2. Copyright © 2004 Pearson Education, Inc. Chapter 1 Equations and Inequalities

3. Copyright © 2004 Pearson Education, Inc. 1.1 Linear Equations

4. Slide 1-4 Copyright © 2004 Pearson Education, Inc. Equations An equation is a statement that two expressions are equal.  x + 3 =11 9x = 3x + 6x x 2 + 4x – 8 = 0 To solve an equation means to find all numbers that make the equation true. The numbers are called solutions or roots of the equation. A number that is a solution of an equation is said to satisfy the equation, and the solutions of an equation make up its solution set. Equations with the same solution set are equivalent equations.

5. Slide 1-5 Copyright © 2004 Pearson Education, Inc. Addition and Multiplication Properties of Equality For real numbers a, b, and c: a = b and a + c = b + c are equivalent That is, the same number may be added to both sides of an equation without changing the solution set. For real numbers a, b, and c: If c  0, then a = b and ac = bc are equivalent. That is, both sides of an equation may be multiplied by the same nonzero number without changing the solution set.

6. Slide 1-6 Copyright © 2004 Pearson Education, Inc. Solving Linear Equations Linear Equations in One Variable  A linear equation in one variable is an equation that can be written in the form ax + b = 0 where a and b are real numbers with a  0  A linear equation is also called a first-degree equation since the greatest degree of the variable is one.

7. Slide 1-7 Copyright © 2004 Pearson Education, Inc. Solving a Linear Equation Example: Solve 6p + 2  10(p + 2) = 12p + 14. Solution: 6p + 2  10(p + 2) = 12p + 14 6p + 2  10p –20 = 12p + 14  4p  18 = 12p + 14  4p + 4p  18 = 12p + 14 + 4p – 18 = 16p + 14  32 = 16p

8. Slide 1-8 Copyright © 2004 Pearson Education, Inc. Solving a Linear Equation continued Check: 6p + 2  10(p + 2) = 12p + 14 6(  2) + 2  10(  2 + 2) = 12(  2) + 14  12 + 2 =  24 + 14  10 =  10 Since replacing p with  2 results in a true statement,  2 is the solution of the given equation. The solution set is{  2}.

9. Slide 1-9 Copyright © 2004 Pearson Education, Inc. Clearing Fractions If an equation has fractions as coefficients, begin by multiplying both sides by the least common denominator to clear the fractions. Example: The solution set is {  11} Check:

10. Slide 1-10 Copyright © 2004 Pearson Education, Inc. Identities, Conditional Equations and Contradictions 1. If solving a linear equation leads to a true statement such as 0 = 0, the equation is an identity. Its solution set is {all real numbers}. 2. If solving a linear equation leads to a single solution such as x = 3, the equation is conditional. The solution set consists of a single element. 3. If solving a linear equation leads to a false statement such as 10 =  8, the equation is a contradiction. Its solution set is .

11. Slide 1-11 Copyright © 2004 Pearson Education, Inc. Examples of Identifying Types of Equations a)  4(x  2) + 6x = 2(x + 4)  4x + 8 + 6x = 2x + 8 2x + 8 = 2x + 8 0 = 0 identity b) 6x + 5 = 23 6x = 18 x = 3 conditional equation c) 5(3x + 2) = 15x  8 15x + 10 = 15x  8 10 =  8 contradiction

12. Slide 1-12 Copyright © 2004 Pearson Education, Inc. Solving for a Specified Variable A formula is an example of a literal equation. The methods used to solve a linear equation can be used to solve some literal equations for a specified variable. Example: Solve p = 6x +2z for z. Solution:

13. Slide 1-13 Copyright © 2004 Pearson Education, Inc. Simple Interest Formula The formula for simple interest is I = Prt.  Example: Stanley has $8000 to invest in a mutual fund. The funds with higher rates of returns also have greater risk. Stanley wants a return of at least $300 every six months. What is the lowest rate of return that will meet his goal?  Solution:  Stanley must chose a rate with a return of at least 7.5%

14. Copyright © 2004 Pearson Education, Inc. Applications and Modeling with Linear Equations 1.2

15. Slide 1-15 Copyright © 2004 Pearson Education, Inc. Solving an Applied Problem Step 1Read the problem carefully until you understand what is given and what is to be found. Step 2Assign a variable to represent the unknown value, using diagrams or tables as needed. Write down what the variable represents. If necessary, express any other unknown values in terms of the variable. Step 3Write an equation using the variable expression(2).

