Essential Question: What is the procedure used to solve an absolute value equation of inequality?

 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)

 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.

 Example 3: Simple Absolute Value Inequality ◦ Solve |3x – 7| < 11  Two equations 1)3x – 7 < 11  Add 7 to both sides  3x < 18  Divide both sides by 3  x < 6 2)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

 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

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

 Ex 5 (continued) – test the critical points ◦ Solve |x 2 – 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,∞)

 Problem #14: Extraneous Roots ◦ Solve ◦ Two equations 1)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

 Problem #14 (continued): Extraneous Roots ◦ Solve ◦ Two equations 2)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

 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

 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