimedes (page 557). Choose one of these mathematicians and ﬁnd out more about him from the library or on the Internet. Write a short essay on your ﬁndings. Include both biographical information and a description of the mathematics for which he is famous. DISCOVERY PR OJECT EQUATIONS THROUGH THE AGES Equations have been used to solve problems throughout recorded history in every civilization. (See, for example
, Exercise 66 on page 85.) Here is a problem from ancient Babylon (ca. 2000 B.C.): I found a stone but did not weigh it. After I added a seventh, and then added an eleventh of the result, I weighed it and found it weighed 1 mina. What was the original weight of the stone? The answer given on the cuneiform tablet is mina, 8 sheqel, and 22 se, where 1 mina 60 sheqel and 1 sheqel 180 se. 1 2 2 3 In ancient Egypt knowing how to solve word problems was a highly prized secret. The Rhind Papyrus (ca. 1850 B.C.) contains many such problems (see page 470). Problem 32 in the Papyrus states: A quantity, its third, its quarter, added together become 2. What is the quantity? The answer in Egyptian notation is much like our notation 41. 1 4 76, where the bar indicates “reciprocal,” The Greek mathematician Diophantus (ca. 250 A.D., see page 49) wrote the book Arithmetica, which contains many word problems and equations. The Indian mathematician Bhaskara (12th century A.D., see page 63) and the Chinese mathematician Chang Ch’iu-Chien (6th century A.D.) also studied and wrote about equations. Of course, equations continue to be important today. 1. Solve the Babylonian problem and show that their answer is correct. 2. Solve the Egyptian problem and show that their answer is correct. 3. The ancient Egyptians and Babylonians used equations to solve practical problems. From the examples given here, do you think that they may have enjoyed posing and solving word problems just for fun? 4. Solve this problem from 12th century India: A peacock is perched at the top of a 15-cubit pillar, and a snake’s hole is at the foot of the pillar. Seeing the snake at a distance of 45 cubits from its hole, the peacock pounces obliquely upon the snake as it slithers home. At how many cubits from the snake’s hole do they meet, assuming that each has traveled an equal distance? 15 x 45 5. Consider this problem from 6th century China. If a rooster is worth 5 coins, a hen 3 coins, and three chicks together one coin, how many roosters, hens, and chicks, totaling
100, can be bought for 100 coins? This problem has several answers. Use trial and error to ﬁnd at least one answer. Is this a practical problem or more of a riddle? Write a short essay to support your opinion. 6. Write a short essay explaining how equations affect your own life in today’s world 86 SE CTIO N 1.3 \| Quadratic Equations 87 1.3 Quadratic Equations LEARNING OBJECTIVES After completing this section, you will be able to: ■ Solve quadratic equations by factoring ■ Solve quadratic equations by completing the square ■ Solve quadratic equations using the Quadratic Formula ■ Model with quadratic equations Linear Equations 4x 7 6x 8 21 2 3x 1 2 3 4 x Quadratic Equations x 2 2x 8 0 3x 10 4x Check Your Answers 15 24 8 1 2 2 5 8 1 2 64 40 24 2x 1 5 In Section 1.1 we learned how to solve linear equations, which are ﬁrst-degree equations. In this section we learn how to solve quadratic equasuch as 2x 2 3 5x. tions, which are second-degree equations such as We will also see that many real-life problems can be modeled using quadratic equations. x 2 2x 3 0 4 3x 2 or or QUADRATIC EQUATIONS A quadratic equation is an equation of the form ax 2 bx c 0 where a, b, and c are real numbers with a 0. ■ Solving Quadratic Equations by Factoring Some quadratic equations can be solved by factoring and using the following basic property of real numbers. ZERO-PRODUCT PROPERTY AB 0 if and only if A 0 or B 0 This means that if we can factor the left-hand side of a quadratic (or other) equation, then we can solve it by setting each factor equal to 0 in turn. This method works only when the right-hand side of the equation is 0. E X AM P L E 1 \| Solving a Quadratic Equation by Factoring Solve the equation x 2 5x 24. ▼ SO LUTI O N We must ﬁrst rewrite the equation so that the right-hand side is 0. ✔ ✔ x 2 5x 24 2 5x 24 0 x 8 0 x 3 0 or
Given equation Subtract 24 Factor Zero-Product Property x 3 x 8 Solve The solutions are x 3 and x 8. ✎ Practice what you’ve learned: Do Exercise 5. ▲ 88 CHAPTER 1 \| Equations and Inequalities Completing the Square Area of blue region is x2 2 b 2 b a x x2 bx Add a small square of area “complete” the square. 1 b/2 2 2 to 24 x 5 # 48, 6 # 4, 1 tion as inﬁnitely many ways, such as Do you see why one side of the equation must be 0 in Example 1? Factoring the equadoes not help us ﬁnd the solutions, since 24 can be factored in # 2 5B A ■ Solving Quadratic Equations by Completing the Square As we saw in Section 1.1, Example 5(b), if a quadratic equation is of the form x a, then we can solve it by taking the square root of each side. In an equation of 1 this form the left-hand side is a perfect square: the square of a linear expression in x. So if a quadratic equation does not factor readily, then we can solve it by completing the square., and so on. 2 c 60 2 1 2 COMPLETING THE SQUARE x 2 bx To make of x. This gives the perfect square a perfect square, add 2, the square of half the coefﬁcient b 2 b a x 2 bx To complete the square, we add a constant to a quadratic expression to make it a perfect square. For example, to make a perfect square, we must add 2 6x x A 6 2B x 2 9. Then 2 6x 9 x 3 2 2 1 is a perfect square. The table gives some more examples of completing the square. Expression Add Complete the square x 2 8x x 2 12x x 2 3x 2 13x x 2 8 2 b a 16 2 8x 16 x x 4 2 2 1 2 36 2 12x 36 x a a a 12 2 b 2 3 2 b 13 2 b 9 4 3 4 2 x 2 3x 9 4 2 13x 13 2 b x 2 E X AM P L E 2 \| Solving Quadratic Equations by Completing the Square Solve each equation. (a) x 2 8x 13 0 (b) 3x 2 12x 6
0 ▼ SO LUTI O N (a) x 2 8x 13 0 2 8x 13 x Given equation Subtract 13 2 8x 16 13 16 x Complete the square: add 1 x 4 2 3 x 4 13 2 Perfect square Take square root x 4 13 Add 4 2 8 2 b a 16 When completing the square, make sure the coefﬁcient of x 2 is 1. If it isn’t, you must factor this coefﬁcient from both terms that contain x: ax2 bx a x2 b a x b a Then complete the square inside the parentheses. Remember that the term added inside the parentheses is multiplied by a François Viète (1540–1603) had a successful political career before taking up mathematics late in life. He became one of the most famous French mathematicians of the 16th century. Viète introduced a new level of abstraction in algebra by using letters to stand for known quantities in an equation. Before Viète’s time, each equation had to be solved on its own. For instance, the quadratic equations 3x 5x 2 2 x 8 0 2 6x 4 0 had to be solved separately by completing the square. Viète’s idea was to consider all quadratic equations at once by writing ax2 bx c 0 where a, b, and c are known quantities. Thus, he made it possible to write a formula (in this case, the Quadratic Formula) involving a, b, and c that can be used to solve all such equations in one fell swoop. Viète’s mathematical genius proved quite valuable during a war between France and Spain. To communicate with their troops, the Spaniards used a complicated code that Viète managed to decipher. Unaware of Viète’s accomplishment, the Spanish king, Philip II, protested to the Pope, claiming that the French were using witchcraft to read his messages. SE CTIO N 1.3 \| Quadratic Equations 89 (b) After subtracting 6 from each side of the equation, we must factor the coefﬁcient of x 2 (the 3) from the left side to put the equation in the correct form for completing the square. 3x 2 12x 6 0 Given equation 3x 2 12x 6 Subtract 6 2 4x x 3 1 2 6 Factor 3 from LHS 2 Now we complete the square by adding
1 everything inside the parentheses is multiplied by 3, this means that we are actually adding to the left side of the equation. Thus, we must add 12 to the right side as well. inside the parentheses. Since 3 # 4 12 2 4 2 x 3 1 2 4x 12 Complete the square: add 4 Perfect square Divide by 3 Take square root x 2 12 ✎ Practice what you’ve learned: Do Exercises 17 and 25. Add 2 ▲ ■ The Quadratic Formula We can use the technique of completing the square to derive a formula for the roots of the general quadratic equation ax 2 bx c 0. THE QUADRATIC FORMULA The roots of the quadratic equation ax 2 bx c 0, where a 0, are x b 2b2 4ac 2a ▼ P RO O F First, we divide each side of the equation by a and move the constant to the right side, giving x2 b a x c a Divide by a We now complete the square by adding b/2a 1 2 2 to each side of the equation: x 2 b a x b 2a b a x b 2a b a 2 2 c a 2 b 2a b a 4ac b2 4a2 x b 2a 2b2 4ac 2a Complete the square: Add 2 b 2a b a Perfect square Take square root x b 2b2 4ac 2a Subtract b 2a ▲ 90 CHAPTER 1 \| Equations and Inequalities The Quadratic Formula could be used to solve the equations in Examples 1 and 2. You should carry out the details of these calculations. E X AM P L E 3 \| Using the Quadratic Formula Find all solutions of each equation. (a) 3x 2 5x 1 0 (b) 4x 2 12x 9 0 (c) x 2 2x 2 0 ▼ SO LUTI O N (a) In this quadratic equation a 3, b 5, and c 1. b –5 3x2 5x 1 0 a 3 c 1 4x Another Method 2 12x By the Quadratic Formula 137 6 If approximations are desired, we can use a calculator to obtain x 5 137 6 1.8471 and x 5 137 6 0.1805 3 2 (b) Using the Quadratic Formula with a 4, b 12, and c 9 gives 12 2 x 2 4 # 4 #
9 12 2 # 4 1 2 12 0 8 3 2 This equation has only one solution, x 3 2. (c) Using the Quadratic Formula with a 1, b 2, and c 2 gives 2 222 4 # 2 2 x 2 14 2 2 211 2 1 11 Since the square of any real number is nonnegative, number system. The equation has no real solution. 11 is undeﬁned in the real ✎ Practice what you’ve learned: Do Exercises 31, 39, and 45. ▲ In the next section we study the complex number system, in which the square roots of negative numbers do exist. The equation in Example 3(c) does have solutions in the complex number system. ■ The Discriminant The quantity b 2 4ac that appears under the square root sign in the Quadratic Formula is called the discriminant of the equation and is given the symbol D. If D 0, then 2b2 4ac is undeﬁned, and the quadratic equation has no real solution, as in Example 3(c). If D 0, then the equation has only one real solution, as in Example 3(b). Finally, if D 0, then the equation has two distinct real solutions, as in Example 3(a). The following box summarizes these observations. ax 2 bx c 0 SE CTIO N 1.3 \| Quadratic Equations 91 THE DISCRIMINANT The discriminant of the general quadratic D b 2 4ac. D 0 D 0 D 0 3. If 1. If 2. If, then the equation has no real solution., then the equation has exactly one real solution., then the equation has two distinct real solutions. ax 2 bx c 0 a 0 1 is 2 E X AM P L E 4 \| Using the Discriminant Use the discriminant to determine how many real solutions each equation has. (a) 4x 2 12x 9 0 x 2 4x 1 0 (b) (c) 1 3x 2 2x 4 0 ▼ SO LUTI O N (a) The discriminant is real solutions. (b) The discriminant is solution. (c) The discriminant is D solution. 1 1 D 42 4 D 12 2 20, so the equation has two distinct, so the equation has exactly one real 2 2 2 4 1 3B A 4 4 3 0, so the equation has no real ✎ Practice
what you’ve learned: Do Exercises 65, 67, and 69. ▲ ■ Modeling with Quadratic Equations Let’s look at some real-life problems that can be modeled by quadratic equations. The principles discussed in Section 1.2 for setting up equations as models are useful here as well. E X AM P L E 5 \| Dimensions of a Building Lot A rectangular building lot is 8 ft longer than it is wide and has an area of 2900 ft 2. Find the dimensions of the lot. ▼ SO LUTI O N We are asked to ﬁnd the width and length of the lot. So let Identify the variable „ width of lot Then we translate the information given in the problem into the language of algebra (see Figure 1). Translate from words to algebra In Words Width of lot Length of lot In Algebra „ „ 8 „ „+8 FIGURE 1 92 CHAPTER 1 \| Equations and Inequalities Now we set up the model. Set up the model width of lot length of lot area of lot Solve „ 2900 „ 8 2 2 8„ 2900 1 „ „ „ 50 2 8„ 2900 0 „ 58 0 1 2 1 2 „ 50 or „ 58 Expand Subtract 2900 Factor Zero-Product Property Since the width of the lot must be a positive number, we conclude that length of the lot is ✎ Practice what you’ve learned: Do Exercise 81. „ 8 50 8 58 ft. „ 50 ft. The ▲ E X AM P L E 6 \| A Distance-Speed-Time Problem A jet ﬂew from New York to Los Angeles, a distance of 4200 km. The speed for the return trip was 100 km/h faster than the outbound speed. If the total trip took 13 hours, what was the jet’s speed from New York to Los Angeles? ▼ SO LUTI O N We are asked for the speed of the jet from New York to Los Angeles. So let Identify the variable Then s speed from New York to Los Angeles s 100 speed from Los Angeles to New York Now we organize the information in a table. We ﬁll in the “Distance” column ﬁrst, since we know that the cities are 4200 km apart. Then we ﬁll in the “Speed�
� column, since we have expressed both speeds (rates) in terms of the variable s. Finally, we calculate the entries for the “Time” column, using time distance rate Translate from words to algebra Distance (km) Speed (km/h) Time (h) N.Y. to L.A. L.A. to N.Y. 4200 4200 s s 100 4200 s 4200 s 100 The total trip took 13 hours, so we have the model Set up the model time from N.Y. to L.A. time from L.A. to N.Y. total time 4200 s 4200 s 100 13 Multiplying by the common denominator, s s 100, we get 4200 1 1 2 s 100 4200s 13s s 100 2 1 2 1300s 8400s 420,000 13s 0 13s 2 7100s 420,000 2 SE CTIO N 1.3 \| Quadratic Equations 93 Although this equation does factor, with numbers this large it is probably quicker to use the Quadratic Formula and a calculator. Solve 7100 2 1 s 7100 2 2 1 2 4 13 2 13 1 2 1 420,000 2 7100 8500 26 s 600 or s 1400 26 53.8 Since s represents speed, we reject the negative answer and conclude that the jet’s speed from New York to Los Angeles was 600 km/h. ✎ Practice what you’ve learned: Do Exercise 91. ▲ E X AM P L E 7 \| The Path of a Projectile This formula depends on the fact that acceleration due to gravity is constant near the earth’s surface. Here, we neglect the effect of air resistance. An object thrown or ﬁred straight upward at an initial speed of √0 ft/s will reach a height of h feet after t seconds, where h and t are related by the formula h 16t 2 √0t ascent descent h FIGURE 2 Suppose that a bullet is shot straight upward with an initial speed of 800 ft/s. Its path is shown in Figure 2. (a) When does the bullet fall back to ground level? (b) When does it reach a height of 6400 ft? (c) When does it reach a height of 2 mi? (d) How high is the highest point the bullet reaches? ▼ SO LUTI O N Since the initial speed in this case is �
�0 800 ft/s, the formula is (a) Ground level corresponds to h 0, so we must solve the equation h 16t 2 800t 0 16t 0 16t 2 800t t 50 1 2 Set h = 0 Factor Thus, t 0 or t 50. This means the bullet starts to ground level after 50 s. t 0 1 2 at ground level and returns (b) Setting h 6400 gives the equation 6400 16t 2 800t Set h = 6400 16t 2 800t 6400 0 2 50t 400 0 0 t 40 t t 10 1 2 1 2 t 10 or t 40 All terms to LHS Divide by 16 Factor Solve The bullet reaches 6400 ft after 10 s (on its ascent) and again after 40 s (on its descent to earth). (c) Two miles is 2 5280 10,560 ft. 10,560 16t2 800t Set h = 10,560 16t2 800t 10,560 0 t2 50t 660 0 All terms to LHS Divide by 16 The discriminant of this equation is, which is nega1 tive. Thus, the equation has no real solution. The bullet never reaches a height of 2 mi. 660 2 1 2 D 50 2 4 140 6400 ft 2 mi 94 CHAPTER 1 \| Equations and Inequalities (d) Each height that the bullet reaches is attained twice—once on its ascent and once on its descent. The only exception is the highest point of its path, which is reached only once. This means that for the highest value of h, the following equation has only one solution for t: 10,000 ft h 16t2 800t 16t2 800t h 0 All terms to LHS This in turn means that the discriminant D of the equation is 0, so 800 D 1 16 2 4 h 0 1 640,000 64h 0 2 2 The maximum height reached is 10,000 ft. ✎ Practice what you’ve learned: Do Exercise 97. ▲ h 10,000 1. ▼ CONCE PTS 1. The Quadratic Formula gives us the solutions of the equation ax2 bx c 0. (a) State the Quadratic Formula: x 2x2 x 4 0 1, and c (b) In the equation b equation is x, a.,. So the solution of the 7. 9. 11. 13. 15. x2 7x 12 0 3x2 5
x 2 0 2y2 7y 3 0 6x2 5x 4 2 5 x 100 x 1 2 8. 10. 12. 14. 16. x2 8x 12 0 4x2 4x 15 0 4„2 4„ 3 3x2 1 4x x 1 21 x 6x 1 2. 17–28 ■ Solve the equation by completing the square. 2. Explain how you would use each method to solve the equation x 2 4x 5 0. (a) By factoring: (b) By completing the square: (c) By using the Quadratic Formula: 3. For the quadratic equation ax2 bx c 0 the discriminant is D tions a quadratic equation has.. The discriminant tells us how many solu- If D 0, the equation has If D 0, the equation has If D 0, the equation has solution(s). solution(s). solution(s). 4. Make up quadratic equations that have the following number of solutions: Two solutions: One solution: No solution:... ▼ SKI LLS 5–16 ■ Solve the equation by factoring. ✎ 5. x 2 x 12 6. x 2 3x 4 ✎ ✎ ✎ ✎ ✎ 17. 19. 21. 23. 25. 27. 0 x2 2x 5 0 x2 6x 11 0 2 x 3 x 4 x2 22x 21 0 2x2 8x 1 0 4x2 x 0 18. 20. 22. 24. 26. 28. x2 4x 2 0 x2 3x 7 0 4 x2 5x 1 0 x2 18x 19 3x2 6x 1 0 2 3 4 x 1 x 8 29–52 ■ Find all real solutions of the equation. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. x2 2x 15 0 x2 7x 10 0 2x2 x 3 0 x2 3x 1 0 x2 12x 27 0 3x2 6x 5 0 z2 3 0 2z 9 4x2 16x 9 0 „ 1 2 3 „ 2 15 x 1 0 x 16 2 1 30. 32. 34. 36. 38. 40. 42. 44. 46. x2 5x 6 0 x2 30x 200 0 3x2 7x 4 0 2x2 8x 4
0 8x2 6x 9 0 x2 6x 1 0 2y2 y 1 0 2 0 x2 4x 1 3 5z z2 0 2 2x 23 2 48. 16 x 0 49. 51. 10y2 16y 5 0 3x2 2x 2 0 50. 52. 25x2 70x 49 0 5x2 7x 5 0 84. Geometry Find the length x if the shaded area is 160 in2. SE CTIO N 1.3 \| Quadratic Equations 95 53. 53–58 ■ Use the quadratic formula and a calculator to ﬁnd all real solutions, correct to three decimals. x2 0.011x 0.064 0 x2 2.450x 1.501 0 2.232x2 4.112x 6.219 x2 2.450x 1.500 0 x2 1.800x 0.810 0 12.714x2 7.103x 0.987 55. 54. 56. 57. 58. 59–64 ■ Solve the equation for the indicated variable. 59. h 1 2 gt 2 √0t; for t 60. S n 1 n 1 2 2 ; for n 61. A 2x 2 4xh; for x 62. A 2pr 2 2prh; for r 63. 1 s a 1 s b 1 c ; for s 64. 1 r 2 1 r 4 r 2; for r ✎ ✎ ✎ 65–72 ■ Use the discriminant to determine the number of real solutions of the equation. Do not solve the equation. 66. x 2 6x 9 65. x 2 6x 1 0 68. x 2 2.21x 1.21 0 67. x 2 2.20x 1.21 0 2 5x 13 2 4x 4 4x 8 9 2 rx s 0 2 rx s 0 s 0 1 s 0, r 2 1s 0 0 71. 70. 69. 72. 9x x x 2 1 2 b a 73. 74. 73–76 ■ Solve the equation for x. a 0 2 2ax 1 0 2 x 1 b 0 2 5bx 4 0 2x 1 a 1 2 2a 1 ax 1 2 2x 1 b 2 b 0 2 0 x 76. 75. bx 77–78 ■ Find all values of k that ensure that the given equation has exactly one solution. 77.