16. Slide 1-16 Copyright © 2004 Pearson Education, Inc. Solving an Applied Problem continued Step 4Solve the equation. Step 5State the answer to the problem. Does it seem reasonable? Step 6Check the answer in the words of the original problem.

17. Slide 1-17 Copyright © 2004 Pearson Education, Inc. Example--Geometry The perimeter of a rectangle is 108 cm. The length is 8 cm longer than the width. Find the dimensions of the rectangle. Step 1 Read the problem. We must find the dimensions of the rectangle. Step 2 Assign a variable. x = width x + 8 = length

18. Slide 1-18 Copyright © 2004 Pearson Education, Inc. Example--Geometry continued Step 3 Write an equation. Perimeter = x + x + x + 8 + x + 8 Step 4 Solve the equation. 108 = x + x + x + 8 + x + 8 108 = 4x + 16 92 = 4x 23 = x

19. Slide 1-19 Copyright © 2004 Pearson Education, Inc. Example--Geometry continued Step 5 State the answer. The width of the rectangle is 23 cm, the length of the rectangle is 23 + 8 = 31cm. Step 6 Check. The perimeter is 108 cm. 31 + 31 + 23 + 23 = 108

20. Slide 1-20 Copyright © 2004 Pearson Education, Inc. Motion Problems In a motion problem, the components distance, rate, and time, are denoted by the letters d, r, and t, respectively. (The rate is also called the speed or velocity. Here, rate is understood to be constant.) These variables are related by the equation and its related forms

21. Slide 1-21 Copyright © 2004 Pearson Education, Inc. Example Nicholas and Catherine are traveling to a math conference. The trip takes 1 ½ hours for Catherine and 2 ½ hours for Nicholas, since he lives 75 miles farther away. Nicholas travels 6 mph faster than Catherine. Find their average rates. Step 1 Read the problem. We are looking for Catherine’s and Nicholas’ average rate.

22. Slide 1-22 Copyright © 2004 Pearson Education, Inc. Example continued Step 2 Assign a variable. Let x = Catherine’s rate. Then x + 6 = Nicholas’ rate. Summarize the information in a table. Step 3 Write an equation. 2.5(x + 6) = 1.5x + 75 2.5(x + 6)2.5x + 6Nicholas 1.5x1.5xCatherine dtr

23. Slide 1-23 Copyright © 2004 Pearson Education, Inc. Example continued Step 4 Solve. Step 5 State the answer. Catherine’s rate is 60 mph and Nicholas’ rate is 60 + 6 = 66 mph. Step 6 Check Distance traveled by Catherine 60(1.5) = 90 Distance traveled by Nicholas 66(2.5) = 165 The difference in the distances 165  90 = 75

24. Slide 1-24 Copyright © 2004 Pearson Education, Inc. Work Rate Problems If a job can be done in t units of time then the rate of work is 1/t of the job per time unit. Therefore, rate  time = portion of job completed. If the letters r, t, and A represent the rate at which work is done, the time, and the amount of work accomplished, respectively, then A = rt. Amounts of work are often measured in terms of the number of jobs accomplished. For instance, if one job is accomplished in t time units, then A = 1 and r = 1/t.

25. Slide 1-25 Copyright © 2004 Pearson Education, Inc. Mixture Problems In mixture problems, the rate (percent) of concentration is multiplied by the quantity to get the amount of pure substance present. Also, the concentration in the final mixture must be between the concentrations of the two solutions making up the mixture. Example: How many gallons of a 10% solution acid solution must be mixed with 5 gallons of a 20% acid solution to obtain a 15% acid solution?

26. Slide 1-26 Copyright © 2004 Pearson Education, Inc. Mixture Problems continued Read the problem. We must find the required number of gallons of 10% solution. Assign a variable. Let x = number of gallons of 10% solution. Write an equation..10x +.20(5) =.15(x + 5).15(x + 5)x + 515%.20(5)520%.10xx10% Liters of pure acidLiters of SolutionStrength

27. Slide 1-27 Copyright © 2004 Pearson Education, Inc. Mixture Problems continued Solve. State the answer. 5 gallons of 10% solution should be mixed with 5 gallons of 20% solution, given 5 + 5 = 10 gallons of 15 percent solution. Check..10(5) +.20(5) =.15(5 + 5) 0.5 + 1.0 = 1.5

28. Slide 1-28 Copyright © 2004 Pearson Education, Inc. Investment Problem Antonia sold a house for $210,000. He invests some of the money in an account that pays 4% interest. He invests the remaining amount in an investment that pays 7%. His total annual interest earned is $12,540. How much was invested in each account? Read. We must find the amount invested at each rate.

