CHAPTER 3: Quadratic Functions and Equations; Inequalities

Transcription

1 171S Completing the square to solve quadratic equation: 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 very good 6 minute video by a textbook author. He shows how to solve two problems: 3y and 3x < 5. Printed instructions for graphing one variable inequality with TI calculator. Maximize area of rectangle with river as one side. 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. 1

2 171S 3.5 Solving Equations and Inequalities with Absolute Value Solve equations with absolute value. Solve inequalities with absolute value. Equations with Absolute Value For a > 0 and an algebraic expression X: X = a is equivalent to Solve: Solution: 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. Solve: Solution: 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. 2

3 171S 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. For a > 0 and an algebraic expression X: 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. Inequalities with Absolute Value For example, x < 3 is equivalent to 3 < x < 3 Solve: Solve and graph the solution set: Solution: y 1 is equivalent to y 1 or y 1 2x is equivalent to 4 < 2x + 3 < 4 3

4 171S 287/2. x = 4.5 Solve: Solve and graph the solution set: Solution: Remember and use the following: For a > 0 and an algebraic expression X: x = 4.5 means x = 4.5 or x = 4.5 X = a is equivalent to X = a or X = a. Y1 = x Y2 = 4.5 Since Y1 and Y2 intersect, there is a common solution. Thus, x = 4.5 has the solution x = 4.5 or x = /14. x 7 = 5 287/7. x = 10.7 Y1 = x Y2 = 10.7 Since Y1 and Y2 do not intersect, there is no common solution. Thus, x = 10.7 has no solution. Y1 = x 7 Y2 = 5 Thus, x 7 = 5 has the solution x = 2 or x = 12. You may explore absolute value equation at sascurriculumpathways.com using the username able7oxygen and QL# 1424 or go directly to herseries/medalg/tool.jsf?resourceid=1424 4

5 171S 287/28. 5x = 5 287/20. (1/3)x 4 = 13 Thus, (1/3)x 4 = 13 has the solution x = 27 or x = 51. (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. Y1 = (1/3)x 4 Y2 = 13 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 = 0.2. Y1 = 5x Y2 = 5 Thus, 5x = 5 has the solution x = 1.4 or x = 0.2. Solve and write interval notation for the solution set. Then graph the solution set. 288/42. 5x < 4 288/ x + 3 = 2 Y1 = 5x Y2 = 4 Thus, 5x 4 has the solution 0.8 x 0.8. Y2 = 2 Y1 = 5 4x + 3 Y1 = Y2 at x = 1.5 or x = 0. Thus, 5 4x + 3 = 2 has the solution x = 1.5 or x = 0. Instructions for graphing one variable inequality with TI calculator. 5

6 171S Solve and write interval notation for the solution set. Then graph the solution set. 288/ x > x > 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, ). Instructions for graphing one variable inequality with TI calculator. 6

7 171S Solve and write interval notation for the solution set. Then graph the solution set. 288/60. (2x 1) / 3 > 5/6 (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, ). Instructions for graphing one variable inequality with TI calculator. 7