4x2 kx 25 0 78. kx2 36x k 0 ▼ APPLICATIONS 79. Number Problem Find two numbers whose sum is 55 and whose product is 684. 80. Number Problem The sum of the squares of two con- secutive even integers is 1252. Find the integers. ✎ 81. Dimensions of a Garden A rectangular garden is 10 ft longer than it is wide. Its area is 875 ft 2. What are its dimensions? 82. Dimensions of a Room A rectangular bedroom is 7 ft longer than it is wide. Its area is 228 ft 2. What is the width of the room? 83. Dimensions of a Garden A farmer has a rectangular garden plot surrounded by 200 ft of fence. Find the length and width of the garden if its area is 2400 ft 2. perimeter=200 ft x 14 in. 13 in. x 85. Geometry Find the length x if the shaded area is 1200 cm2. x x 1 cm 86. Proﬁt A small-appliance manufacturer ﬁnds that the proﬁt P (in dollars) generated by producing x microwave ovens per week is given by the formula 0 x 200. How many ovens must be manufactured in a given week to generate a proﬁt of $1250? provided that 300 x P 1 10 x 1 2 87. Dimensions of a Box A box with a square base and no top is to be made from a square piece of cardboard by cutting 4-in. squares from each corner and folding up the sides, as shown in the ﬁgure. The box is to hold 100 in3. How big a piece of cardboard is needed? 4 in. 4 in. 88. Dimensions of a Can A cylindrical can has a volume of 40p cm3 and is 10 cm tall. What is its diameter? [Hint: Use the volume formula listed on the inside back cover of this book.] 10 cm 96 CHAPTER 1 \| Equations and Inequalities 89. Dimensions of a Lot A parcel of land is 6 ft longer than it is wide. Each diagonal from one corner to the opposite corner is 174 ft long. What are the dimensions of the parcel? 90. Height of a Flagpole A ﬂagpole is secured on opposite sides by two guy wires, each of which is 5 ft longer than the pole. The distance between the points where the wires are ﬁxed to the ground is
equal to the length of one guy wire. How tall is the ﬂagpole (to the nearest inch)? 95–96 ■ Falling-Body Problems Suppose an object is dropped from a height h0 above the ground. Then its height after t seconds is given by h 16t2 h0, where h is measured in feet. Use this information to solve the problem. 95. If a ball is dropped from 288 ft above the ground, how long does it take to reach ground level? 96. A ball is dropped from the top of a building 96 ft tall. (a) How long will it take to fall half the distance to ground level? (b) How long will it take to fall to ground level? 97–98 ■ Falling-Body Problems Use the formula h 16t 2 √0t discussed in Example 7. 97. A ball is thrown straight upward at an initial speed of ✎ 40 ft/s. √0 (a) When does the ball reach a height of 24 ft? (b) When does it reach a height of 48 ft? (c) What is the greatest height reached by the ball? (d) When does the ball reach the highest point of its path? (e) When does the ball hit the ground? ✎ 91. Distance, Speed, and Time A salesman drives from Ajax to Barrington, a distance of 120 mi, at a steady speed. He then increases his speed by 10 mi/h to drive the 150 mi from Barrington to Collins. If the second leg of his trip took 6 min more time than the ﬁrst leg, how fast was he driving between Ajax and Barrington? 92. Distance,Speed,and Time Kiran drove from Tortula to Cactus, a distance of 250 mi. She increased her speed by 10 mi/h for the 360-mi trip from Cactus to Dry Junction. If the total trip took 11 h, what was her speed from Tortula to Cactus? 93. Distance, Speed, and Time It took a crew 2 h 40 min to row 6 km upstream and back again. If the rate of ﬂow of the stream was 3 km/h, what was the rowing speed of the crew in still water? 98. How fast would a ball have to be thrown upward to reach a [Hint: Use the discriminant of maximum height of 100 ft? the equation 16t 2 √0t h 0.] 99. Fish Population
The ﬁsh population in a certain lake rises and falls according to the formula F 1000 1 30 17t t 2 2 Here F is the number of ﬁsh at time t, where t is measured in years since January 1, 2002, when the ﬁsh population was ﬁrst estimated. (a) On what date will the ﬁsh population again be the same as it was on January 1, 2002? (b) By what date will all the ﬁsh in the lake have died? 94. Speed of a Boat Two ﬁshing boats depart a harbor at the 100. Comparing Areas A wire 360 in. long is cut into two same time, one traveling east, the other south. The eastbound boat travels at a speed 3 mi/h faster than the southbound boat. After two hours the boats are 30 mi apart. Find the speed of the southbound boat. pieces. One piece is formed into a square, and the other is formed into a circle. If the two ﬁgures have the same area, what are the lengths of the two pieces of wire (to the nearest tenth of an inch)? W N S E 0 m i 3 101. Width of a Lawn A factory is to be built on a lot measuring 180 ft by 240 ft. A local building code speciﬁes that a lawn of uniform width and equal in area to the factory must surround the factory. What must the width of this lawn be, and what are the dimensions of the factory? 102. Reach of a Ladder A 19 1 2 ing. The base of the ladder is high up the building does the ladder reach? 7 -foot ladder leans against a buildft from the building. How 1 2 1 19 ft2 1 7 ft2 103. Sharing a Job Henry and Irene working together can wash all the windows of their house in 1 h 48 min. Working alone, it takes Henry h more than Irene to do the job. How long does it take each person working alone to wash all the windows? 1 1 2 104. Sharing a Job Jack, Kay, and Lynn deliver advertising ﬂyers in a small town. If each person works alone, it takes Jack 4 h to deliver all the ﬂyers, and it takes Lynn 1 h longer than it takes Kay. Working together, they can deliver all the ﬂyers in 40% of the time it takes Kay working alone. How
long does it take Kay to deliver all the ﬂyers alone? 105. Gravity If an imaginary line segment is drawn between the centers of the earth and the moon, then the net gravitational force F acting on an object situated on this line segment is F K 2 x 0.012K 239 x 1 2 2 SE CT IO N 1.4 \| Complex Numbers 97 where K 0 is a constant and x is the distance of the object from the center of the earth, measured in thousands of miles. How far from the center of the earth is the “dead spot” where no net gravitational force acts upon the object? (Express your answer to the nearest thousand miles.) x ▼ DISCOVE RY • DISCUSSION • WRITI NG 106. Relationship Between Roots and Coefﬁcients The Quadratic Formula gives us the roots of a quadratic equation from its coefﬁcients. We can also obtain the coefﬁcients from the roots. For example, ﬁnd the roots of the equation x 2 9x 20 0 and show that the product of the roots is the constant term 20 and the sum of the roots is 9, the negative of the coefﬁcient of x. Show that the same relationship between roots and coefﬁcients holds for the following equations: x x 2 2x 8 0 2 4x 2 0 Use the Quadratic Formula to prove that in general, if the equation x 2 bx c 0 has roots r1 and r2, then c r1r2 and b. r1 1 r22 1.4 Complex Numbers LEARNING OBJECTIVES After completing this section, you will be able to: ■ Add and subtract complex numbers ■ Multiply and divide complex numbers ■ Simplify expressions with roots of negative numbers ■ Find complex roots of quadratic equations See the note on Cardano (page 340) for an example of how complex numbers are used to ﬁnd real solutions of polynomial equations. In Section 1.3 we saw that if the discriminant of a quadratic equation is negative, the equation has no real solution. For example, the equation 2 4 0 x has no real solution. If we try to solve this equation, we get x 2 4, so x 14 But this is impossible, since the square of any real number is positive. [For example, 2, a positive number.] Thus, negative numbers don�
�t have real square roots. 2 4 1 2 98 CHAPTER 1 \| Equations and Inequalities To make it possible to solve all quadratic equations, mathematicians invented an expanded number system, called the complex number system. First they deﬁned the new number i 11 This means that i 2 1. A complex number is then a number of the form a bi, where a and b are real numbers. DEFINITION OF COMPLEX NUMBERS A complex number is an expression of the form a bi where a and b are real numbers and i 2 1. The real part of this complex number is a and the imaginary part is b. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Note that both the real and imaginary parts of a complex number are real numbers. E X AM P L E 1 \| Complex Numbers The following are examples of complex numbers. 3 4i 2 3i 1 2 1 2 Real part 3, imaginary part 4 Real part, imaginary part 2 3 Real part 0, imaginary part 6 Real part 7, imaginary part 0 ✎ Practice what you’ve learned: Do Exercises 5 and 9. 6i 7 ▲ A number such as 6i, which has real part 0, is called a pure imaginary number. A real number like 7 can be thought of as a complex number with imaginary part 0. In the complex number system every quadratic equation has solutions. The numbers 2i and 2i are solutions of x 2 4 because 2 4 4 and 2 2 1 2i 2i 1 2 1 2 2 2i 1 2 2 1 2 2i 2 4 1 1 2 4 Although we use the term imaginary in this context, imaginary numbers should not be thought of as any less “real” (in the ordinary rather than the mathematical sense of that word) than negative numbers or irrational numbers. All numbers (except possibly the positive integers) are creations of the human mind—the numbers 1 and as well as the number i. We study complex numbers because they complete, in a useful and elegant fashion, our study of the solutions of equations. In fact, imaginary numbers are useful not only in algebra and mathematics, but in the other sciences as well. To give just one example, in electrical theory the reactance of a circuit is a quantity whose measure is an imaginary number. 12 ■ Arithmetic Operations on Complex Numbers Complex numbers are added, subtracted, multiplied, and divided just
as we would any a b 1c. The only difference that we need to keep in mind is that number of the form i 2 1. Thus, the following calculations are valid. a bi 1 2 1 c di 2 ac ac 1 ac bd 2 ad bc 1 ad bc i bdi i bd 1 2 ad bc 2 1 2 1 2 Multiply and collect like terms 1 i 2 1 2 i Combine real and imaginary parts We therefore deﬁne the sum, difference, and product of complex numbers as follows. SE CT IO N 1.4 \| Complex Numbers 99 ADDING, SUBTRACTING, AND MULTIPLYING COMPLEX NUMBERS Deﬁnition Addition a bi 1 2 Subtraction a bi 2 Description c di c di To add complex numbers, add the real parts and the imaginary parts. To subtract complex numbers, subtract the real parts and the imaginary parts. 1 2 ac bd 1 2 ad bc i 2 Multiply complex numbers like binomials, using i 2 1. 1 1 Multiplication a bi # c di 2 1 Graphing calculators can perform arithmetic operations on complex numbers. (3+5i)+(4-2i) (3+5i)*(4-2i) 7+3i 22+14i Complex Conjugates Number Conjugate 3 2i 1 i 4i 5 3 2i 1 i 4i 5 E X AM P L E 2 \| Adding, Subtracting, and Multiplying Complex Numbers Express the following in the form a bi. 4 2i (b) (a) 1 (d) i 23 1 4 2i 3 5i 3 5i 3 5i 4 2i (c ▼ SO LUTI O N (a) According to the deﬁnition, we add the real parts and we add the imaginary parts. (b) (c) 3 5i 3 5i 23 i 1 1 i 2 1 4 2i 2 1 221 2 3 5i 1 4 2i 2 2 2 3 11i 2 4 2i 11i 3i 2 1 i 1 7i 5 # 4 4 i 22 14i (d) 1 ✎ Practice what you’ve learned: Do Exercises 15, 19, 25, and 33. i 2 2 2 1 1 ▲ Division of complex numbers is much like rationalizing the denominator of a radical expression, which we considered in
Section P.8. For the complex number z a bi we deﬁne its complex conjugate to be z # z z a bi a bi a bi. Note that a 2 1 2 1 2 b 2 So the product of a complex number and its conjugate is always a nonnegative real number. We use this property to divide complex numbers. DIVIDING COMPLEX NUMBERS To simplify the quotient a bi c di, multiply the numerator and the denominator by the complex conjugate of the denominator: a bi c di a bi c di b a c di c di b a 1 ac bd 1 2 c2 d bc ad 2 i 2 100 CHAPTER 1 \| Equations and Inequalities Rather than memorizing this entire formula, it is easier to just remember the ﬁrst step and then multiply out the numerator and the denominator as usual. E X AM P L E 3 \| Dividing Complex Numbers Express the following in the form a bi. (a) 3 5i 1 2i (b) 7 3i 4i ▼ SO LUTI O N We multiply both the numerator and denominator by the complex conjugate of the denominator to make the new denominator a real number. (a) The complex conjugate of 1 2i is 1 2i 1 2i. 3 5i 1 2i 3 5i 1 2i b a 1 2i 1 2i b a 7 11i 5 7 5 11 5 i (b) The complex conjugate of 4i is 4i. Therefore, 7 3i 4i 7 3i 4i ✎ Practice what you’ve learned: Do Exercises 37 and 43. 12 28i 16 4i 4i b 3 4 b a a 7 4 i ▲ ■ Square Roots of Negative Numbers Just as every positive real number r has two square roots, every negative number has two square roots as well. If r is a negative number, then its square roots are i 1r i 1r 2 i2r r 2i2r r, because 1 1 1r i 1r and and 2 1r. 2 1 2 1 2 1 2 SQUARE ROOTS OF NEGATIVE NUMBERS If r is negative, then the principal square root of r is The two square roots of r are i 1r 1r i 1r i 1r and. We usually write i 1b instead of
1b i to avoid confusion with.1bi Leonhard Euler (1707–1783) was born in Basel, Switzerland, the son of a pastor. When Euler was 13, his father sent him to the University at Basel to study theology, but Euler soon decided to devote himself to the sciences. Besides theology he studied mathematics, medicine, astronomy, physics, and Asian languages. It is said that Euler could calculate as effortlessly as “men breathe or as eagles ﬂy.” One hundred years before Euler, Fermat (see page 159) had conjec- 22n 1 is a prime number for all n. The ﬁrst ﬁve of tured that these numbers are 5, 17, 257, 65537, and 4,294,967,297. It is easy to show that the ﬁrst four are prime. The ﬁfth was also thought to be prime until Euler, with his phenomenal calculating ability, showed that it is the product 641 6,700,417 and so is not prime. Euler published more than any other mathematician in history. His collected works comprise 75 large volumes. Although he was blind for the last 17 years of his life, he continued to work and publish. In his writings he popularized the use of the symbols p, e, and i, which you will ﬁnd in this textbook. One of Euler’s most lasting contributions is his development of complex numbers. E X AM P L E 4 \| Square Roots of Negative Numbers SE CTIO N 1.4 \| Complex Numbers 101 11 i 11 i (a) (c) ✎ Practice what you’ve learned: Do Exercises 47 and 49. 116 i 116 4i (b) 13 i 13 ▲ Special care must be taken in performing calculations that involve square roots of negwhen a and b are positive, this is not true when 1a # 1b 1ab ative numbers. Although both are negative. For example, but so 12 # 13 i 12 # i 13 i2 2 16 1 3 16 16 2 1 2 1 12 # 13 1 2 1 2 1 3 2 When multiplying radicals of negative numbers, express them ﬁrst in the form r 0) to avoid possible errors of this type. i 1r (where E X AM P L E 5 \| Using Square Roots of Negative Numbers 3 14 and express in the