29. Slide 1-29 Copyright © 2004 Pearson Education, Inc. Assign a variable. Let x = the dollar amount invested at 4% Let 210,000  x = amount invested at 7% Write an equation..04x +.07(210,000  x) = 12,540 Solve. Investment Problem continued

30. Slide 1-30 Copyright © 2004 Pearson Education, Inc. Investment Problem continued State the answer. Antonio invested $72,000 at 4% and $210,000 - $72,000 = $138,000 at 7%. Check. 72,000(.04) + 138,000(.07) = 12,540 2880 + 9660 = 12,540

31. Copyright © 2004 Pearson Education, Inc. Complex Numbers 1.3

32. Slide 1-32 Copyright © 2004 Pearson Education, Inc. Basic Concepts i 2 =  1 i is the imaginary unit a + bi are called complex numbers  a is the real part  b is the imaginary part

33. Slide 1-33 Copyright © 2004 Pearson Education, Inc. Examples a)b) c) d) e)

34. Slide 1-34 Copyright © 2004 Pearson Education, Inc. Addition and Subtraction of Complex Numbers For complex numbers a + bi and c + di, Examples (10  4i)  (5  2i) = (10  5) + [  4  (  2)]i = 5  2i (4  6i) + (  3 + 7i) = [4 + (  3)] + [  6 + 7]i = 1 + i

35. Slide 1-35 Copyright © 2004 Pearson Education, Inc. Multiplication of Complex Numbers For complex numbers a + bi and c + di, The product of two complex numbers is found by multiplying as if the numbers were binomials and using the fact that i 2 =  1.

36. Slide 1-36 Copyright © 2004 Pearson Education, Inc. Examples (2  4i)(3 + 5i) (7 + 3i) 2

37. Slide 1-37 Copyright © 2004 Pearson Education, Inc. Powers of i i 1 = ii 5 = ii 9 = i i 2 =  1i 6 =  1i 10 =  1 i 3 =  ii 7 =  i i 11 =  i i 4 = 1 i 8 = 1 i 12 = 1 and so on.

38. Slide 1-38 Copyright © 2004 Pearson Education, Inc. Simplifying Examples i 17 i 4 = 1 i 17 = (i 4 ) 4 i = 1 i = i i  4

39. Slide 1-39 Copyright © 2004 Pearson Education, Inc. Property of Complex Conjugates For real numbers a and b, (a + bi)(a  bi) = a 2 + b 2. The product of a complex number and its conjugate is always a real number. Example

40. Copyright © 2004 Pearson Education, Inc. Quadratic Equations 1.4

41. Slide 1-41 Copyright © 2004 Pearson Education, Inc. Quadratic Equation in One Variable An equation that can be written in the form ax 2 + bx + c = 0, where a, b, and c are real numbers with a  0, is a quadratic equation. Examples: x 2 = 365x 2 + 5x  7 = 0

42. Slide 1-42 Copyright © 2004 Pearson Education, Inc. Zero-Factor Property If a and b are complex numbers with ab = 0, then a = 0 or b = 0 or both. Example:

43. Slide 1-43 Copyright © 2004 Pearson Education, Inc. Square Root Property If x 2 = k, then Examples

44. Slide 1-44 Copyright © 2004 Pearson Education, Inc. Completing the Square To solve ax 2 + bx + c = 0, a  0, by completing the square: Step 1If a  1, multiply both sides of the equation by 1/a. Step 2Rewrite the equation so that the constant term is alone on one side of the equal sign. Step 3Square half the coefficient of x, and add this square to both sides of the equation.

45. Slide 1-45 Copyright © 2004 Pearson Education, Inc. Completing the Square continued Step 4Factor the resulting trinomial as a perfect square and combine terms on the other side. Step 5Use the square root property to complete the solution.

