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

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

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• Slide 1
• Essential Question: What is the procedure used to solve an absolute value equation of inequality?
• Slide 2
• Just like equations, absolute value inequalities can be solved either by graphing or algebraically Graphing Methods Intersection Method Wont 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 (, 2) and (2, 5)
• Slide 3
• 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.
• Slide 4
• 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
• Slide 5
• 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
• Slide 6
• 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
• Slide 7
• 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,)
• Slide 8
• 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
• Slide 9
• 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
• Slide 10
• 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
• Slide 11
• 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