form a bi. 112 13 Evaluate 2 1 2 1 ▼ SO LUTI O N 112 13 3 14 2 1 1 2 2 1 112 i 13 1 2 13 i 13 2 1 6 13 2 13 1 3 i 14 3 2i i 2 2 2 # 2 13 3 13 1 2 1 8 13 i 13 ✎ Practice what you’ve learned: Do Exercise 51. 2 ▲ ■ Complex Solutions of Quadratic Equations We have already seen that if a 0, ax 2 bx c 0 are then the solutions of the quadratic equation x b 2b2 4ac 2a If b 2 4ac 0, then the equation has no real solution. But in the complex number system, this equation will always have solutions, because negative numbers have square roots in this expanded setting. E X AM P L E 6 \| Quadratic Equations with Complex Solutions Solve each equation. (a) x 2 9 0 (b) x 2 4x 5 0 ▼ SO LUTI O N (a) The equation x 2 9 0 means x 2 9, so x 19 i 19 3i The solutions are therefore 3i and 3i. 102 CHAPTER 1 \| Equations and Inequalities (b) By the Quadratic Formula we have 4 242 4 # 5 2 x 4 14 2 4 2i 2 2 1 2 i 2 2 2 i So the solutions are 2 i and 2 i. ✎ Practice what you’ve learned: Do Exercises 57 and 59. ▲ We see from Example 6 that if a quadratic equation with real coefﬁcients has complex solutions, then these solutions are complex conjugates of each other. So if a bi is a solution, then a bi is also a solution. E X AM P L E 7 \| Complex Conjugates as Solutions of a Quadratic Show that the solutions of the equation 4x 2 24x 37 0 are complex conjugates of each other. ▼ SO LUTI O N We use the Quadratic Formula to get x 1 24 2 24 2 2 1 24 116 8 3 1 2 i 37 2 1 4 2 1 2 4 4 2 24 4i 8 3 1 2 i So the solutions are and ✎ Practice what you’ve learned: Do Exercise 65. 3 1 2 i, and these are complex conjugates. ▲ 1. ▼ CONCE PTS 1. The imaginary number i has the property
that i 2 2. For the complex number 3 4i the real part is the imaginary part is. ✎ 9. 3. and 11. 13. 2 3 i 13 14 10. 12. 14. 1 2 i 13 2 15 3. (a) The complex conjugate of 3 4i is 3 4i. (b) 3 4i 1 3 4i 4. If 3 4i 2 21. is a solution of a quadratic equation with real coefﬁ- cients, then is also a solution of the equation. ▼ SKI LLS 5–14 ■ Find the real and imaginary parts of the complex number. 5. 5 7i ✎ 7. 2 5i 3 6. 6 4i 4 7i 2 8. 15–46 ■ Evaluate the expression and write the result in the form a bi. 3 4i 4i 1 ✎ ✎ ✎ 15. 17. 19. 21. 23. 25. 27. 29. 2 1 1 A B 2 5i 2 6 6i 7 1 2 i 12 8i 1 1 2i 4 1 7 i 2 1 3 4i 6 5i 2i 5 12i 2 3i 2 2 2 16. 18. 20. 22. 24. 26. 28. 30. 2 1 1 2 5i 3 2i 2 4 i 1 6i 1 i 1 2i 2 A 5 3i 2 1 12i 6i 2 5 1 3 i 2 5i 24i 1 6 3 7i BA 2 1 B 2 31. i 3 ✎ 33. i 100 35. ✎ 37. 39. 41. ✎ 43. 1 i 2 3i 1 2i 26 39i 2 3i 10i 1 2i 4 6i 3i 45. 1 1 i 1 1 i 4 2i 32. 2 1 34. i 1002 36. 38. 40. 1 1 i 5 i 3 4i 25 4 3i 42. 1 2 3i 1 2 44. 46. 3 5i 15i 1 2i 47–56 ■ Evaluate the radical expression and express the result in the form a bi. ✎ ✎ ✎ 47. 125 49. 51. 53. 55. 13 112 3 15 1 2 18 1 12 2 1 136 12 19 1 11 2 48. 50. 52. 54. 56. B 21 9 4 3 127 13 14 1 1 11 1 11 17149 128 16 18 2 2 1 57
–72 ■ Find all solutions of the equation and express them in the form a bi. 57. x 2 49 0 58. 9x 2 4 0 ✎ SE CTI ON 1.5 \| Other Types of Equations 103 59. x 2 4x 5 0 61. x 2 2x 5 0 63. x 2 x 1 0 65. 2x 2 2x 1 0 t 3 3 t 0 67. 60. x 2 2x 2 0 62. x 2 6x 10 0 64. x 2 3x 3 0 66. 2x 2 3 2x z 4 12 z 0 68. 69. 6x 2 12x 7 0 70. 4x 2 16x 19 0 71. 1 2 x 2 x 5 0 73–80 ■ Recall that the symbol of z. If z a bi and „ c di, prove each statement. z 72. x 2 1 2 x 1 0 represents the complex conjugate z„ z # „ z z 74. 76. 73. z „ z „ 2 z2 75. 77. 78. 79. 80 is a real number is a pure imaginary number is a real number if and only if z is real ▼ DISCOVE RY • DISCUSSION • WRITI NG 81. Complex Conjugate Roots Suppose that the equation ax 2 bx c 0 has real coefﬁcients and complex roots. Why must the roots be complex conjugates of each other? (Think about how you would ﬁnd the roots using the Quadratic Formula.) 82. Powers of i Calculate the ﬁrst 12 powers of i, that is, i, i 2, i 3,..., i 12. Do you notice a pattern? Explain how you would calculate any whole number power of i, using the pattern that you have discovered. Use this procedure to calculate i 4446. 1.5 Other Types of Equations LEARNING OBJECTIVES After completing this section, you will be able to: ■ Solve basic polynomial equations ■ Solve equations involving radicals ■ Solve equations of quadratic type ■ Model with equations So far, we have learned how to solve linear and quadratic equations. In this section we study other types of equations, including those that involve higher powers, fractional expressions, and radicals. ■ Polynomial Equations Some equations can be solved by factoring and using the Zero-Product Property, which says
that if a product equals 0, then at least one of the factors must equal 0. 104 CHAPTER 1 \| Equations and Inequalities E X AM P L E 1 \| Solving an Equation by Factoring Solve the equation x 5 9x 3. ▼ SO LUTI O N We bring all terms to one side and then factor. 5 9x or x 3 0 or Subtract 9x3 Factor x3 Difference of squares Zero-Product Property Solve The solutions are x 0, x 3, and x 3. You should check that each of these satisﬁes the original equation. ✎ Practice what you’ve learned: Do Exercise 5. ▲ To divide each side of the equation in Example 1 by the common factor x 3 would be wrong, because in doing so, we would lose the solution x 0. Never divide both sides of an equation by an expression that contains the variable unless you know that the expression cannot equal 0. E X AM P L E 2 \| Factoring by Grouping Solve the equation x 3 3x 2 4x 12 0. ▼ SO LUTI O N The left-hand side of the equation can be factored by grouping the terms in pairs. 2 1 x 3 3x or x 2 0 or x 3 0 4x 12 Group terms Factor x2 and 4 Factor x + 3 Difference of squares Zero-Product Property x 2 x 2 The solutions are x 2, 2, and 3. x 3 Solve ✎ Practice what you’ve learned: Do Exercise 15. ▲ E X AM P L E 3 \| An Equation Involving Fractional Expressions Solve the equation 3 x 5 x 2 2. ▼ SO LUTI O N To simplify the equation, we multiply each side by the common denominator 2x 1 2 x 2 2 Multiply by LCD x(x + 2) 3 1 x 2 5x 2x 2 8x 6 2x 2 4x 2 4x Expand Expand LHS Check Your Answers x 3 : LHS 3 3 RHS 2 LHS RHS 5 3 2 2 x 1 : LHS 3 1 5 1 2 2 RHS 2 LHS RHS Check Your Answers x 1 4 : 1 4 B A LHS 2 1 4 B 1 2 A RHS 1 22 1 2 9 4 1 3 1 2 2 LHS RHS x 1 : LHS 2 1 2 1 RHS 1 12 1 1
1 0 2 LHS RHS ✔ ✔ ✔ SE CTI ON 1.5 \| Other Types of Equations 105 0 2x 0 x 0 x 1 2 x 3 0 or x 1 0 2 4x 6 2 2x 3 x 3 2 1 1 Subtract 8x + 6 Divide both sides by 2 Factor Zero-Product Property x 3 x 1 Solve We must check our answers because multiplying by an expression that contains the variable can introduce extraneous solutions (see the Warning on pages 68–69). From Check Your Answers we see that the solutions are x 3 and 1. ✎ Practice what you’ve learned: Do Exercise 19. ▲ ■ Equations Involving Radicals When you solve an equation that involves radicals, you must be especially careful to check your ﬁnal answers. The next example demonstrates why. E X AM P L E 4 \| An Equation Involving a Radical 2x 1 12 x Solve the equation. ▼ SO LUTI O N To eliminate the square root, we ﬁrst isolate it on one side of the equal sign, then square. 1 4x 2x 1 2x 1 12 x 2 2 x 2 2 4x 1 2 x 2 3x 1 0 0 x 1 4x 4x 1 2 1 4x 1 0 or Subtract 1 Square each side Expand LHS Add 2 + x Factor Zero-Product Property 1 4 x Solve and x 1 are only potential solutions. We must check them The values to see whether they satisfy the original equation. From Check Your Answers we see that x is a solution but x 1 is not. The only solution is ✎ Practice what you’ve learned: Do Exercise 29. x ▲ 1 4 1 4. When we solve an equation, we may end up with one or more extraneous solutions, that is, potential solutions that do not satisfy the original equation. In Example 4 the value x 1 is an extraneous solution. Extraneous solutions may be introduced when we square each side of an equation because the operation of squaring can turn a false equation into a true one. For example, 1 1, but. Thus, the squared equation may be true for more values of the variable than the original equation. That is why you must always check your answers to make sure that each satisﬁes the original equation. 2 12 1 2 1 ■ Equations of Quadratic Type An equation of the form aW 2
bW c 0, where W is an algebraic expression, is an equation of quadratic type. We solve equations of quadratic type by substituting for the algebraic expression, as we see in the next two examples. 106 CHAPTER 1 \| Equations and Inequalities E X AM P L E 5 \| An Equation of Quadratic Type Solve the equation. ▼ SO LUTI O N We could solve this equation by multiplying it out ﬁrst. But it’s easier to as the unknown in this equation and give it a new name W. think of the expression This turns the equation into a quadratic equation in the new variable W Given equation W 2 6W Let W 1 1 x Factor W 4 0 or W 2 0 Zero-Product Property W 4 W 2 Solve Now we change these values of W back into the corresponding values of x Subtract 1 Take reciprocals and x 1. The solutions are ✎ Practice what you’ve learned: Do Exercise 37. x 1 3 ▲ MATHEMATICS IN THE MODERN WORLD Error-Correcting Codes A S A N The pictures sent back by the Pathﬁnder spacecraft from the surface of Mars on July 4, 1997, were astoundingly clear. But few watching these pictures were aware of the complex mathematics used to accomplish that feat. The distance to Mars is enormous, and the background noise (or static) is many times stronger than the original signal emitted by the spacecraft. So when scientists receive the signal, it is full of errors. To get a clear picture, the errors must be found and corrected. This same problem of errors is routinely encountered in transmitting bank records when you use an ATM or voice when you are talking on the telephone. To understand how errors are found and corrected, we must ﬁrst understand that to transmit pictures, sound, or text, we transform them into bits (the digits 0 or 1; see page 175). To help the receiver recognize errors, the message is “coded” by inserting additional bits. For example, suppose you want to transmit the message “10100.” A very simple-minded code is as follows: Send each digit a million times. The person receiving the message reads it in blocks of a million digits. If the ﬁrst block is mostly 1’s, the person concludes that you are probably trying to transmit a 1, and so on. To say
that this code is not efﬁcient is a bit of an understatement; it requires sending a million times more data than the original message. Another method inserts “check digits.” For example, for each block of eight digits insert a ninth digit; the inserted digit is 0 if there is an even number of 1’s in the block and 1 if there is an odd number. So if a single digit is wrong (a 0 changed to a 1 or vice versa), the check digits allow us to recognize that an error has occurred. This method does not tell us where the error is, so we can’t correct it. Modern error-correcting codes use interesting mathematical algorithms that require inserting relatively few digits but that allow the receiver to not only recognize, but also correct, errors. The ﬁrst error-correcting code was developed in the 1940s by Richard Hamming at MIT. It is interesting to note that the English language has a built-in error correcting mechanism; to test it, try reading this error-laden sentence: Gve mo libty ox giv ne deth. SE CTI ON 1.5 \| Other Types of Equations 107 E X AM P L E 6 \| A Fourth-Degree Equation of Quadratic Type Find all solutions of the equation x 4 8x 2 8 0. ▼ SO LUTI O N If we set W x 2, then we get a quadratic equation in the new variable W: 2 8x Write x4 as (x2)2 1 2 2 8 0 x W 2 8W 8 0 2 2 4 # 8 Let W = x2 4 212 Quadratic Formula W = x2 Take square roots 12 x x 24 2 12 So, there are four solutions: 24 2 12, 24 2 12, 24 2 12, 24 2 12 Using a calculator, we obtain the approximations x 2.61, 1.08, 2.61, 1.08. ✎ Practice what you’ve learned: Do Exercise 39. ▲ E X AM P L E 7 \| An Equation Involving Fractional Powers Find all solutions of the equation x 1/3 x 1/6 2 0. ▼ SO LUTI O N This equation is of quadratic type because if we let W x 1/6, then W 2 x 2 1/3. 1/6 x 1 2 x 1/3 x 1/ or
W 2 0 2 1 1 2 Given equation Let W = x1/6 Factor Zero-Product Property W 1 x 1/6 1 W 2 x 1/6 2 x 2 1 From Check Your Answers we see that x 1 is a solution but x 64 is not. The only solution is x 1. x 16 1 Take the 6th power 6 64 W = x1/6 Solve 2 Check Your Answers x 1 : x 64 : LHS 11/3 11/6 2 0 LHS 641/3 641/6 2 RHS 0 LHS RHS ✔ 4 2 2 4 RHS 0 LHS RHS ✎ Practice what you’ve learned: Do Exercise 45. ▲ ■ Applications Many real-life problems can be modeled with the types of equations that we have studied in this section. 108 CHAPTER 1 \| Equations and Inequalities E X AM P L E 8 \| Dividing a Lottery Jackpot A group of people come forward to claim a $1,000,000 lottery jackpot, which the winners are to share equally. Before the jackpot is divided, three more winning ticket holders show up. As a result, each person’s share is reduced by $75,000. How many winners were in the original group? ▼ SO LUTI O N We are asked for the number of people in the original group. So let Identify the variable x number of winners in the original group We translate the information in the problem as follows: Translate from words to algebra In Words Number of winners in original group Number of winners in ﬁnal group Winnings per person, originally Winnings per person, ﬁnally In Algebra x x 3 1,000,000 x 1,000,000 x 3 Now we set up the model. Set up the model winnings per person, originally $75,000 winnings per person, ﬁnally Solve 1,000,000 1 x 3 75,000x 1,000,000 x 75,000 1,000,000 x 3 1,000,000x x 3 1 2 2 x 3 40 1 x 3 3x 2 3x 40 0 40x or x 5 0 2 Multiply by LCD x(x + 3) Divide by 25,000 Expand, simplify, and divide by 3 Factor Zero-Product Property x 8 x 5 Solve Since we can’t have a negative number of people
, we conclude that there were ﬁve winners in the original group. Check Your Answer winnings per person, originally $1,000,000 5 winnings per person, finally $1,000,000 8 $200,000 $125,000 $200,000 $75,000 $125,000 ✔ ✎ Practice what you’ve learned: Do Exercise 75. ▲ E X AM P L E 9 \| Energy Expended in Bird Flight Ornithologists have determined that some species of birds tend to avoid ﬂights over large bodies of water during daylight hours, because air generally rises over land and falls over water in the daytime, so ﬂying over water requires more energy. A bird is released from point A on an island, 5 mi from B, the nearest point on a straight shoreline. The bird ﬂies to a point C on the shoreline and then ﬂies along the shoreline to its nesting area D, as SE CTI ON 1.5 \| Other Types of Equations 109 Island A 5 mi B shown in Figure 1. Suppose the bird has 170 kcal of energy reserves. It uses 10 kcal/mi ﬂying over land and 14 kcal/mi ﬂying over water. (a) Where should the point C be located so that the bird uses exactly 170 kcal of energy during its ﬂight? C D (b) Does the bird have enough energy reserves to ﬂy directly from A to D? ▼ SO LUTI O N (a) We are asked to ﬁnd the location of C. So let x 12 mi Nesting area Identify the variable x distance from B to C From the ﬁgure, and from the fact that FIGURE 1 energy used energy per mile miles flown we determine the following: In Words In Algebra Translate from words to algebra Distance from B to C Distance ﬂown over water (from A to C) Distance ﬂown over land (from C to D) Energy used over water Energy used over land Now we set up the model. 2 25 x 2x 12 x 142x 10 1 2 25 12 x 2 Pythagorean Theorem Set up the model total energy used energy used over water energy used over land 170 142x 2 25 10 12 x 1 2 To solve this equation, we eliminate the square root by ﬁrst bringing all other terms to the left of the equal sign
and then squaring each side. Solve 170 10 12 x 142x 2 25 50 10x 2 1 2500 1000x 100x 2 1 50 10x 142x 2 14 2 2 25 2 25 x 1 2 4900 2 1000x 2400 2 2 1 2 196x 0 96x Isolate square root term on RHS Simplify LHS Square each side Expand All terms to RHS This equation could be factored, but because the numbers are so large, it is easier to use the Quadratic Formula and a calculator: 1 1000 2 1000 2 96 2 2 or 96 1 2 1 2400 2 1000 280 192 Point C should be either mi or mi from B so that the bird uses exactly 170 kcal 6 of energy during its ﬂight. 3 2 3 3 4 (b) By the Pythagorean Theorem (see page 284), the length of the route directly from 252 122 13 A to D is 14 13 182 kcal. This is more energy than the bird has available, so it can’t use this route. mi, so the energy the bird requires for that route is ✎ Practice what you’ve learned: Do Exercise 83. ▲ 110 CHAPTER 1 \| Equations and Inequalities 1. ▼ CONCE PTS 1. (a) The solutions of the equation (b) To solve the equation hand side. x 4 x2 x3 4x2 0 1 2, we 0 are. the left- 35. 36 10. Solve the equation 12x x 0 by doing the following steps. (a) Isolate the radical: (b) Square both sides:.. (c) The solutions of the resulting quadratic equation are. (d) The solution(s) that satisfy the original equation are. x 1 3. The equation 2 5 x 1 6 0 is of 2 type. To solve the equation, we set W 1 1 2. The result- 47. 4 1 ✎ 37 38. x x 2 b 4x x 2 39. x 4 13x 2 40 0 41. 2x 4 4x 2 1 0 43. x 6 26x 3 27 0 45. x 4/3 5x 2/3 6 0 ✎ ✎ 4 40. x 4 5x 2 4 0 42. x 6 2x 3 3 0 44. x 8 15x 4 16 46. 1x 3 14 x 4 0 x 1 x 4 2 1/2 5 7/3 x 1 1 x