46. Slide 1-46 Copyright © 2004 Pearson Education, Inc. x 2  6x  12 = 0 Step 1Not necessary since a = 1. Step 2 x 2  6x = 12 Step 3x 2  6x + 9 = 12 + 9 Step 4 (x  3) 2 = 21 Step 5 Example a = 1

47. Slide 1-47 Copyright © 2004 Pearson Education, Inc. Example a  1 Complete the square.

48. Slide 1-48 Copyright © 2004 Pearson Education, Inc. Example a  1 continued The solution set is

49. Slide 1-49 Copyright © 2004 Pearson Education, Inc. Quadratic Formula The solutions of the quadratic equation ax 2 + bx + c = 0, where a  0 are

50. Slide 1-50 Copyright © 2004 Pearson Education, Inc. Example Solve: 3x 2 = x  5.

51. Slide 1-51 Copyright © 2004 Pearson Education, Inc. Solving for a Variable that is Squared Solve the given formula for r, V =  r 2 h Disregard the negative solution since r can not be negative.

52. Slide 1-52 Copyright © 2004 Pearson Education, Inc. Discriminant The quantity under the radical in the quadratic formula b 2  4ac, is called the discriminant. Nonreal complexTwoNegative RationalOne (a double solution)Zero IrrationalTwoPositive, but not a perfect square RationalTwoPositive, perfect square Kind of SolutionsNumber of SolutionsDiscriminant

53. Slide 1-53 Copyright © 2004 Pearson Education, Inc. Examples Determine the number of distinct solutions and tell whether they are rational, irrational, or nonreal complex numbers. 3x 2  x + 2 = 0 b 2  4ac = (  1) 2  (4)(2)(3)=  23 Two nonreal complex solutions x 2  12x =  36 x 2  12x + 36 = 0 b 2  4ac = (  12) 2  (4)(36)= 0 One rational double solution 3x 2 + x  5 = 0 b 2  4ac = 1 2  (4)(3)(  5)= 61 Two irrational solutions

54. Copyright © 2004 Pearson Education, Inc. Applications and Modeling with Quadratic Equations 1.5

55. Slide 1-55 Copyright © 2004 Pearson Education, Inc. Problem Solving When solving problems that lead to quadratic equations, we may get a solution that does not satisfy the physical constraints of the problem. For example, if x represents a width and two solutions of the quadratic equation are  9 and 1, the value  9 must be rejected since a width must be a positive number.

56. Slide 1-56 Copyright © 2004 Pearson Education, Inc. Example A piece of machinery is capable of producing rectangular sheets of metal such that the length is three times the width. Furthermore, equal- sized squares measuring 5 in. on a side can be cut from the corners so that the resulting piece of metal can be shaped into an open box by folding up the flaps. If specifications call for the volume of the box to be 1435 in. 3, what should the dimensions of the original piece of metal be?

57. Slide 1-57 Copyright © 2004 Pearson Education, Inc. Solution Step 1 Read the problem. We must find the dimensions of the original piece of metal. Step 2Assign a variable. The length is 3 times the width, let x = width and 3x = length The box is formed by cutting 5 + 5 = 10 inches from both the length and width. The length and width of the box are shown on the figures.

58. Slide 1-58 Copyright © 2004 Pearson Education, Inc. Solution continued Step 3Write an equation. V = lwh Volume = length  width  height 1435 = (3x  10)(x  10)(5) Step 4Solve the equation.

59. Slide 1-59 Copyright © 2004 Pearson Education, Inc. Solution continued Step 5State the answers. Only 17 satisfies the restriction. Thus, the dimensions of the original piece of metal should be 17 in. by 3(17) = 51 in. Step 6Check. The length of the bottom of the box is 51  2(5) = 41 in. The width is 17  2(5) = 7 in. The height is 5 in. So, the volume of the box is 41(7)(5) = 1435 in. 3

60. Slide 1-60 Copyright © 2004 Pearson Education, Inc. The Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. a 2 + b 2 = c 2 Leg b Leg a Hypotenuse c

61. Slide 1-61 Copyright © 2004 Pearson Education, Inc. Example A kite is flying on 70 feet of string. Its vertical distance from the ground is 15 feet more than its horizontal distance from the person flying it. Assuming that the string is being held at ground level, find its horizontal distance from the person and its vertical distance from the ground.

