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