4 3/2 2 4/3 x 1 1 x 4 5/2 0 2 1/3 0 2 1 2 48. 49. x 3/2 8x 1/2 16x1/2 0 50. x 1/2 3x1/2 10x3/2 51. x 1/2 3x 1/3 3x 1/6 9 x 51x 6 0 52. 2 1 1 2 53. 55. 57 1x 5 x 5 2 1x 3 x 3 x 1 54. 4x4 16x2 4 0 56. 23 4x 2 4x x 3/2 2 58. 211 x 2 2 211 x 2 1 59. 2x 1x 2 2 60. 31 2x 12x 1 25 1x 61–70 ■ Find all solutions, real and complex, of the equation. 61. x 3 1 63. x 3 x 2 x 0 62. x 4 16 0 64. x 4 x 3 x 2 x 0 65. x 4 6x 2 8 0 67. x 6 9x 3 8 0 66. x 3 3x 2 9x 27 0 68. x 6 9x 4 4x 2 36 0 69. 2x 2 1 8 2 1 2x 2x 2 9 70. 1 2x 2 7 6 x 2 71–74 ■ Solve the equation for the variable x. The constants a and b represent positive real numbers. 71. x 4 5ax 2 4a 2 0 72. a 3x 3 b 3 0 73. 74. 1x a 1x a 121x 6 1x a 13 x b16 x ab 0 ✎ ▼ APPLICATIONS 75. Chartering a Bus A social club charters a bus at a cost of $900 to take a group of members on an excursion to Atlantic City. At the last minute, ﬁve people in the group decide not to go. This raises the transportation cost per person by $2. How many people originally intended to take the trip? ing quadratic equation is. 4. The equation x6 7x3 8 0 is of type. To solve the equation, we set W. The resulting quadratic equation is. ✎ ✎ ▼ SKI LLS 5–60 ■ Find all real solutions of the equation. 5. x 3 16x 7. x 6 81x 2 0 9. x 5 8x 2 0 11. x 3 5x 2 6x 0 13. x 4 4
x 3 2x 2 0 15. x 3 5x 2 2x 10 0 17 18. 7x 3 x 1 x 3 3x 2 x 6. x 5 27x 2 8. x 5 16x 0 10. x 4 64x 0 12. x 4 x 3 6x 2 0 x 2 x 2 5 9 14. 16. 2x 3 x 2 18x 9 0 3 0 2 1 1 2 ✎ 19. 1 x 1 1 x 2 5 4 20. 10 x 12 x 3 4 0 21. 2 x x 100 50 2 x 3 4 x 4 x 4 22. 1 2x 23. 2 28 2 4 x x 1 x 3 1 5x 24. 26. x 2x 7 x 2 x 3 4 x 28. 30. 14 6x 2x x 19 3x 0 2x 1x 1 8 32. 34. x 2 1x 7 10 25. 1 x 1 2 2 x 0 ✎ 27. 29. 31. 33. 1x 2 x 12x 1 1 x 15 x 1 x 2 x 1x 3 x 2 76. Buying a Cottage A group of friends decides to buy a vacation home for $120,000, sharing the cost equally. If they can ﬁnd one more person to join them, each person’s contribution will drop by $6000. How many people are in the group? 77. Fish Population A large pond is stocked with ﬁsh. The ﬁsh population P is modeled by the formula P 3t 10 1t 140, where t is the number of days since the ﬁsh were ﬁrst introduced into the pond. How many days will it take for the ﬁsh population to reach 500? 78. The Lens Equation If F is the focal length of a convex lens and an object is placed at a distance x from the lens, then its image will be at a distance y from the lens, where F, x, and y are related by the lens equation 1 F 1 x 1 y Suppose that a lens has a focal length of 4.8 cm and that the image of an object is 4 cm closer to the lens than the object itself. How far from the lens is the object? 79. Volume of Grain Grain is falling from a chute onto the ground, forming a conical pile whose diameter is always three times its height. How high is the pile (to the nearest hundredth of a foot)
when it contains 1000 ft 3 of grain? 80. Radius of a Tank A spherical tank has a capacity of 750 gallons. Using the fact that 1 gallon is about 0.1337 ft 3, ﬁnd the radius of the tank (to the nearest hundredth of a foot). 81. Radius of a Sphere A jeweler has three small solid spheres made of gold, of radius 2 mm, 3 mm, and 4 mm. He decides to melt these down and make just one sphere out of them. What will the radius of this larger sphere be? 82. Dimensions of a Box A large plywood box has a volume of 180 ft 3. Its length is 9 ft greater than its height, and its width is 4 ft less than its height. What are the dimensions of the box? x+9 x x-4 SE CTI ON 1.5 \| Other Types of Equations 111 mile, while building a new road would cost $200,000 per mile. How much of the abandoned road should be used (as indicated in the ﬁgure) if the ofﬁcials intend to spend exactly $6.8 million? Would it cost less than this amount to build a new road connecting the towns directly? Grimley F oxton New road 10 mi Abandoned road 40 mi 84. Distance, Speed, and Time A boardwalk is parallel to and 210 ft inland from a straight shoreline. A sandy beach lies between the boardwalk and the shoreline. A man is standing on the boardwalk, exactly 750 ft across the sand from his beach umbrella, which is right at the shoreline. The man walks 4 ft/s on the boardwalk and 2 ft/s on the sand. How far should he walk on the boardwalk before veering off onto the sand if he wishes to reach his umbrella in exactly 4 min 45 s? 750 ft 210 ft boardwalk 85. Dimensions of a Lot A city lot has the shape of a right triangle whose hypotenuse is 7 ft longer than one of the other sides. The perimeter of the lot is 392 ft. How long is each side of the lot? 86. TV Monitors Two television monitors sitting beside each other on a shelf in an appliance store have the same screen height. One has a conventional screen, which is 5 in. wider than it is high. The other has a wider, high-deﬁnition screen, which is 1.8 times as wide as it is high. The diagonal
measure of the wider screen is 14 in. more than the diagonal measure of the smaller. What is the height of the screens, correct to the nearest 0.1 in.? ✎ 83. Construction Costs The town of Foxton lies 10 mi north of an abandoned east-west road that runs through Grimley, as shown in the ﬁgure. The point on the abandoned road closest to Foxton is 40 mi from Grimley. County ofﬁcials are about to build a new road connecting the two towns. They have determined that restoring the old road would cost $100,000 per 87. Depth of a Well One method for determining the depth of a well is to drop a stone into it and then measure the time it takes until the splash is heard. If d is the depth of the well (in feet) 112 CHAPTER 1 \| Equations and Inequalities t1, so 1d/4 and t1 the time (in seconds) it takes for the stone to fall, then d 16t 2. Now if t2 is the time it takes for the 1 sound to travel back up, then d 1090t 2 because the speed of d/1090. Thus, the total time elapsed sound is 1090 ft/s. So t 2 between dropping the stone and hearing the splash is t 2 How deep is the well if this total time is 3 s? 1d/4 d/1090 t 1 Time stone falls: t⁄=œ∑d 4 Time sound rises: t¤= d 1090 ▼ DISCOVE RY • DISCUSSION • WRITI NG 88. Solving an Equation in Different Ways We have learned several different ways to solve an equation in this section. Some equations can be tackled by more than one method. For x 1x 2 0 is of quadratic type: example, the equation and x u 2 and factoring. 1x u We can solve it by letting 1x, square each side, and then solve the Or we could solve for resulting quadratic equation. Solve the following equations using both methods indicated, and show that you get the same ﬁnal answers. (a) x 1x 2 0 quadratic type; solve for the radical, and square (b) 12 x 3 1 2 2 10 x 3 1 0 quadratic type; multiply by LCD 1.6 Inequalities LEARNING OBJECTIVES After completing this
section, you will be able to: ■ Solve linear inequalities ■ Solve nonlinear inequalities ■ Model with inequalities x 1 2 3 4 5 4x 7 c 19 11 19 ✔ 15 19 ✔ 19 19 ✔ 23 19 27 19 Some problems in algebra lead to inequalities instead of equations. An inequality looks just like an equation, except that in the place of the equal sign is one of the symbols,,, or. Here is an example of an inequality: 4x 7 19 The table in the margin shows that some numbers satisfy the inequality and some numbers do not. To solve an inequality that contains a variable means to ﬁnd all values of the variable that make the inequality true. Unlike an equation, an inequality generally has inﬁnitely many solutions, which form an interval or a union of intervals on the real line. The following illustration shows how an inequality differs from its corresponding equation: Equation: 4x 7 19 Solution x 3 Inequality: 4x 7 19 x 3 Graph 0 0 3 3 To solve inequalities, we use the following rules to isolate the variable on one side of the inequality sign. These rules tell us when two inequalities are equivalent (the symbol 3 means “is equivalent to”). In these rules the symbols A, B, and C stand for real numbers or SECTION 1.6 \| Inequalities 113 algebraic expressions. Here we state the rules for inequalities involving the symbol, but they apply to all four inequality symbols. RULES FOR INEQUALITIES Rule 1. A B 3 A C B C 2. A B 3 A C B C,, then A B 3 CA CB then A B 3 CA CB 3. If C 0 4. If C 0 5. If A 0 then B 0, and A B 3 1 A 1 B then 6. If A B and C D, Description Adding the same quantity to each side of an inequality gives an equivalent inequality. Subtracting the same quantity from each side of an inequality gives an equivalent inequality. Multiplying each side of an inequality by the same positive quantity gives an equivalent inequality. Multiplying each side of an inequality by the same negative quantity reverses the direction of the inequality. Taking reciprocals of each side of an inequality involving positive quantities reverses the direction of the inequality. A C B D Inequalities can be added. Pay special attention to Rules 3 and 4. Rule 3 says that we can multiply (or divide) each side of an inequality by a positive
number, but Rule 4 says that if we multiply each side of an inequality by a negative number, then we reverse the direction of the inequality. For example, if we start with the inequality and multiply by 2, we get but if we multiply by 2, we get 3 5 6 10 6 10 ■ Solving Linear Inequalities An inequality is linear if each term is constant or a multiple of the variable. To solve a linear inequality, we isolate the variable on one side of the inequality sign. E X AM P L E 1 \| Solving a Linear Inequality Solve the inequality 3x 9x 4 and sketch the solution set. ▼ SO LUTI O N 3x 9x 4 3x 9x 9x 4 9x 6x 4 6x 1 A 2 x 2 3 Given inequality Subtract 9x Simplify Simplify 2 3 1 Multiplying by the negative number 6 reverses the direction of the inequality. 1 6B1 A 4 2 Multiply by 1 6 (or divide by – 6) 6B1 _ 2 3 0 FIGURE 1 The solution set consists of all numbers greater than 3, q inequality is the interval. It is graphed in Figure 1. ✎ Practice what you’ve learned: Do Exercise 19. B A 2. In other words, the solution of the ▲ 114 CHAPTER 1 \| Equations and Inequalities E X AM P L E 2 \| Solving a Pair of Simultaneous Inequalities Solve the inequalities 4 3x 2 13. 0 2 5 FIGURE 2 ▼ SO LUTI O N The solution set consists of all values of x that satisfy both of the inequalities Using Rules 1 and 3, we see that the following and inequalities are equivalent: 3x 2 13. 4 3x 2 4 3x 2 13 6 3x 15 2 x 5 Given inequality Add 2 Divide by 3 Therefore, the solution set is, as shown in Figure 2. 3 ✎ Practice what you’ve learned: Do Exercise 29. 2, 5 2 ▲ ■ Solving Nonlinear Inequalities To solve inequalities involving squares and other powers of the variable, we use factoring, together with the following principle. THE SIGN OF A PRODUCT OR QUOTIENT If a product or a quotient has an even number of negative factors, then its value is positive. If a product or a quotient has an odd number of negative factors, then its value is negative. For example, to solve
the inequality x2 5x 6, we ﬁrst move all terms to the left- hand side and factor to get This form of the inequality says that the product must be negative or zero, so to solve the inequality, we must determine where each factor is negative or positive (because the sign of a product depends on the sign of the factors). The details are explained in Example 3, in which we use the following guidelines. 2 1 1 2 x 3 GUIDELINES FOR SOLVING NONLINEAR INEQUALITIES 1. Move All Terms to One Side. If necessary, rewrite the inequality so that all nonzero terms appear on one side of the inequality sign. If the nonzero side of the inequality involves quotients, bring them to a common denominator. 2. Factor. Factor the nonzero side of the inequality. 3. Find the Intervals. Determine the values for which each factor is zero. These numbers will divide the real line into intervals. List the intervals that are determined by these numbers. 4. Make a Table or Diagram. Use test values to make a table or diagram of the signs of each factor on each interval. In the last row of the table determine the sign of the product (or quotient) of these factors. 5. Solve. Determine the solution of the inequality from the last row of the sign table. Be sure to check whether the inequality is satisﬁed by some or all of the endpoints of the intervals. (This may happen if the inequality involves or.) SECTION 1.6 \| Inequalities 115 The factoring technique that is described in these guidelines works only if all nonzero terms appear on one side of the inequality symbol. If the inequality is not written in this form, ﬁrst rewrite it, as indicated in Step 1. E X AM P L E 3 \| Solving a Quadratic Inequality x 2 5x 6. Solve the inequality ▼ SO LUTI O N We will follow the guidelines on page 114. Move all terms to one side. We move all the terms to the left-hand side. x2 5x 6 Given inequality x2 5x 6 0 Subtract 5x, add 6 (_`, 2) (2, 3) (3, `) 0 2 3 FIGURE 3 Test value x = 1 Test value x = 2 1 2 Test value x = 4 0 2 3 FIGURE 4 Factor. Factoring the left
-hand side of the inequality, we get x 3 x 2 0 Factor 1 2 1 Find the intervals. The factors of the left-hand side are. These factors are zero when x is 2 and 3, respectively. As shown in Figure 3, the numbers 2 and 3 divide the real line into the three intervals x 2 x 3 and 2 2, 3 1 The factors x 2 and x 3 change sign only at 2 and 3, respectively. So these factors maintain their sign on each of these three intervals.,, 1 2 2 1 2 q, 2 3, q Make a table or diagram. To determine the sign of each factor on each of the intervals that we found, we use test values. We choose a number inside each interval and check the sign of the factors x 2 and x 3 at the number we chose. For the interval let’s choose the test value 1 (see Figure 4). Substituting 1 for x in the factors x 2 and x 3, we get q So both factors are negative on this interval. Notice that we need to check only one test value for each interval because the factors x 2 and x 3 do not change sign on any of the three intervals we found. Using the test values and x 4 for the intervals (see Figure 4), respectively, we construct the following sign table. The ﬁnal row of the table is obtained from the fact that the expression in the last row is the product of the two factors. x 2 3, q 2, 3 and 1 2 2 2 1 1 Interval Sign of x 2 Sign of x 3 Sign of x 2 1 21 x 3 2 qq, 2 1 2 2, 3 1 2 3, qq 2 1 If you prefer, you can represent this information on a real line, as in the following sign diagram. The vertical lines indicate the points at which the real line is divided into intervals: 2 3 Sign of x-2 Sign of x-3 Sign of (x-2)(x-3) - - + + - - + + + 116 CHAPTER 1 \| Equations and Inequalities 0 2 3 FIGURE 5 Solve. We read from the table or the diagram that terval. Thus, the solution of the inequality 2 is 2 x 2 x 3 is negative on the in, 3 4 3 We have included the endpoints 2 and 3 because we seek values of x such that the product is either less than or equal to zero. The solution is illustrated in Figure