62. Slide 1-62 Copyright © 2004 Pearson Education, Inc. Example continued Step 1Read the problem. We are finding the horizontal and vertical distance. Step 2Assign a variable. Let x = the horizontal distance Let x + 15 = the vertical distance Step 3Write an equation. x 2 + (x + 15) 2 = 70 2 70 ft of string x + 15 x

63. Slide 1-63 Copyright © 2004 Pearson Education, Inc. Example continued Step 4Solve:

64. Slide 1-64 Copyright © 2004 Pearson Education, Inc. Example continued Step 5State the answer. Since x represents a length, the horizontal distance could only be 41.43 feet. The vertical distance would be 41.43 + 15 = 56.43 feet. Step 6Check. The lengths 41.43, 56.43 and 70 satisfy the Pythagorean theorem.

65. Slide 1-65 Copyright © 2004 Pearson Education, Inc. Modeling with Quadratic Equations The bar graph shows the sales of SUV’s in the United States, in millions. The quadratic equation S =.016x 2 +.124x +.787 models sales of SUV’s from 1990 to 2001, where S represents sales in millions and x = 0 represents 1990, x = 1 represents 1991, and so on. a) Use this model to predict sales in 2000 and 2001. Compare the results to the actual figures of 3.4 million and 3.8 million from the graph.

66. Slide 1-66 Copyright © 2004 Pearson Education, Inc. Modeling with Quadratic Equations continued b) According to the model, in what year do sales reach 3 million? (Round down to the nearest year.) Is the result accurate?

67. Slide 1-67 Copyright © 2004 Pearson Education, Inc. Solution a) The predictions are greater than the actual figures of 3.4 and 3.8 million. For 2000, x = 10 For 2001, x = 11

68. Slide 1-68 Copyright © 2004 Pearson Education, Inc. Solution continued b) Reject the negative solution and round 8.5 down to 8 The year 1998 corresponds to x = 8. The model is a bit misleading, since SUV sales did not reach 3 million until 2000.

69. Copyright © 2004 Pearson Education, Inc. Other Types of Equations 1.6

70. Slide 1-70 Copyright © 2004 Pearson Education, Inc. Rational Equations A rational equation is an equation that has a rational expression for one or more terms. Since a rational expression is not defined when its denominator is 0, values of the variable for which any denominator equals 0 cannot be solutions of the equation. To solve a rational equation, begin by multiplying both sides by the least common denominator of the terms of the equation.

71. Slide 1-71 Copyright © 2004 Pearson Education, Inc. Example Solve: Solution: The least common denominator is 4(x + 3),which is equal to 0 if x =  3. Therefore,  3 cannot possibly be a solution of this equation. Check the result we find that the solution set is {21}.

72. Slide 1-72 Copyright © 2004 Pearson Education, Inc. Another Example Solve: Solution:

73. Slide 1-73 Copyright © 2004 Pearson Education, Inc. Another Example continued Neither proposed solution is valid, so the solution set is 0.

74. Slide 1-74 Copyright © 2004 Pearson Education, Inc. Equations with Radicals To solve an equation in which the variable appears in the radicand, we use the following power property to eliminate the radical. Power Property  If P and Q are algebraic expressions, then every solution of the equation P = Q is also a solution of the equation P n = Q n, for any positive integer n. When using the power property to solve equations, the new equation may have more solutions than the original equation. When an equation contains radicals (or rational exponents), it is essential to check all proposed solutions in the original equations.

75. Slide 1-75 Copyright © 2004 Pearson Education, Inc. Radicals Solving an Equation Involving Radicals Isolate the radical on one side of the equation. Raise each side of the equation to a power that is the same as the index of the radical so that the radical is eliminated. If the equation still contains a radical, repeat Steps 1 and 2. Solve the resulting equation. Check each proposed solution in the original equation.

76. Slide 1-76 Copyright © 2004 Pearson Education, Inc. Example Solve Isolate the radical. Square both sides of the equation. Simplify. The result is a quadratic. Put the quadratic in standard form. Factor. Solve for x.

77. Slide 1-77 Copyright © 2004 Pearson Education, Inc. Example continued Check: If x = 9, then ? As the check shows, only 9 is a solution, giving the solution set {9} Check If x = 2, then ?

