Chapter 2: Equations and Inequalities 2.5A: Absolute-Value Inequalities

Chapter 2: Equations and Inequalities2.5A: Absolute-Value Inequalities Essential Question: What is the procedure used to solve an absolute value equation of inequality?

2.5A: Absolute-Value Inequalities • Just like equations, absolute value inequalities can be solved either by graphing or algebraically • Graphing Methods • Intersection Method • Won’t be covered – less accurate than below • x-Intercept Method • Example 2: Solving using the x-intercept method • Solve • Get inequality to compare with 0: • Take great care with parenthesis in the calculator (can split the absolute values) • Solution are the intervals (½, 2) and (2, 5)

2.5A: Absolute-Value Inequalities • Algebraic Methods • Just like regular inequalities, get the absolute value stuff by itself, and create two inequalities. • The first one works like normal • The second inequality flips the inequality and the sign of everything the absolute value was equal to • Also like inequalities, all critical points (real and extraneous solutions) must be found, and intervals are tested between those critical points.

2.5A: Absolute-Value Inequalities • Example 3: Simple Absolute Value Inequality • Solve |3x – 7| < 11 • Two equations • 3x – 7 < 11 • Add 7 to both sides • 3x < 18 • Divide both sides by 3 • x < 6 • 3x – 7 > -11 • Add 7 to both sides • 3x > -4 • Divide both sides by 3 • x >-4/3 • Two critical points found: -4/3 and 6

2.5A: Absolute-Value Inequalities • Ex 3 (continued) – test the critical points • Solve |3x – 7| < 11 • 2 critical points → 3 intervals to be tested • (-∞, -4/3] • Test x = -2, result is 13 < 11 FAIL • [-4/3, 6] • Test x = 0, result is 7 < 11 SUCCESS • [6, ∞) • Test x = 7, result is 14 < 11 FAIL • Solution is the interval [-4/3, 6] • Note: Graphing would also reveal the interval solution

2.5A: Absolute-Value Inequalities • Example 5: Quadratic absolute value inequality • Solve |x2 – x – 4| > 2 • Two equations • x2 – x – 4 > 2 • Subtract 2 from both sides • x2 – x – 6 > 0 • (x – 3)(x + 2) > 0 • Critical Points: x = 3 or x = -2 • x2 – x – 4 < -2 • Add 2 to both sides • x2 – x – 2 < 0 • (x – 2)(x + 1) < 0 • Critical Points: x = 2 or x = -1 • Four critical points found: -2, -1, 2 and 3

2.5A: Absolute-Value Inequalities • Ex 5 (continued) – test the critical points • Solve |x2 – x – 4| > 2 • Four critical points → 5 intervals to test • (-∞,-2] • Use x = -3, result is 8 > 2 SUCCESS • [-2,-1] • Use x = -1.5, result is 0.25 > 2 FAIL • [-1,2] • Use x = 0, result is 4 > 2 SUCCESS • [2,3] • Use x = 2.5, result is 0.25> 2 FAIL • [3,∞) • Use x = 4, result is 8 > 2 SUCCESS • Solution are the intervals (-∞,-2], [-1,2], and [3,∞)

2.5A: Absolute-Value Inequalities • Problem #14: Extraneous Roots • Solve • Two equations • Normal: • Subtract 2 from both sides • Find a common denominator • Simplify numerator • Real solution: -x-3 = 0, x = -3 • Extraneous solution: x+2 = 0, x = -2

2.5A: Absolute-Value Inequalities • Problem #14 (continued): Extraneous Roots • Solve • Two equations • Flip: • Add 2 to both sides • Find a common denominator • Simplify numerator • Real solution: 3x+5 = 0, x = -5/3 • Extraneous solution: x+2 = 0, x = -2

2.5A: Absolute-Value Inequalities • Problem #14 (testing): Extraneous Roots • Solve • Three critical points → -3, -2, -5/3→ 4 intervals • (-∞, -3) • Test x = -4, result is 1.5 < 2 SUCCESS • (-3, -2) • Test x = -2.5, result is 3 < 2 FAIL • (-2, -5/3) • Test x = -1.8, result is 4 < 2 FAIL • (-5/3, ∞) • Test x = 0, result is 0.5 < 2 SUCCESS • Solution are the intervals (-∞,-3) and (-5/3,∞) • Note: Graphing helps provide a faster check, especially as you increase the number of critical points

2.5A: Absolute-Value Inequalities • Assignment (Oct 26) • Page 131 • 1-27, odd problems • Note #1: Show work • Note #2: 25 and 27 are going to have to be solved by graphing

Chapter 2: Equations and Inequalities 2.5A: Absolute-Value Inequalities