CHAPTER 3: Quadratic Functions and Equations; Inequalities

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1 171S MAT 171 Precalculus Algebra Dr. Claude Moore Cape Fear Community College CHAPTER 3: Quadratic Functions and Equations; Inequalities 3.1 The Complex Numbers 3.2 Quadratic Equations, Functions, Zeros, and Models 3.3 Analyzing Graphs of Quadratic Functions 3.4 Solving Rational Equations and Radical Equations This is a good 6 minute video to solve two problems: 3y and 3x < 5. Some Media for this Section 1. Absolute Value Equation short video (by Dr. Moore) demonstrating the solution of absolute value equations Absolute Value Inequality 1 short video (by Dr. Moore) demonstrating the solution of simple inequalities Absolute Value Inequality 2 short video (by Dr. Moore) demonstrating the solution more complex inequalities. NOTE: These videos are in the Technology on the Important Links webpage. 3.5 Solving Equations and Inequalities with Absolute Value Solve equations with absolute value. Solve inequalities with absolute value. Equations with Absolute Value X = a is equivalent to Solve: The solutions are 5 and 5. To check, note that 5 and 5 are both 5 units from 0 on the number line. X = a or X = a. 1

2 171S Solve: First, add one to both sides to get the expression in the form X = a. (continued) The possible solutions are 2 and 8. Check x = 2: Check x = 8: TRUE TRUE Let s check the possible solutions 2 and 8. The solutions are 2 and 8. More About Absolute Value Equations When a = 0, X = a is equivalent to X = 0. Note that for a < 0, X = a has no solution, because the absolute value of an expression is never negative. The solution is the empty set, denoted Inequalities with Absolute Value Inequalities sometimes contain absolute value notation. The following properties are used to solve them. X < a is equivalent to a < X < a. X > a is equivalent to X < a or X > a. Similar statements hold for X < a and X > a. 2

3 171S Inequalities with Absolute Value For example, x < 3 is equivalent to 3 < x < 3 Solve: Solve and graph the solution set: y 1 is equivalent to y 1 or y 1 2x is equivalent to 4 < 2x + 3 < 4 287/2. x = 4.5 Solve: Solve and graph the solution set: Remember and use the following: x = 4.5 means x = 4.5 or x = 4.5 X = a is equivalent to X = a or X = a. 3

4 171S 287/14. x 7 = 5 287/7. x = /20. (1/3)x 4 = /28. 5x = 5 (1/3)x 4 = 13 or (1/3)x 4 = 13 (1/3)x = 9 or (1/3)x = 17 x = 3( 9) or x = 3(17) x = 27 or x = 51 The solution is X = 27 or X = 51. 5x = 5 yields 5x + 4 = 3 5x + 4 = 3 or 5x + 4 = 3 5x = 7 or 5x = 1 x = 7/5 = 1.4 or x = 1/5 = 0.2 The solution is X = 1.4 or X =

5 171S 288/42. 5x < 4 288/ x + 3 = 2 Y1 = Y2 at x = 1.5 or x = / x > /60. (2x 1) / 3 > 5/6 5 2x > x < 10 or 5 2x > 10 2x < 15 or 2x > 5 Remember to reverse the inequality sign when multiplying or dividing by a negative. x > 15/2 = 7.5 or x < 5/2 = 2.5 The solution is 2.5 < x < 7.5. Interval notation: ( 2.5, 7.5) Since x = 2.5 gives y = 0 and x = 7.5 gives y = 0, the inequality is false for 2.5 and 7.5. Thus, we have open circles at these two values. So, the solution is (, 2.5) U (7.5, ). (2x 1) / 3 5/6 (2x 1) / 3 5/6 or (2x 1) / 3 5/6 Multiply by 6 to remove fractions. 2(2x 1) 5 or 2(2x 1) 5 Continue the solution to get the following: x 3/4 or x 7/4 The solution is x 0.75 or x Interval notation: (, 0.75] U [1.75, ) Since x = 0.75 gives y = 1 and x = 1.75 gives y = 1, the inequality is true for 0.75 and Thus, we have closed circles at these two values. So, the solution is (, 0.75] U [1.75, ). 5