78. Slide 1-78 Copyright © 2004 Pearson Education, Inc. Solving an Equation Containing Two Radicals Example: Solve Solution:

79. Slide 1-79 Copyright © 2004 Pearson Education, Inc. Solving an Equation Containing Two Radicals continued

80. Slide 1-80 Copyright © 2004 Pearson Education, Inc. Solving an Equation Containing Two Radicals continued Check: x = 7 ? Check: x =  1 ? The solution set is {  1}

81. Slide 1-81 Copyright © 2004 Pearson Education, Inc. Quadratics Equations Quadratic in Form An equation is said to be quadratic in form if it can be written as au 2 + bu + c = 0, where a  0, and u is some algebraic expression.

82. Slide 1-82 Copyright © 2004 Pearson Education, Inc. Example x 4  2x 2 + 1 = 0 x 4 = (x 2 ) 2 (x 2 ) 2  2x 2 + 1 = 0 Let u = x 2 u 2  2u + 1 = 0 (u  1)(u  1) = 0 Solve the quadratic equation. u  1= 0 Zero-factor Property u = 1 To find x, replace u with x 2. x 2 = 1 x = x =  1 Checking in the original problem, the solution set is {  1}.

83. Slide 1-83 Copyright © 2004 Pearson Education, Inc. Another Example Solve: (x + 1) 2/3  (x + 1) 1/3  2 = 0. Solution: Since (x + 1) 2/3 = [(x + 1) 1/3 ] 2, let u = (x + 1) 1/3. u 2  u  2 = 0 Substitute. (u  2)(u + 1) = 0 Factor. u  2 = 0 or u + 1 = 0 u = 2 or u =  1 Now replace u with (x + 1) 1/3 (x + 1) 1/3 = 2 or (x + 1) 1/3 =  1 [(x + 1) 1/3 ] 3 = 2 3 or [(x + 1) 1/3 ] 3 = (  1) 3 Cube each side. x + 1 = 8 or x + 1 =  1 x = 7 or x =  2

84. Slide 1-84 Copyright © 2004 Pearson Education, Inc. Another Example continued Check: (x + 1) 2/3  (x + 1) 1/3  2 = 0 Both check, so the solution set is {  2, 7} If x =  2, then, (  2 + 1) 2/3  (  2 + 1) 1/3  2 = 0 ? (  1) 2/3  (  1) 1/3  2 = 0 ? 1 + 1  2 = 0 ? 0 = 0 True If x = 7, then (7 + 1) 2/3   (7 + 1) 1/3  2 = 0. 8 2/3  8 1/3  2 = 0 ? 4  2  2 = 0 ? 0 = 0 True

85. Copyright © 2004 Pearson Education, Inc. Inequalities 1.7

86. Slide 1-86 Copyright © 2004 Pearson Education, Inc. Inequality An inequality says that one expression is greater than, greater than or equal to, less than, or less than or equal to, another. Properties of Inequality For real numbers a, b, and c: 1. If a < b, then a + c < b + c 2.If a 0, then ac < bc, 3.If a bc.

87. Slide 1-87 Copyright © 2004 Pearson Education, Inc. Linear Inequalities A linear inequality in one variable is an inequality that can be written in the form ax + b > 0, where a and b are real numbers with a  0. (Any of the symbols , <, or  may also be used.) Always remember to reverse the direction of the inequality symbol when multiplying or dividing by a negative number.

88. Slide 1-88 Copyright © 2004 Pearson Education, Inc. Example Solve ) The original inequality is satisfied by any real number less than 4. The solution set can be written {x|x < 4}. 0–55

89. Slide 1-89 Copyright © 2004 Pearson Education, Inc. Open Interval Notation

90. Slide 1-90 Copyright © 2004 Pearson Education, Inc. Half-open interval

91. Slide 1-91 Copyright © 2004 Pearson Education, Inc. Closed Interval and All Real Numbers

92. Slide 1-92 Copyright © 2004 Pearson Education, Inc. Example Solve 9x  3x + 8  3(x + 4) ] In interval notation the solution set is ( , 4/3]. 0–55

93. Slide 1-93 Copyright © 2004 Pearson Education, Inc. Three-Part Inequalities Solve  5 < 7 + 2x < 21 ( ) –5–100 510 The solution set is the interval (  6, 7).

94. Slide 1-94 Copyright © 2004 Pearson Education, Inc. Quadratic Inequalities A quadratic inequality is an inequality that can be written in the form ax 2 + bc + c, , or .) Solving a Quadratic Inequality  Step 1Solve the corresponding quadratic equation.  Step 2Identify the intervals determined by the solutions of the equation.  Step 3Use a test value from each interval to determine which intervals form the solution set.