5. ✎ Practice what you’ve learned: Do Exercise 39. ▲ E X AM P L E 4 \| Solving an Inequality Solve the inequality 2x2 x 1. ▼ SO LUTI O N We will follow the guidelines on page 114. Move all terms to one side. We move all the terms to the left-hand side. 2x2 x 1 2x2 x 1 0 Given inequality Subtract 1 Factor. Factoring the left-hand side of the inequality, we get 0 Find the intervals. The factors of the left-hand side are 2x 1 and x 1. These factors are zero when x is and 1. These numbers divide the real line into the intervals 2x 1 x 1 Factor 2 1 1 2 1 2 q, 1 2 B A, A 1 2, 1, A B 1, q B Make a diagram. We make the following diagram, using test points to determine the sign of each factor in each interval. - 1 2 1 Sign of 2x+1 Sign of x-1 Sign of (2x+1)(x-1) - - + + - - + + + Solve. From the diagram we see that q, 1 2 B or for x in 1, q A 2 1 x 1. So the solution set is the union of these two intervals: for x in the interval 2x FIGURE 6 q, 1 2 B A 1 1, q 2 The solution set is graphed in Figure 6. ✎ Practice what you’ve learned: Do Exercise 41. ▲ E X AM P L E 5 \| Solving an Inequality with Repeated Factors x 3 Solve the inequality ▼ SO LUTI O N All nonzero terms are already on one side of the inequality, and the nonzero side of the inequality is already factored. So we begin by ﬁnding the intervals for this inequality. SECTION 1.6 \| Inequalities 117 Find the intervals. The factors of the left-hand side are x, zero when x 0, 1, 3. These numbers divide the real line into the intervals 2 1 x 1 2, and x 3. These are 2 Make a diagram. We make the following diagram, using test points to determine the sign of each factor in each interval. 1 1 1 2 2 2 1 q, 0, 0, 1, 1, 3, 3, q 0 1 3 Sign of x Sign of (x-1
)2 Sign of (x-3) Sign of x(x-1)2(x-3) - + - + + + - - + + - - + + + + Solve. From the diagram we see that or for x in 1. So the solution set is the union of these two intervals: 2 1 for x in the interval 0, 1 1 2 1, 3 1 2 The solution set is graphed in Figure 7. ✎ Practice what you’ve learned: Do Exercise 53. E X AM P L E 6 \| Solving an Inequality Involving a Quotient Solve the inequality ▼ SO LUTI O N 1 x 1 x 1. 0, 1 1 2 ▲ Move all terms to one side. We move the terms to the left-hand side and simplify using a common denominator. 0 1 3 FIGURE 7 It is tempting to multiply both sides of the inequality by 1 x (as you would if this were an equation). But this doesn’t work because we don’t know whether 1 x is positive or negative, so we can’t tell whether the inequality needs to be reversed. (See Exercise 97.) 1 x 1 x 1 1 0 Given inequality Subtract Common denominator 1 – x Combine the fractions 2x 1 x 0 Simplify Find the intervals. x is 0 and 1. These numbers divide the real line into the intervals The factors of the left-hand side are 2x and 1 x. These are zero when q, 0 0, 1, 1 2, 1 2 1, q 2 1 Make a diagram. We make the following diagram using test points to determine the sign of each factor in each interval. Sign of 2x Sign of 1-x 2x 1-x Sign of 0 1 - + - + + + + - - 118 CHAPTER 1 \| Equations and Inequalities 0 1 FIGURE 8 Solve. From the diagram we see that 2x 1 x 0 for x in the interval. We include 0, 1 2 3 the endpoint 0 because the original inequality requires that the quotient be greater than or equal to 1. However, we do not include the other endpoint 1 because the quotient in the inequality is not deﬁned at 1. So the solution set is the interval The solution set is graphed in Figure 8. ✎ Practice what you’ve learned: Do Exercise 61. 0, 1 3 2 ▲ Example 6 shows
that we should always check the endpoints of the solution set to see whether they satisfy the original inequality. ■ Modeling with Inequalities Modeling real-life problems frequently leads to inequalities because we are often interested in determining when one quantity is more (or less) than another. E X AM P L E 7 \| Carnival Tickets A carnival has two plans for tickets. Plan A: $5 entrance fee and 25¢ each ride Plan B: $2 entrance fee and 50¢ each ride How many rides would you have to take for Plan A to be less expensive than Plan B? ▼ SO LUTI O N We are asked for the number of rides for which Plan A is less expensive than Plan B. So let Identify the variable x number of rides The information in the problem may be organized as follows. Translate from words to algebra In Words Number of rides Cost with Plan A Cost with Plan B In Algebra x 5 0.25x 2 0.50x Set up the model Now we set up the model. cost with Plan A cost with Plan B 5 0.25x 2 0.50x Solve 3 0.25x 0.50x Subtract 2 3 0.25x 12 x Subtract 0.25x Divide by 0.25 So if you plan to take more than 12 rides, Plan A is less expensive. ✎ Practice what you’ve learned: Do Exercise 83. ▲ 30 5 86 41 * C * F E X AM P L E 8 \| Relationship Between Fahrenheit and Celsius Scales SECTION 1.6 \| Inequalities 119 The instructions on a bottle of medicine indicate that the bottle should be stored at a tem5 °C perature between. What range of temperatures does this correspond to on the Fahrenheit scale? 30 °C and ▼ SO LUTI O N (F ) is given by the equation terms of inequalities, we have C 5 91 The relationship between degrees Celsius (C ) and degrees Fahrenheit. Expressing the statement on the bottle in F 32 2 So the corresponding Fahrenheit temperatures satisfy the inequalities 5 C 30 9 5 F 32 5 5 91 2 # 5 F 32 9 9 F 32 54 30 # 30 5 9 32 F 54 32 41 F 86 Substitute C (F 32) 5 9 Multiply by 9 5 Simplify Add 32 Simplify The medicine should be stored at a temperature between 41 °F and 86 °F. ✎ Practice what you’ve learned: Do Exercise 81
. ▲ 1. ▼ CONCE PTS 1. Fill in the blank with an appropriate inequality sign. (a) If x 5, then x 3 (b) If x 5, then 3x (c) If x 2, then 3x (d) If x 2, then x 2. 15. 6. 2. (a) If x 2. True or false? x 1 2 1 or both negative. x 1 1 (b) If x 2 0, then x and x 1 are either both positive 5, then x and x 1 are each greater than 5. 2, 1, 12, 2, 4. Determine which ele- 6 ▼ SKI LLS S 3–10 ■ Let ments of S satisfy the inequality. 3. x 3 0 2, 1, 0, 1 5 3 2 x 1 5. 2 7. 1 2x 4 7 9. 1 x 1 2 4. x 1 2 6. 2x 1 x 8. 2 3 x 2 10. x 2 2 4 11–34 ■ Solve the linear inequality. Express the solution using interval notation and graph the solution set. 11. 2x 7 13. 2x 5 3 12. 4x 10 14. 3x 11 5 ✎ ✎ ✎ 15. 7 x 5 17. 2x 1 0 19. 3x 11 6x 8 ✎ 2 1 3 1 1 8x 23. 21 3x 25. 1 27. 2 x 5 4 29. 1 2x 5 7 31. 2 8 2x 1 2 x 3 12 1 6 33. 2 3 2 x x 12 x 16 16. 5 3x 16 18. 0 5 2x 20. 6 x 2x 9 5 x 1 1 1 2 x 1 24. 22. 2 3 5 2 6 2 1 2 7x 3 26. 28. 5 3x 4 14 30. 1 3x 4 16 3 3x 7 1 2 4 3x 5 1 2 32. 34. 1 4 35–74 ■ Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. 0 0 2 1 1 x 35 37. 39. x 2 3x 18 0 41. 2x 2 x 1 43. 3x 2 3x 2x 2 4 x 6 1 2 2 3 x 45. 47. x 2 4 x 2 49 36 3x 38. 40. x 2 5x 6 0 42. x 2 x 2 44. 5x 2 3
x 3x 2 2 46. x 2 2x 3 48. x 2 9 120 CHAPTER 1 \| Equations and Inequalities 50. 51. 53 x2 2 1 x2 1 1 2 57. 59. 2 0 54. 55. x 3 4x 0 x 3 x 1 4x 63. 65. 61. x 1 2 x 67. 69. 71 73. x 4 x 2 52 58. 56. 16x 60. 62. 64. 66. 68. 70. 72 3x 2x 74. x 5 x 2 75–78 ■ Determine the values of the variable for which the expression is deﬁned as a real number. 75. 216 9x 2 77. 1 1/2 a 2 5x 14 b x 76. 78. 2 5x 2 23x 4 1 x 2 x B 79. Solve the inequality for x, assuming that a, b, and c are positive constants. (a) a 1 bx c 2 bc (b) a bx c 2a 80. Suppose that a, b, c, and d are positive numbers such that Show that ✎ ▼ APPLICATIONS 81. Temperature Scales Use the relationship between C and F given in Example 8 to ﬁnd the interval on the Fahrenheit scale corresponding to the temperature range 20 C 30. 82. Temperature Scales What interval on the Celsius scale cor- responds to the temperature range 50 F 95? ✎ 83. Car Rental Cost A car rental company offers two plans for renting a car. Plan A: $30 per day and 10¢ per mile Plan B: $50 per day with free unlimited mileage For what range of miles will Plan B save you money? 84. Long-Distance Cost A telephone company offers two long- distance plans. Plan A: $25 per month and 5¢ per minute Plan B: $5 per month and 12¢ per minute For how many minutes of long-distance calls would Plan B be ﬁnancially advantageous? 85. Driving Cost It is estimated that the annual cost of driving a certain new car is given by the formula C 0.35m 2200 where m represents the number of miles driven per year and C is the cost in dollars. Jane has purchased such a car and decides to budget between $6400 and $7100 for next year’s driving costs. What is the corresponding range of miles that she can drive her new car? 86. Air Temperature As dry air moves upward
, it expands and, in so doing, cools at a rate of about rise, up to about 12 km. (a) If the ground temperature is temperature at height h. 1°C for each 100-meter 20 °C, write a formula for the (b) What range of temperatures can be expected if a plane takes off and reaches a maximum height of 5 km? 87. Airline Ticket Price A charter airline ﬁnds that on its Saturday ﬂights from Philadelphia to London all 120 seats will be sold if the ticket price is $200. However, for each $3 increase in ticket price, the number of seats sold decreases by one. (a) Find a formula for the number of seats sold if the ticket price is P dollars. (b) Over a certain period the number of seats sold for this ﬂight ranged between 90 and 115. What was the corresponding range of ticket prices? 88. Accuracy of a Scale A coffee merchant sells a customer 3 lb of Hawaiian Kona at $6.50 per pound. The merchant’s scale is accurate to within 0.03 lb. By how much could the customer have been overcharged or undercharged because of possible inaccuracy in the scale? 89. Gravity The gravitational force F exerted by the earth on an object having a mass of 100 kg is given by the equation F 4,000,000 d 2 where d is the distance (in km) of the object from the center of the earth, and the force F is measured in newtons (N). For what distances will the gravitational force exerted by the earth on this object be between 0.0004 N and 0.01 N? 90. Bonﬁre Temperature In the vicinity of a bonﬁre the temperature T in C at a distance of x meters from the center of the ﬁre was given by T 600,000 2 300 x At what range of distances from the ﬁre’s center was the temperature less than 500 C? where d is measured in feet. Kerry wants her stopping distance not to exceed 240 ft. At what range of speeds can she travel? SE CTI ON 1. 7 \| Absolute Value Equations and Inequalities 121 91. Falling Ball Using calculus, it can be shown that if a ball is thrown upward with an initial velocity of 16 ft/s from the top of a building 128 ft high, then its height h above the ground t seconds later will be h 128
16t 16t 2 During what time interval will the ball be at least 32 ft above the ground? 92. Gas Mileage The gas mileage g (measured in mi/gal) for a particular vehicle, driven at √ mi/h, is given by the formula g 10 0.9√ 0.01√ 2, as long as √ is between 10 mi/h and 75 mi/h. For what range of speeds is the vehicle’s mileage 30 mi/gal or better? 93. Stopping Distance For a certain model of car the distance d required to stop the vehicle if it is traveling at √ mi/h is given by the formula d √ √ 2 20 240 ft 94. Manufacturer’s Proﬁt If a manufacturer sells x units of a certain product, revenue R and cost C (in dollars) are given by R 20x C 2000 8x 0.0025x 2 Use the fact that profit revenue cost to determine how many units the manufacturer should sell to enjoy a proﬁt of at least $2400. 95. Fencing a Garden A determined gardener has 120 ft of deer-resistant fence. She wants to enclose a rectangular vegetable garden in her backyard, and she wants the area that is enclosed to be at least 800 ft 2. What range of values is possible for the length of her garden? ▼ DISCOVE RY • DISCUSSION • WRITI NG If a b, is a 2 b2? (Check 96. Do Powers Preserve Order? both positive and negative values for a and b.) If a b, is a 3 b3? On the basis of your observations, state a general rule about the relationship between a n and b n when a b and n is a positive integer. 97. What’s Wrong Here? It is tempting to try to solve an in- equality like an equation. For instance, we might try to solve 1 3/x by multiplying both sides by x, to get x 3, so the. But that’s wrong; for example, solution would be x 1 lies in this interval but does not satisfy the original inequality. Explain why this method doesn’t work (think about the sign of x). Then solve the inequality correctly. q, 3 2 1 1.7 Absolute Value Equations and Inequalities LEARNING OBJECTIVES After completing this section, you will be able to: ■ Sol
ve absolute value equations ■ Solve absolute value inequalities Recall from Section P.3 that the absolute value of a number a is given by a 0 0 a a e if a 0 if a 0 122 CHAPTER 1 \| Equations and Inequalities \| _5 \|=5 \| 5\|=5 _5 0 5 FIGURE 1 \| 5-2 \|=\| 2-5\|=3 0 2 5 FIGURE 2 and that it represents the distance from a to the origin on the real number line (see is the distance between x and a on the real number line. Figure 1). More generally, Figure 2 illustrates the fact that the distance between 2 and 5 is 3. x a 0 0 ■ Absolute Value Equations We use the following property to solve equations that involve absolute value. C is equivalent to x C x 0 0 This property says that to solve an absolute value equation, we must solve two separate is equivalent to the two equations x 5 and 5 equations. For example, the equation x 5. x 0 0 E X AM P L E 1 \| Solving an Absolute Value Equation Check Your Answers x 1: LHS RHS x 4: LHS 0 0 2 # 4 5 3 0 3 RHS 0 Solve the equation 2x 5 0 0 ▼ SO LUTI O N The equation 3. 2x 5 0 3 0 is equivalent to two equations: 2x 5 3 or 2x 5 3 2x 8 x 4 2x 2 x 1 Add 5 Divide by 2 The solutions are 1 and 4. ✎ Practice what you’ve learned: Do Exercise 13. ✔ ✔ E X AM P L E 2 \| Solving an Absolute Value Equation Solve the equation 3 x 7 5 14. 0 0 ▼ SO LUTI O N We ﬁrst isolate the absolute value on one side of the equal sign. x 7 5 14 Subtract 5 Divide by 3 x 7 3 or x 7 3 Take cases x 10 x 4 Add 7 The solutions are 4 and 10. ✎ Practice what you’ve learned: Do Exercise 17. ■ Absolute Value Inequalities We use the following properties to solve inequalities that involve absolute value. ▲ ▲ These properties hold when x is replaced by any algebraic expression. c c _c x 0 c \| x \| FIGURE 3 2 2 0 3 5 7 FIGURE 4 _2 0 2 3 FIGURE 5 SE CTI ON 1. 7 \| Absolute Value Equations and In
equalities 123 PROPERTIES OF ABSOLUTE VALUE INEQUALITIES Inequality c x 1. 0 0 2. 3. 4 Equivalent form c x c c x c x c or c x x c or c x Graph _c _c _c _c 0 0 0 0 c c c c These properties can be proved by using the deﬁnition of absolute value. To prove says that the distance from x to Property 1, for example, note that the inequality 0 is less than c, and from Figure 3 you can see that this is true if and only if x is between c and c. c x 0 0 0 0 E X AM P L E 3 \| Solving an Absolute Value Inequality 2. Solve the inequality x 5 0 ▼ SO LUTI O N 1 The inequality is equivalent to Property 1 Add 5 The solution set is the open interval 3, 7. 1 2 ▼ SO LUTI O N 2 Geometrically, the solution set consists of all numbers x whose distance from 5 is less than 2. From Figure 4 we see that this is the interval 3, 7 1. 2 ✎ Practice what you’ve learned: Do Exercise 27. E X AM P L E 4 \| Solving an Absolute Value Inequality Solve the inequality 3x 2 4. 0 ▼ SO LUTI O N By Property 4 the inequality 3x 2 4 is equivalent to or 3x 2 4 3x 2 x 2 3 So the solution set is 0 0 3x 2 4 3x 6 x 2 x x 2 or x 2 36 0 5 The solution set is graphed in Figure 5. ✎ Practice what you’ve learned: Do Exercise 33. 1 q, 2 Subtract 2 Divide by 3 2 3, q 3 2 4 ▲ ▲ E X AM P L E 5 \| Piston Tolerances The speciﬁcations for a car engine indicate that the pistons have diameter 3.8745 in. with a tolerance of 0.0015 in. This means that the diameters can vary from the indicated speciﬁcation by as much as 0.0015 in. and still be acceptable. 124 CHAPTER 1 \| Equations and Inequalities d (a) Find an inequality involving absolute values that describes the range of possible diameters for the pistons. (b) Solve the inequality. ▼ SO LUTI O
N (a) Let d represent the actual diameter of a piston. Since the difference between the actual diameter (d) and the speciﬁed diameter (3.8745) is less than 0.0015, we have (b) The inequality is equivalent to d 3.8745 0.0015 0 0 0.0015 d 3.8745 0.0015 3.8730 d 3.8760 Property 1 Add 3.8745 Acceptable piston diameters may vary between 3.8730 in. and 3.8760 in. ✎ Practice what you’ve learned: Do Exercise 57. ▲ 1. ▼ CONCE PTS 1. The equation x 3 0 0 has the two solutions. 2. The solution of the inequality 0 3. The solution of the inequality vals. x 0 x 0 0 3 3 and ✎. ✎ is the interval is a union of two inter- 4. (a) The set of all points on the real line whose distance from zero is less than 3 can be described by the absolute value inequality x 0 0. (b) The set of all points on the real line whose distance from zero is greater than 3 can be described by the absolute value inequality x 0 0. ▼ SKI LLS 5–22 ■ Solve the equation. 5. 7. 9. 11. 13. 15. 17. 19. 21. ✎ ✎ 0 0 4x 24 3 28 2 0.5 9 0 3x 4x 15 0 33 3 x 5 6 0 3x 2 0 0 0 0 6. 8. 10. 12. 14. 16. 18. 20. 22. 0 0 0 6x 15 7 2 7 0 3 1 6 14 15 2x 4 0 4 1 2 2x 1 0 1 x 2 0 0 2x 3 x 4 0 2 x 2 1 5 2x 0 20 23–48 ■ Solve the inequality. Express the answer using interval notation. 23. 25. x 0 2x 0 0 4 7 0 24. 0 26. 1 2 0 3x x 15 1 0 0 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 3 1 2 0. 2x.001 3 13 2x 1 0 5 6 0 4x 28. 30. 32. 34. 36. 38. 40. 44. 46. 48 5x 2x 51 1 2 0 0 1 2x
3 0 5 0 42. 7 0 49–52 ■ A phrase that describes a set of real numbers is given. Express the phrase as an inequality involving an absolute value. 49. All real numbers x less than 3 units from 0 50. All real numbers x more than 2 units from 0 51. All real numbers x at least 5 units from 7 52. All real numbers x at most 4 units from 2 53–56 ■ A set of real numbers is graphed. Find an inequality involving an absolute value that describes the set. 53. 54. 55. 56. _5 _4 _3 _2 _1 0 _5 _4 _3 _2 _1 0 _5 _4 _3 _2 _1 0 _5 _4 _3 _2 _1 ✎ ▼ APPLICATIONS 57. Thickness of a Laminate A company manufactures industrial laminates (thin nylon-based sheets) of thickness 0.020 in., with a tolerance of 0.003 in. (a) Find an inequality involving absolute values that describes the range of possible thickness for the laminate. (b) Solve the inequality that you found in part (a). 0.020 in. CHAPTER 1 \| REVIEW ▼ P R O P E RTI LAS Properties of Equality (p. 66 CA CB C 0 1 2 Linear Equations (p. 66) A linear equation is an equation of the form ax b 0 Power Equations (p. 69) A power equation is an equation of the form tions are Xn a. Its solu- X 1n a X 1n a if n is odd if n is even If n is even and a 0, the equation has no real solution. Simple Interest (p. 75) If a principal P is invested at simple annual interest rate r for t years, then the interest I is given by I Prt Zero-Product Property (p. 87) If AB 0 then A 0 or B 0. Completing the Square (p. 88) x2 bx To make perfect square a perfect square, add 2. This gives the b 2 b a x 2 bx CHAPTER 1 \| Review 125 58. Range of Height The average height of adult males is 68.2 in., and 95% of adult males have height h that satisﬁes the inequality h 68.2 2.9 ` ` 2 Solve the inequality to ﬁnd the range of heights. ▼ DISCOVE R
Y • DISCUSSION • WRITI NG 59. Using Distances to Solve Absolute Value Inequalities is the distance between a and b on the Recall that 0 number line. For any number x, what do 0 represent? Use this interpretation to solve the inequality x 3 x 1 0 the solution of the inequality 0 geometrically. In general, if a b, what is and? Quadratic Formula (pp. 89–91) A quadratic equation is an equation of the form ax2 bx c 0 Its solutions are given by the Quadratic Formula: x b 2b2 4ac 2a b2 4ac. The discriminant is D If D 0, the equation has two real solutions. If D 0, the equation has one solution. If D 0, the equation has two complex solutions. Complex Numbers (pp. 98–101) A complex number is a number of the form a bi, where i 11. The complex conjugate of a bi is a bi a bi To multiply complex numbers, treat them as binomials and use i2 1 to simplify the result. To divide complex numbers, multiply numerator and denominator by the complex conjugate of the denominator: a bi c di a bi c di # c di c di 1 ¢ ≤ ¢ ≤ a bi c di 2 1 c2 d2 2 Absolute Value Equations (p. 122) To solve an absolute value equation, we use C 3 x C or x C x 0 0 Absolute Value Inequalities (p. 123) To solve an absolute value inequality, we use C 3 C x C C 3 x C or x C x x 0 0 0 0 126 CHAPTER 1 \| Equations and Inequalities Inequalities (p. 113) Adding the same quantity to each side of an inequality gives an equivalent inequality: A B 3 A C B C Multiplying each side of an inequality by the same positive quantity gives an equivalent inequality. Multiplying each side by the same negative quantity reverses the direction of the inequality: A B 3 CA CB A B 3 CA CB if C 0 if MARY Section 1.1 ■ Solve linear equations ■ Solve power equations ■ Solve for one variable in terms of another Section 1.2 ■ Make equations that model real-world situations ■ Use equations to answer questions about real-world situations Section 1.3 ■ Solve quadratic equations by factoring, completing the square, or
using the quadratic formula ■ Model with quadratic equations Section 1.4 ■ Add and subtract complex numbers ■ Multiply and divide complex numbers ■ Simplify expressions with roots of negative numbers ■ Find complex roots of quadratic equations Section 1.5 ■ Solve basic polynomial equations ■ Solve equations involving radicals or fractional powers ■ Solve equations of quadratic type ■ Model with equations Section 1.6 ■ Solve linear inequalities ■ Solve nonlinear inequalities ■ Model with inequalities Section 1.7 ■ Solve absolute value equations ■ Solve absolute value inequalities ▼ E X E RC I S E S 1–18 ■ Find all real solutions of the equation. 1. 5x 11 36 3. 3x 12 24 2. 3 x 5 3x 4. 5x 7 42 5. 7x 6 4x. 9. 2 1 4 1 2 x 5 2 Review Exercises 1–12 13–18, 23–28 19–22 Review Exercises 49–56 49–56 Review Exercises 31–38 52, 53, 55, 56 Review Exercises 57–58 59–64 65–66 67–70 Review Exercises 69–74 25–28, 42–44 41–44 51, 53, 54 Review Exercises 75–78 79–84, 89 90 Review Exercises 45–48 85–88 6. 8 2x 14 x 2x 3 x 3 1 5 2 5 8. 8 5x 5 6 the following formula, where x is the speed of the car (in mi/h): CHAPTER 1 \| Review 127 11. 10. 2x 5 3 2x 1 2x 1 x 5 2 x 1 x 1 13. x 2 144 15. 5x 4 16 0 17. 5x 3 15 0 3 1 x 2 12. x x 2 14. 4x 2 49 16. x 3 27 0 18. 6x 4 15 0 19–22 ■ Solve the equation for the indicated variable. solve for x 19. 20. 21. ; A x y 2 V xy yz xz ; J 1 t 1 2t 1 3t ; solve for y solve for t 22. F k q1q2 r 2 ; solve for r 23–48 ■ Find all real solutions of the equation. 2 23. 3 64 x 1 1 13 x 3 25. 27. 4x 3/4 500 0 x 1 x 1 3x 29. 3x 6
31. x 2 9x 14 0 33. 2x 2 x 1 35. 4x 3 25x 0 37. 3x 2 4x 1 0 3 2 39. 1 x x 1 41. x 4 8x 2 9 0 43. x1/2 2x 1/2 x 3/2 0 1 1x 44. 1 1 1x x 7 0 2x 5 2 2 2 4 9 1 0 45. 47 24. 26. x 2/3 4 0 x 2 28. 1/5 2 1 2 30 32. x 2 24x 144 0 34. 3x 2 5x 2 0 36. x 3 2x 2 5x 10 0 38. x 2 3x 9 0 1 x x 2 x 2 x 41x 32 40. 42. 8 x 2 4 15 0 2 46. 48. 18 3x 0 3 x 0 4 0 3 15 0 49. A shopkeeper sells raisins for $3.20 per pound and nuts for $2.40 per pound. She decides to mix the raisins and nuts and sell 50 lb of the mixture for $2.72 per pound. What quantities of raisins and nuts should she use? 50. Anthony leaves Kingstown at 2:00 P.M. and drives to Queensville, 160 mi distant, at 45 mi/h. At 2:15 P.M. Helen leaves Queensville and drives to Kingstown at 40 mi/h. At what time do they pass each other on the road? 51. A woman cycles 8 mi/h faster than she runs. Every morning 1 she cycles 4 mi and runs mi, for a total of 1 hour of exer2 cise. How fast does she run? 2 52. The approximate distance d (in feet) that drivers travel after noticing that they must come to a sudden stop is given by 2 d x x 20 If a car travels 75 ft before stopping, what was its speed before the brakes were applied? 53. The hypotenuse of a right triangle has length 20 cm. The sum of the lengths of the other two sides is 28 cm. Find the lengths of the other two sides of the triangle. 54. Abbie paints twice as fast as Beth and three times as fast as Cathie. If it takes them 60 min to paint a living room with all three working together, how long would it take Abbie if she works alone? 55. A rectangular swimming pool is 8 ft deep everywhere and twice as long
as it is wide. If the pool holds 8464 ft 3 of water, what are its dimensions? 56. A gardening enthusiast wishes to fence in three adjoining garden plots, one for each of his children, as shown in the ﬁgure. If each plot is to be 80 ft 2 in area and he has 88 ft of fencing material at hand, what dimensions should each plot have? 57–66 ■ Evaluate the expression and write the result in the form a bi. 6 4i 3i 58. 1 60. 3 1 5 2i 2 2 i 5 57. 59. 61. 3 5i 2 7i 1 1 2 2 1 2 3i 2 3i 63. i 45 2 i 4 3i 3 i 2 1 15 # 120 3 62. 64. 66. 65. 1 1 13 2 14 2 2 1 67–74 ■ Find all real and imaginary solutions of the equation. 67. x 2 16 0 69. x 2 6x 10 0 71. x 4 256 0 2 4x 68. x 2 12 70. 2x 2 3x 2 0 72. x 3 2x 2 4x 8 0 74. x 3 125 2x 1 73. x 2 1 2 75–88 ■ Solve the inequality. Express the solution using interval notation, and graph the solution set on the real number line. 75. 3x 2 11 77. 1 2x 5 3 76. 12 x 7x 78. 3 x 2x 7 128 CHAP TER 1 \| Equations and Inequalities 79. x 2 4x 12 0 80. x 2 1 89. For what values of x is the algebraic expression deﬁned as a 82. 2x 2 x 3 real number? (a) 224 x 3x 2 (b) 5 84. 86. 3 x x x 4 2 4x 4 0.02 0 90. The volume of a sphere is given by 3, where r is the radius. Find the interval of values of the radius so that the volume is between 8 ft 3 and 12 ft 3, inclusive. 1 24 x x V 4 3pr 4 2x 81. 83. 85. 87. 88. 0 0 0 2x 1 x 1 0 [Hint: Interpret the quantities as distances.] 1 0 x 3 0 0 0 0 0 ■ CHAPTER 1 \| TEST 1. Find all real solutions of each equation. (a) 4x 3 2x 7 (b) 8x3 125 (
c) x2/3 64 0 (d) x 2x 5 x 3 2x 1 2. Einstein’s famous equation E mc2 gives the relationship between energy E and mass m. In this equation c represents the speed of light. Solve the equation to express c in terms of E and m. 3. Natasha drove from Bedingﬁeld to Portsmouth at an average speed of 100 km/h to attend a job interview. On the way back she decided to slow down to enjoy the scenery, so she drove at just 75 km/h. Her trip involved a total of 3.5 hours of driving time. What is the distance between Bedingﬁeld and Portsmouth? 4. Calculate, and write the result in the standard form for complex numbers: a bi. (a) (b) (c) (d) (e) (f) 2 1 3 5i 3 5i 2i 2 2 1 1 1 2i 3 4i i25 2 12 1 18 14 2 2 1 5. Find all solutions, real and complex, of each equation. (a) x 2 x 12 0 (b) 2x 2 4x 3 0 23 1x 5 2 (c) (d) x 1/2 3x 1/4 2 0 (e) x 4 16x 2 0 x 4 (f) 10 0 3 0 0 6. A rectangular parcel of land is 70 ft longer than it is wide. Each diagonal between opposite corners is 130 ft. What are the dimensions of the parcel? 7. Solve each inequality. Sketch the solution on a real number line, and write the answer using interval notation. (a) 1 5 2x 10 (b) x 0 x 1 1 x 3 0 2x c) (d) 8. A bottle of medicine must be stored at a temperature between 5C and 10C. What range does [Note: The Fahrenheit (F ) and Celsius (C ) scales this correspond to on the Fahrenheit scale? C 5 satisfy the relation 91 F 32.] 2 9. For what values of x is the expression 24x x 2 deﬁned as a real number? 129 MAKING THE BEST DECISIONS When you buy a car, subscribe to a cell phone plan, or put an addition on your house, you need to make decisions. Such decisions are usually difﬁcult because they require you to choose between several good alternatives. For example, there are
many good car models, but which one has the optimal combination of features for the amount of money you want to spend? In this Focus we explore how to construct and use algebraic models of real-life situations to help make the best (or optimal) decisions. E X AM P L E 1 \| Buying a Car Ben wants to buy a new car, and he has narrowed his choices to two models. Model A sells for $12,500, gets 25 mi/gal, and costs $350 a year for insurance. Model B sells for $16,100, gets 48 mi/gal, and costs $425 a year for insurance. Ben drives about 36,000 miles a year, and gas costs about $1.50 a gallon. (a) Find a formula for the total cost of owning model A for any number of years. (b) Find a formula for the total cost of owning model B for any number of years. (c) Make a table of the total cost of owning each model from 1 year to 6 years, in 1-year increments. (d) If Ben expects to keep the car for 3 years, which model is more economical? What if he expects to keep it for 5 years? Thinking About the Problem Model A has a smaller initial price and costs less in insurance per year but is more costly to operate (uses more gas) than model B. Model B has a larger initial price and costs more to insure but is cheaper to operate (uses less gas) than model A. If Ben drives a lot, then what he will save in gas with model B could make up for the initial cost of buying the car and the higher yearly insurance premiums. So how many years of driving does it take before the gas savings make up for the initial higher price? To ﬁnd out, we must write formulas for the total cost for each car: cost price insurance cost gas cost The insurance costs and gas costs depend on the number of years Ben drives the car. ▼ SO LUTI O N The cost of operating each model depends on the number of years of ownership. So let n number of years Ben expects to own the car (a) For model A we have the following. In Words Price of car Insurance cost for n years Cost of gas per year Cost of gas for n years In Algebra 12,500 350n 36,000/25 1 2160n 2 $1.50 $2160 130 Making the Best Decisions 131 Let C represent the cost of
owning model A for n years. Then cost of ownership initial cost insurance cost gas cost C 12,500 350n 2160n C 12,500 2510n (b) For model B we have the following. In Words Price of car Insurance cost for n years Cost of gas per year Cost of gas for n years In Algebra 16,100 425n 36,000/48 1 1125n 2 $1.50 $1125 Let C represent the cost of owning model B for n years. cost of ownership initial cost insurance cost gas cost C 16,100 425n 1125n C 16,100 1550n (c) If Ben keeps the car for 2 years, the cost of ownership can be calculated from the formulas we found by substituting 2 for n. For model A: C 12,500 2510 17,520 2 1 2 For model B: C 16,100 1550 2 2 The other entries in the table are calculated similarly. 1 19,200 Years Cost of ownership model A Cost of ownership model B 1 2 3 4 5 6 15,010 17,520 20,030 22,540 25,050 27,560 17,650 19,200 20,750 22,300 23,350 25,400 (d) If Ben intends to keep the car 3 years, then model A is a better buy (see the table), but if he intends to keep the car 5 years, model B is the better buy. ▲ E X AM P L E 2 \| Equal Ownership Cost Find the number of years of ownership for which the cost to Ben (from Example 1) of owning model A equals the cost of owning model B. 132 Focus On Modeling Thinking About the Problem We see from the table that the cost of owning model A starts lower but then exceeds that for model B. We want to ﬁnd the value of n for which the two costs are equal. ▼ SO LUTI O N We equate the cost of owning model A to that of model B and solve for n. 12,500 2510n 16,100 1550n 960n 3600 n 3.75 Set the two costs equal Subtract 12,500 and 1550n Divide by 960 If Ben keeps the car for about 3.75 years, the cost of owning either model would be the ▲ same. E X AM P L E 3 \| Dividing Assets Fairly When high-tech company A goes bankrupt, it owes $120 million to company B
and $480 million to company C. Unfortunately, company A only has $300 million in assets. How should the court divide these assets between companies B and C? Explore the following methods and determine which are fair. (a) Companies B and C divide the assets equally. (b) The two companies share the losses equally. (c) The two companies get an amount that is proportional to the amount they are owed. Thinking About the Problem It might seem fair for companies B and C to divide the assets equally between them. Or it might seem fair that they share the loss equally between them. To be certain of the fairness of each plan, we should calculate how much each company loses under each plan. ▼ SO LUTI O N (a) Under this method, company B gets $150 million and company C gets $150 million. Because B is owed only $120 million, it will get $30 million more than it is owed. This doesn’t seem fair to C, which will still lose $330 million. (b) We want B and C to each lose the same amount. Let x be the amount of money com. We can organize the in- 300 x pany B gets. Then company C would get the rest formation as follows. 1 2 In Words Amount B gets Amount C gets Amount B loses Amount C loses In Algebra x 300 x 120 x 480 1 300 x 180 x 2 Because we want B and C to lose equal amounts, we must have 180 x 120 x 2x 60 x 30 Amounts B and C lose are equal Add x, subtract 180 Divide by 2 Making the Best Decisions 133 Check Your Answer B loses 120 30 150 million C loses 480 330 150 million They lose equal amounts. ✔ Thus, B gets 30 million dollars. The negative sign means that B must give up an additional $30 million and pay it to C. So C gets all of the $300 million plus $30 million from B for a total of $330 million. Doing this would ensure that the two companies lose the same amount (see Check Your Answer). This method is clearly not fair. (c) The claims total $120 million $480 million $600 million. The assets total $300 million. Because company B is owed $120 million out of the total claim of $600 million, it would get 120 million 600 million 300 million $60 million Because company C is owed 480 million, it would get 480 million 600 million 300 million $240 million This seems like the fairest alternative
. ▲ Problems 1. Renting Versus Buying a Photocopier A certain ofﬁce can purchase a photocopier for $5800 with a maintenance fee of $25 a month. On the other hand, they can rent the photocopier for $95 a month (including maintenance). If they purchase the photocopier, each copy would cost 3¢; if they rent, the cost is 6¢ per copy. The ofﬁce estimates that they make 8000 copies a month. (a) Find a formula for the cost C of purchasing and using the copier for n months. (b) Find a formula for the cost C of renting and using the copier for n months. (c) Make a table of the cost of each method for 1 year to 6 years of use, in 1-year increments. (d) After how many months of use would the cost be the same for each method? 2. Car Rental A businessman intends to rent a car for a 3-day business trip. The rental is $65 a day and 15¢ per mile (Plan 1) or $90 a day with unlimited mileage (Plan 2). He is not sure how many miles he will drive but estimates that it will be between 400 and 800 miles. (a) For each plan, ﬁnd a formula for the cost C in terms of the number x of miles driven. (b) Which rental plan is cheaper if the businessman drives 400 miles? 800 miles? (c) At what mileage do the two plans cost the same? 3. Cost and Revenue A tire company determines that to manufacture a certain type of tire, it costs $8000 to set up the production process. Each tire that is produced costs $22 in material and labor. The company sells this tire to wholesale distributors for $49 each. (a) Find a formula for the total cost C of producing x tires. (b) Find a formula for the revenue R from selling x tires. (c) Find a formula for the proﬁt P from selling x tires. [Note: proﬁt revenue cost.] (d) How many tires must the company sell to break even? 4. Enlarging a Field A farmer has a rectangular cow pasture with width 100 ft and length 180 ft. An increase in the number of cows requires the farmer to increase the area of her pasture. She has two options: Option 1: Increase the length of the ﬁe
ld. Option 2: Increase the width of the ﬁeld. 134 Focus On Modeling It costs $10 per foot to install new fence. Moving the old fence costs $6 per linear foot of fence to be moved. (a) For each option, ﬁnd a formula for A, the area gained, in terms of the cost C. (b) Complete the table for the area gained in terms of the cost for each option. Cost Area gain (option 1) Area gain (option 2) 2500 ft2 180 ft2 $1100 $1200 $1500 $2000 $2500 $3000 (c) If the farmer has $1200 for this project, which option gives her the greatest gain in area for her money? What if she had $2000 for the project? 5. Edging a Planter A woman wants to make a small planter and surround it with edging material. She is deciding between two designs. Design 1: A square planter Design 2: A circular planter Edging material costs $3 a foot for the straight variety, which she would use for design 1, and $4 a foot for the ﬂexible variety, which she would use for design 2. (a) If she decides on a perimeter of 24 ft, which design would give her the larger planting area? (b) If she decides to spend $120 on edging material, which design would give her the larger planting area? 6. Planting Crops A farmer is considering two plans of crop rotation on his 100-acre farm. Plan A: Plant tomatoes every season. Plan B: Alternate between soybeans and tomatoes each season. The revenue from tomatoes is $1600 an acre, and the revenue from soybeans is $1200 an acre. Tomatoes require fertilizing the land, which costs about $300 an acre. Soybeans do not require fertilizer; moreover, they add nitrogen to the soil so tomatoes can be planted the following season without fertilizing. (a) Find a formula for the proﬁt if plan A is used for n years. (b) Find a formula for the proﬁt if plan B is used for 2n years (starting with soybeans). (c) If the farmer intends to plant these crops for 10 years, which plan is more proﬁtable? Proﬁt Revenue Cost 7. Cell Phone Plan Genevieve is mulling over the three cell phone plans shown in the table. Making the Best
Decisions 135 Minutes included Monthly cost Each additional minute Plan A Plan B Plan C 500 500 500 $30 $40 $60 $0.50 $0.30 $0.10 From past experience, Genevieve knows that she will always use more than 500 minutes of cell phone time every month. (a) Make a table of values that shows the cost of each plan for 500 to 1100 minutes, in 100-minute increments. (b) Find formulas that give Genevieve’s monthly cost for each plan, assuming that she uses x minutes per month (where x 500). (c) What is the charge from each plan when Genevieve uses 550 minutes? 975 minutes? 1200 minutes? (d) Use your formulas from part (b) to determine the number of usage minutes for which: i(i) Plan A and Plan B give the same cost. (ii) Plan A and Plan C give the same cost. 8. Proﬁt Sharing To form a new enterprise, company A invests $1.4 million and company B invests $2.6 million. The enterprise is sold a year later for $6.4 million. Explore the following methods of dividing the $6.4 million, and comment on their fairness. (a) Companies A and B divide the $6.4 million equally. (b) Companies A and B get their original investment back and share the proﬁt equally. (c) Each company gets a fraction of the $6.4 million proportional to the amount it invested. This page intentionally left blank 2.1 The Coordinate Plane 2.2 Graphs of Equations in Two Variables 2.3 Graphing Calculators: Solving Equations and Inequalities Graphically 2.4 Lines 2.5 Making Models Using Variation CHAPTER 2 Coordinates and Graphs Global Warming? Is the world getting hotter or are we just having a temporary warm spell? To answer this question, scientists collect huge amounts of data on global temperature. A graph of the data can help to reveal long-term changes in temperature, but more precise algebra methods are used to detect any trend that is different from the normal temperature ﬂuctuations. Signiﬁcant global warming could have drastic consequences for the survival of many species. For example, melting polar ice eliminates the ice paths that polar bears need to reach their feeding grounds, resulting in the bears’ drowning. (See Focus on Modeling: F
itting Lines to Data, Problem 6, page 199.) Of course, any changes in the global ecology also have implications for our own well-being. 137137 137 138 CHAPTER 2 \| Coordinates and Graphs 2.1 The Coordinate Plane LEARNING OBJECTIVES After completing this section, you will be able to: ■ Graph points and regions in the coordinate plane ■ Use the Distance Formula ■ Use the Midpoint Formula The Cartesian plane is named in honor of the French mathematician René Descartes (1596–1650), although another Frenchman, Pierre Fermat (1601–1665), also invented the principles of coordinate geometry at the same time. (See their biographies on pages 245 and 159.) The coordinate plane is the link between algebra and geometry. In the coordinate plane we can draw graphs of algebraic equations. The graphs, in turn, allow us to “see” the relationship between the variables in the equation. Just as points on a line can be identiﬁed with real numbers to form the coordinate line, points in a plane can be identiﬁed with ordered pairs of numbers to form the coordinate plane or Cartesian plane. To do this, we draw two perpendicular real lines that intersect at 0 on each line. Usually, one line is horizontal with positive direction to the right and is called the x-axis; the other line is vertical with positive direction upward and is called the y-axis. The point of intersection of the x-axis and the y-axis is the origin O, and the two axes divide the plane into four quadrants, labeled I, II, III, and IV in Figure 1. (The points on the coordinate axes are not assigned to any quadrant.) I I y b O P (a, b) I a x (_2, 2) y 1 0 (1, 3) ) (5, 0_3, _2) (2, _4) FIGURE 1 FIGURE 2 a, b Although the notation for a point is the same as the notation for an open, the context should make interval clear which meaning is intended. a, b 1 1 2 2 1 a, b Any point P in the coordinate plane can be located by a unique ordered pair of num, as shown in Figure 1. The ﬁrst number a is called the x-coordinate of P; the bers second number b is called the y-coordinate of P. We can think
of the coordinates of P as its “address,” because they specify its location in the plane. Several points are labeled with their coordinates in Figure 2. 2 E X AM P L E 1 \| Graphing Regions in the Coordinate Plane Describe and sketch the regions given by each set. (a) y 1 x 0 x, y x, y (b) 51 2 0 6 51 2 0 6 (c) x, y y 0 2 @ 0 51 1 6 ▼ SO LUTI O N (a) The points whose x-coordinates are 0 or positive lie on the y-axis or to the right of it, as shown in Figure 3(a). (b) The set of all points with y-coordinate 1 is a horizontal line one unit above the x-axis, as in Figure 3(b). Coordinates as Addresses The coordinates of a point in the xy-plane uniquely determine its location. We can think of the coordinates as the “address” of the point. In Salt Lake City, Utah, the addresses of most buildings are in fact expressed as coordinates. The city is divided into quadrants with Main Street as the vertical (NorthSouth) axis and S. Temple Street as the horizontal (East-West) axis. An address such as 1760 W 2100 S indicates a location 17.6 blocks west of Main Street and 21 blocks south of S. Temple Street. (This is the address of the main post ofﬁce in Salt Lake City.) With this logical system it is possible for someone who is unfamiliar with the city to locate any address immediately, as easily as one locates a point in the coordinate plane. 500 North St. S. Temple St. 4th South St. 9th South St. 13th South St. 17th South St. 21st South St. Post Office 1760 W 2100 S SE CTI ON 2.1 \| The Coordinate Plane 139 (c) Recall from Section 1.7 that 1 if and only if 1 y 1 y 0 0 So the given region consists of those points in the plane whose y-coordinates lie between 1 and 1. Thus, the region consists of all points that lie between (but not on) the horizontal lines y 1 and y 1. These lines are shown as broken lines in Figure 3(c) to indicate that the points on these lines do not lie in the set=1 x y=_1 (a) x≥
0 (b) y=1 (c) \| y \| <1 FIGURE 3 ✎ Practice what you’ve learned: Do Exercises 7 and 9. ▲ ■ The Distance Formula A, B d x2, y2 2 We now ﬁnd a formula for the distance in the plane. Recall from Section P.3 that the distance between points a and b on a number d line is. So from Figure 4 we see that the distance between the points 1 x1 0 x1, y12 A and the distance between x2, y22 B. on a horizontal line must be y2 on a vertical line must be 0 C x2, y12 1 x2, y12 C 1 between two points a, b 2 and and x2 0 y1 0 x1, y12 b a and ¤ d ( A, B ) B(x¤, y ¤) \| y¤-y⁄ \| y⁄ A(x⁄, y⁄) \| x¤-x⁄ \| C(x¤, y⁄) 0 x⁄ x¤ x FIGURE 4 Since triangle ABC is a right triangle, the Pythagorean Theorem gives A, B d 1 2 2 x2 0 x1 0 2 y2 0 y1 0 2 2 x2 1 x12 2 y2 1 2 y12 DISTANCE FORMULA The distance between the points A 1 x1, y12 x2 1 and B 1 x2, y22 y2 1 A, B d 1 2 2 2 x12 2 y12 in the plane is 140 CHAPTER 2 \| Coordinates and Graphs y 5 4 3 2 1 0 _1 A(2, 5) E X AM P L E 2 \| Finding the Distance Between Two Points Find the distance between the points A 2, 5 1 2 and B 1 4, 1. 2 ▼ SO LUTI O N Using the Distance Formula, we have d(A, B)Å6.32 A 222 2 6 2 1 5 2 2 1 2 24 36 240 6.32 1 1 2 3 4 5 x B(4, _1) See Figure 5. ✎ Practice what you’ve learned: Do Exercise 25(b). FIGURE 5 y 8 6 4 2 0 _2 FIGURE 6 Q(8, 9) A(5, 3) P(1, _2
) 8 x E X AM P L E 3 \| Applying the Distance Formula 1, 2 Which of the points 8, 9 or Q P is closer to the point 1 2 1 2 5, 3 A 1? 2 ▼ SO LUTI O N By the Distance Formula we have d d 1 1 P, A 2 Q 242 52 141 2 3 1 2 6 1 2 2 145 2 1 d d This shows that 1 ✎ Practice what you’ve learned: Do Exercise 39. Q, A P, A 2 2 1, so P is closer to A (see Figure 6). ▲ ▲ ■ The Midpoint Formula of the midpoint M of the line segment that joins the x, y Now let’s ﬁnd the coordinates 2 1. In Figure 7, notice that triangles APM and MQB are x1, y1 2 x2, y22 B point 1 d congruent because 1 and the corresponding angles are equal. to the point A, M M Midpoint M(x, y) A(x⁄, y⁄) P x-x⁄ B(x¤, y¤) Q x¤-x x FIGURE 7 d It follows that A, P 1 2 d 1 M, Q, so 2 x x1 x2 x Solving for x, we get 2x x1 x2, so x x1 x2 2. Similarly, y y1 y2 2. SE CTI ON 2.1 \| The Coordinate Plane 141 MIDPOINT FORMULA A The midpoint of the line segment from x2 2 x1 a B x2, y2 2 1 is 1, x1, y1 2 y1 to y2 2 b E X AM P L E 4 \| Finding the Midpoint The midpoint of the line segment that joins the points 2, 5 1 and 4, 9 is 2 1 2 See Figure 8, 7 1 2 (4, 9) (1, 7) y 8 4 (_2, 5) FIGURE 8 _4 0 4 x ✎ Practice what you’ve learned: Do Exercise 25(c). ▲ E X AM P L E 5 \| Applying the Midpoint Formula R Show that the quadrilateral with vertices lelogram by proving that its two diagonals bisect each other. 1, 2 4, 9 1 2, and S 2, 7 1 2 is a paral-
y 8 4 0 S P FIGURE 9 R Q 4 x ▼ SO LUTI O N other. The midpoint of the diagonal PR is If the two diagonals have the same midpoint, then they must bisect each 1 5 2, 2 9 2 a b 11 2 b 3, a and the midpoint of the diagonal QS is 4 2 2, 4 7 2 a b 11 2 b 3, a so each diagonal bisects the other, as shown in Figure 9. (A theorem from elementary geometry states that the quadrilateral is therefore a parallelogram.) ✎ Practice what you’ve learned: Do Exercise 43. ▲ 142 CHAPTER 2 \| Coordinates and Graphs 2. ▼ CONCE PTS 1. The point that is 3 units to the right of the y-axis and 5 units below the x-axis has coordinates, 1. 2 2. If x is negative and y is positive, then the point (x, y) is in Quadrant. 3. The distance between the points a, b 1 and 2 1, 2 1 2 c, d 1 and is 2 7, 10 2 1. So the distance between. 4. The point midway between So the point midway between a, b 1 and is c, d 1 and 2 7, 10 1 2 1, 2 1 2 is 2 is.. 17. 18. 19. 20. x, y x, y x, y x and 2 and 21–24 ■ A pair of points is graphed. (a) Find the distance between them. (b) Find the midpoint of the segment that joins them. 21. 1 x 22. 24 ▼ SKI LLS 5. Plot the given points in a coordinate plane: 2, 3, 1 2 1 2, 3 4, 5, 1 2, 1 2 4, 5 4, 5, 1 2, 1 2 4, 5 2 23. 6. Find the coordinates of the points shown in the ﬁgure 25–34 ■ A pair of points is graphed. (a) Plot the points in a coordinate plane. (b) Find the distance between them. (c) Find the midpoint of the segment that joins them. 7–20 ■ Sketch the region given by the set. ✎ ✎ 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. x, y x, y x, y x, y
x, y x, y x, y x, y x, y x, y 51 51 51 51 51 51 51 51 51 51 xy 0 6 xy 0 6 x 1 and y 3 6 2 x 2 and y 3 6 ✎ 25. 27. 29. 31. 33. 1 1 1 1 1 0, 8, 2 1 3, 6 6, 2 2, 7, 3 3, 4,, 2 2 1 1 6, 16 2 4, 18, 2 1 1, 3 2 1 11, 6 2 3, 4 2 2 26. 28. 30. 32. 34. 1 1 1 1, 2, 5 2 1, 1 1 2, 10, 0 2 9, 9 1, 6, 2, 1, 2, 13 5, 0 1 A 2 1, 3 2, 1 0, 6 2 5, 3 2 1 1, 3 2 1 7, 1 2 35. Draw the rectangle with vertices 1, 3 1 on a coordinate plane. Find the area of the, 3 D and rectangle. 1 2 36. Draw the parallelogram with vertices C on a coordinate plane. Find the area of the 1, 2 5, 6 and 1 parallelogram. 2 3, 6, 2 1 37. Plot the points A 1, 0, B 5, 0, C 4, 3, and D 2, 3 2 coordinate plane. Draw the segments AB, BC, CD, and DA. What kind of quadrilateral is ABCD, and what is its area? 2 1 2 2 1 1 1 on a 38. Plot the points on a coordinate 2 plane. Where must the point S be located so that the quadrilateral PQRS is a square? Find the area of this square., and 5, 1 0, 1 SE CTI ON 2.1 \| The Coordinate Plane 143 is closer to the origin? is closer to the point 53. Plot the points Q 1 nate plane. Where should the point S be located so that the ﬁgure PQRS is a parallelogram?, and on a coordi- 1, 1 4, 4 6, 7 2 6, 3 or B 5, 8 1 D or 2 3, 0 2 1 2 1 1 ✎ 39. Which of the points A 40. Which of the points C E 2, 1? 1 2 41. Which of the points R 1, 1? 1 2 42. (a) Show that the
points 7, 3 distance from the origin. 1 3, 1 P 1 2 or Q 1 1, 3 2 is closer to the point and 3, 7 1 2 2 are the same (b) Show that the points a, b 1 and b, a 1 2 2 are the same distance from the origin. ✎ 43. Show that the triangle with vertices is isosceles. 4, 3 C 1 2 0, 2 A 1, B 1 2 3, 1, and 2 other. is the midpoint of the line segment AB, and if A has 54. If M 6, 8 1 coordinates 2 2, 3 1 2, ﬁnd the coordinates of B. 55. (a) Sketch the parallelogram with vertices A B 4, 2, and (b) Find the midpoints of the diagonals of this 1, 4 7, 1 1, 2 parallelogram. (c) From part (b), show that the diagonals bisect each 56. The point M in the ﬁgure is the midpoint of the line segment AB. Show that M is equidistant from the vertices of triangle ABC. y B(0, b) M C (0, 0) A(a, 0) x ▼ APPLICATIONS 57. Distances in a City A city has streets that run north and south and avenues that run east and west, all equally spaced. Streets and avenues are numbered sequentially, as shown in the ﬁgure. The walking distance between points A and B is 7 blocks—that is, 3 blocks east and 4 blocks north. To ﬁnd the straight-line distances d, we must use the Distance Formula. (a) Find the straight-line distance (in blocks) between A and B. (b) Find the walking distance and the straight-line distance between the corner of 4th St. and 2nd Ave. and the corner of 11th St. and 26th Ave. (c) What must be true about the points P and Q if the walking distance between P and Q equals the straight-line distance between P and Q blocks. 5th Ave. 4th Ave. s k c o l b 4 3rd Ave. 2nd Ave. 1st Ave. 44. Find the area of the triangle shown in the ﬁgure. y 4 2 C A B _2 0 2 4 6 8 x _2 45. Refer to
triangle ABC in the ﬁgure. (a) Show that triangle ABC is a right triangle by using the converse of the Pythagorean Theorem (see page 284). (b) Find the area of triangle ABC. y 2 A _4 _2 0 2 4 B 6 x _2 C 2, 2 46. Show that the triangle with vertices A 1 is a right triangle by using the converse of the B,, 2 2 1 1 C and Pythagorean Theorem. Find the area of the triangle. 2, 9 47. Show that the points 4, 6 1, 0 C B A, and,, 2 11, 3 6, 7 1 2 1 2 1 2 5, 3 D 1 2 are the vertices of a square. 48. Show that the points A collinear by showing that 1, 3 d B, 2 A, B 1 1 3, 11 d, and 2 B, C C 1 d 5, 15 are. 2 A, C 1 49. Find a point on the y-axis that is equidistant from the points 1 2 2 2 1 5, 5 1 and 1, 1. 2 1 2 50. Find the lengths of the medians of the triangle with vertices, B 1, 0 A C 1 a vertex to the midpoint of the opposite side.), and 3, 6 8, 2 1 2 2 1 2. (A median is a line segment from 51. Find the point that is one-fourth of the distance from the point P 1, 3 1 2 Q to the point 2, 1 1 7, 5 2 and 52. Plot the points 12, 1 P Q 1 5, 7 A Which (if either) of the points perpendicular bisector of the segment PQ? 2 2 1 1 2 on a coordinate plane. lies on the and 6, 7 B 1 2 along the segment PQ. 61. Reﬂecting in the Coordinate Plane Suppose that the y-axis acts as a mirror that reﬂects each point to the right of it into a point to the left of it. (a) The point is reﬂected to what point? 3, 7 1 2 a, b (b) The point 1 (c) What point is reﬂected to 1 (d) Triangle ABC in the ﬁgure is reﬂected to triangle ABC. is reﬂected to what point
? 4, 1? 2 2 Find the coordinates of the points A, B, and C. A(3, 3) A' y 0 B' B(6, 1) x C'C(1, _4) 1 2, 3 62. Completing a Line Segment Plot the points and on a coordinate plane. If M is the midpoint of the line A segment AB, ﬁnd the coordinates of B. Write a brief description of the steps you took to ﬁnd B and your reasons for taking them. 6, 8 M 2 1 2 63. Completing a Parallelogram Plot the points P 0, 3, 1 1 2 R 5, 3 2, 2, and Q on a coordinate plane. Where should the point S be located so that the ﬁgure PQRS is a parallelogram? Write a brief description of the steps you took and your reasons for taking them. 2 1 2 144 CHAPTER 2 \| Coordinates and Graphs 58. Halfway Point Two friends live in the city described in Exercise 57, one at the corner of 3rd St. and 7th Ave. and the other at the corner of 27th St. and 17th Ave. They frequently meet at a coffee shop halfway between their homes. (a) At what intersection is the coffee shop located? (b) How far must each of them walk to get to the coffee shop? 59. Pressure and Depth The graph shows the pressure experienced by an ocean diver at two different depths. Find and interpret the midpoint of the line segment shown in the graph. y 90 60 30 ) 33 99 x Depth (ft) ▼ DISCOVE RY • DISCUSSION • WRITI NG 60. Shifting the Coordinate Plane Suppose that each point in the coordinate plane is shifted 3 units to the right and 2 units upward. (a) The point is shifted to what new point? 5, 3 1 2 (b) The point a, b 1 (c) What point is shifted to 2 3, 4? is shifted to what new point? 1 (d) Triangle ABC in the ﬁgure has been shifted to triangle 2 ABC. Find the coordinates of the points A, B, and C. y B' 0 B(_3, 2) A' A(_5, _1) C(2, 1) C'x DISCOVERY PR OJECT VISUALIZING DATA When scientists analyze data
, they look for a trend or pattern from which they can draw a conclusion about the process they are studying. It is often hard to look at lists of numbers and see any kind of pattern. One of the best ways to reveal a hidden pattern in data is to draw a graph. For instance, a biologist measures the levels of three different enzymes (call them A, B, and C) in 20 blood samples taken from expectant mothers, yielding the data shown in the table (enzyme levels in milligrams per deciliter). Sample 1 2 3 4 5 6 7 8 9 10 A 1.3 2.6 0.9 3.5 2.4 1.7 4.0 3.2 1.3 1.4 B 1.7 6.8 0.6 2.4 3.8 3.3 6.7 4.3 8.4 5.8 C 49 22 53 15 25 30 12 17 45 47 Sample 11 12 13 14 15 16 17 18 19 20 A 2.2 1.5 3.1 4.1 1.8 2.9 2.1 2.7 1.4 0.8 B 0.6 4.8 1.9 3.1 7.5 5.8 5.1 2.5 2.0 2.3 C 25 32 20 10 31 18 30 20 39 56 The biologist wishes to determine whether there is a relationship between the serum levels of these enzymes, so she makes some scatter plots of the data. The scatter plot in Figure 1 shows the levels of enzymes A and B. Each point represents the results for one sample—for instance, sample 1 had 1.3 mg/dL of enzyme A and 1.7 mg/dL of enzyme B, so we plot the point (1.3, 1.7) to represent this pair of data. Similarly, Figure 2 shows the levels of enzymes A and C. From Figure 1 the biologist sees that there is no obvious relationship between enzymes A and B, but from Figure 2 it appears that when the level of enzyme A goes up, the level of enzyme C goes down. B 7 5 3 1 0 C 50 30 10 FIGURE 1 Enzymes A and B are not related. FIGURE 2 Enzymes A and C are strongly related. (CONTINUES) 145 VISUALIZING DATA (CONTINUED) 1. Make a scatter plot of the levels of enzymes B and C in the blood samples. Do you detect any relationship from your
graph? 2. For each scatter plot, determine whether there is a relationship between the two variables in the graphs. If there is, describe the relationship; that is, explain what happens to y as x increases. (a) y (b) y (c) y (d. In each scatter plot below, the value of y increases as x increases. Explain how the relationship between x and y differs in the two cases. y 0 y x 0 x 4. For the data given in the following table, make three scatter plots: one for enzymes A and B, one for B and C, and one for A and C. Determine whether any of these pairs of variables are related. If so, describe the relationship. Sample 1 2 3 4 5 6 7 8 9 10 11 12 A 58 39 15 30 46 59 22 7 41 62 10 6 B 4.1 5.2 7.6 6.0 4.3 3.9 6.3 8.1 4.7 3.7 7.9 8.3 C 51.7 15.4 2.0 7.3 34.2 72.4 4.1 0.5 22.6 96.3 1.3 0.2 146 SE CTI ON 2. 2 \| Graphs of Equations in Two Variables 147 2.2 Graphs of Equations in Two Variables LEARNING OBJECTIVES After completing this section, you will be able to: ■ Graph equations by plotting points ■ Find intercepts of the graph of an equation ■ Identify the equation of a circle ■ Graph circles in a coordinate plane ■ Determine symmetry properties of an equation Fundamental Principle of Analytic Geometry x, y lies on the graph of an A point equation if and only if its coordinates satisfy the equation. 2 1 y=2x-3 4 x y 4 0 FIGURE 1 An equation in two variables, such as y x 2 1, expresses a relationship between two satisﬁes the equation if it makes the equation true when the values quantities. A point for x and y are substituted into the equation. For example, the point satisﬁes the equa1 does not, because 3 12 1. tion y x 2 1 because 10 32 1, but the point 3, 10 1, 3 x, y 1 2 2 1 2 THE GRAPH OF AN EQUATION The graph of an equation in x and y is the set of all points nate plane that satisfy the equation. 1 x
, y 2 in the coordi- ■ Graphing Equations by Plotting Points The graph of an equation is a curve, so to graph an equation, we plot as many points as we can, then connect them by a smooth curve. E X AM P L E 1 \| Sketching a Graph by Plotting Points Sketch the graph of the equation 2x y 3. ▼ SO LUTI O N We ﬁrst solve the given equation for y to get y 2x 3 This helps us to calculate the y-coordinates in the following table 2x 3 5 3 1 1 3 5 x, y 2 1 1, 5 0, 3 1, 1 2, 1 3, 3 4 Of course, there are inﬁnitely many points on the graph, and it is impossible to plot all of them. But the more points we plot, the better we can imagine what the graph represented by the equation looks like. We plot the points that we found in Figure 1; they appear to lie on a line. So we complete the graph by joining the points by a line. (In Section 2.4 we verify that the graph of this equation is indeed a line.) ✎ Practice what you’ve learned: Do Exercise 15. ▲ E X AM P L E 2 \| Sketching a Graph by Plotting Points Sketch the graph of the equation y x 2 2. 148 CHAPTER 2 \| Coordinates and Graphs A detailed discussion of parabolas and their geometric properties is presented in Chapter 8 FIGURE 2 _ 4 _2 FIGURE 3 y=≈-2 4 x y=\| x \| 2 4 x ▼ SO LUTI O N We ﬁnd some of the points that satisfy the equation in the table below. In Figure 2 we plot these points and then connect them by a smooth curve. A curve with this shape is called a parabola, y 2 1 3, 7 2 1 2, 2 2 1 1, 1 0, 2 1, 1 2, 2 3 ✎ Practice what you’ve learned: Do Exercise 19. 2 ▲ E X AM P L E 3 \| Graphing an Absolute Value Equation Sketch the graph of the equation y x. 0 0 ▼ SO LUTI O N We make a table of values, y 2 1 3, 3 2, 2 1, 1 0, 0 1, 1 2, 2 3 In Figure 3 we plot these points and
use them to sketch the graph of the equation. ✎ Practice what you’ve learned: Do Exercise 31. ▲ ■ Intercepts The x-coordinates of the points where a graph intersects the x-axis are called the x-intercepts of the graph and are obtained by setting y 0 in the equation of the graph. The y-coordinates of the points where a graph intersects the y-axis are called the y-intercepts of the graph and are obtained by setting x 0 in the equation of the graph. DEFINITION OF INTERCEPTS Intercepts How to ﬁnd them Where they are on the graph x-intercepts: The x-coordinates of points where the graph of an equation intersects the x-axis Set y 0 and solve for x y-intercepts: The y-coordinates of points where the graph of an equation intersects the y-axis Set x 0 and solve for y y 0 y 0 x x SE CTI ON 2. 2 \| Graphs of Equations in Two Variables 149 E X AM P L E 4 \| Finding Intercepts Find the x- and y-intercepts of the graph of the equation y x 2 2. ▼ SO LUTI O N To ﬁnd the x-intercepts, we set y 0 and solve for x. Thus, y 2 0 _2 y=≈-2 x-intercepts 2 x _2 y -intercept Set y = 0 Add 2 to each side 0 x 2 2 x 2 2 x 12 12 To ﬁnd the y-intercepts, we set x 0 and solve for y. Thus, y 0 2 2 y 2 Set x = 0 and 12. Take the square root The x-intercepts are The y-intercept is 2. FIGURE 4 The graph of this equation was sketched in Example 2. It is repeated in Figure 4 with the x- and y-intercepts labeled. ✎ Practice what you’ve learned: Do Exercise 43. ▲ ■ Circles So far, we have discussed how to ﬁnd the graph of an equation in x and y. The converse problem is to ﬁnd an equation of a graph, that is, an equation that represents a given curve in the xy-plane. Such an equation is satisﬁed by the coordinates of the points on the curve
and by no other point. This is the other half of the fundamental principle of analytic geometry as formulated by Descartes and Fermat. The idea is that if a geometric curve can be represented by an algebraic equation, then the rules of algebra can be used to analyze the curve. As an example of this type of problem, let’s ﬁnd the equation of a circle with whose disis r (see Figure 5). Thus, P is on the circle if and only if. By deﬁnition, the circle is the set of all points h, k x, y P 2 1 h, k radius r and center 2 C tance from the center r d P, C 1 1 1 2 2. From the Distance Formula we have Square each side This is the desired equation. FIGURE 5 EQUATION OF A CIRCLE y 0 P(x, y) r C(h, k) x An equation of the circle with center h, k 1 2 and radius r is This is called the standard form for the equation of the circle. If the center of, then the equation is the circle is the origin x 2 y 2 r 2 0, 0 1 2 150 CHAPTER 2 \| Coordinates and Graphs E X AM P L E 5 \| Graphing a Circle Graph each equation. (a) x 2 y 2 25 (b) x 2 2 y 1 2 25 1 2 1 2 ▼ SO LUTI O N (a) Rewriting the equation as x 2 y 2 5 2, we see that this is an equation of the circle of radius 5 centered at the origin. Its graph is shown in Figure 6. (b) Rewriting the equation as x 2 2 y 1 2 52, we see that this is an equation of the circle of radius 5 centered at. Its graph is shown in Figure 7. 1 2 1 2 ≈+¥=25 y 0 (2, _1) x (x-2)™+(y+1)™=25 FIGURE 6 FIGURE 7 ✎ Practice what you’ve learned: Do Exercises 49 and 51. ▲ E X AM P L E 6 \| Finding an Equation of a Circle (a) Find an equation of the circle with radius 3 and center (b) Find an equation of the circle that has the points 1, 8 P 2, 5. 2 Q and 1 2 5, 6 2 as the end- 1

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