95. Slide 1-95 Copyright © 2004 Pearson Education, Inc. Example Solve x 2  x  20 < 0 Step 1 Find the value that satisfy x 2  x  20 = 0. (x + 4)(x  5) = 0 x + 4 = 0orx  5 = 0 x =  4or x = 5 Step 2 Determine the intervals 0 –5 5 Interval A ( ,  4) Interval B (  4, 5) Interval C (5,  )

96. Slide 1-96 Copyright © 2004 Pearson Education, Inc. Example continued Step 3 Choose a test value in each interval to see if it satisfies the original inequality. If the test value makes the interval true, then the entire interval belongs to the solution set. Since the values in interval B make the inequality true, the solution set is (  4, 5). (10) 2  (10)  20 < 0 70 < 0 False 10 (5,  ) 0  0  20 < 0 True 0 (  4, 5) (  5) 2  (  5)  20 < 0 10 < 0 False 55( ,  4) True/False x 2  x  20 < 0 Test ValueInterval

97. Slide 1-97 Copyright © 2004 Pearson Education, Inc. Rational Inequalities Step 1Rewrite the inequality, if necessary, so that 0 is on one side and there is a single fraction on the other side. Step 2Determine the values that will cause either the numerator or denominator of the rational expression to equal 0. These values determine the intervals on the number line to consider. Step 3Use a test value from each interval to determine which intervals form the solution set.

98. Slide 1-98 Copyright © 2004 Pearson Education, Inc. Example Solve Step 2  The quotient possibly changes sign only where x- values make the numerator or denominator 0. This occurs at  x  2 = 0 or x + 8 = 0  x = 2 or x =  8 x =  2 These form the intervals ( ,  8), (  8,  2) and (  2,  ).

99. Slide 1-99 Copyright © 2004 Pearson Education, Inc. Example continued Step 3 Choose test values. The solution set is (  8,  2]. False 0 (  2,  ) True 44(  8,  2) False  10( ,  8) True/FalseTest ValueInterval

100. Copyright © 2004 Pearson Education, Inc. Absolute Value Equations and Inequalities 1.8

101. Slide 1-101 Copyright © 2004 Pearson Education, Inc. Properties of Absolute Value |a| gives the distance from a to 0 on a number line 1. For b > 0, |a| = b if and only if a = b or a =  b. 2. |a| = |b| if and only if a = b or a =  b. For any positive number b: 3. |a| < b if and only if  b < a < b. 4. |a| > b if and only if a b.

102. Slide 1-102 Copyright © 2004 Pearson Education, Inc. Solving Absolute Value Equations Solve |4  7x| = 10 Check the solutions by substituting them in the original absolute value equation. The solution set is {  6/7, 2}.

103. Slide 1-103 Copyright © 2004 Pearson Education, Inc. Example Solve |3x  5| = |x  6| or Check these solutions. The solutions are

104. Slide 1-104 Copyright © 2004 Pearson Education, Inc. Absolute Value Inequalities Solve: |3x + 4| > 8 or The solution set of the final compound inequality is ( ,  4)  (4/3,  )

105. Slide 1-105 Copyright © 2004 Pearson Education, Inc. Example Solve |3x + 4| < 8 The final inequality gives the solution set (  4, 4/3).

106. Slide 1-106 Copyright © 2004 Pearson Education, Inc. Example Solve |2x + 1|  3 < 10 The final inequality gives the solution set (  7, 6).

107. Slide 1-107 Copyright © 2004 Pearson Education, Inc. Special Cases Solve |2  5x|   4 Solution: Since the absolute value of a number is always nonnegative, the inequality is always true. The solution set includes all real numbers, written ( ,  ). Solve |4x  7| <  3 Solution: There is no number whose absolute value is less than  3 (or less than any negative number). The solution set is .

108. Slide 1-108 Copyright © 2004 Pearson Education, Inc. Special Cases Solve |5x + 15| = 0 Solution: The absolute value of a number will be 0 only if that number is 0. Therefore, |5x + 15| = 0 is equivalent to 5x + 15 = 0, which has a solution set {  3}.

109. Slide 1-109 Copyright © 2004 Pearson Education, Inc. Using Absolute Value to Model Tolerance Suppose y = 2x + 1 and we want y to be within.01 unit of 4. For what values of x will this be true? Solution: