171S3.5 Solving Equations and Inequalities with Absolute Value
February 29, 2012
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
3.2
3.3
3.4
3.5
The Complex Numbers
Quadratic Equations, Functions, Zeros, and Models
Analyzing Graphs of Quadratic Functions
Solving Rational Equations and Radical Equations
Solving Equations and Inequalities with Absolute Value
This is a very good 6­minute video by a textbook author. He shows how to solve two problems: | 3y + 9 | ≥ 6 and | 3x + 5 | ­ 8 < 5. http://www.youtube.com/watch?v=Jad08Q4puOc
Printed instructions for graphing one­variable inequality with TI calculator.
http://cfcc.edu/faculty/cmoore/ti­inequality­1.htm
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. http://cfcc.edu/faculty/cmoore/AbsoluteValueEquations1.wmv
2. Absolute Value Inequality 1 short video (by Dr. Moore) demonstrating the solution of simple inequalities. http://cfcc.edu/faculty/cmoore/AbsoluteInequality1.wmv
3. Absolute Value Inequality 2 short video (by Dr. Moore) demonstrating the solution more complex inequalities. http://cfcc.edu/faculty/cmoore/AbsoluteInequality2.wmv
NOTE: These videos are in the Technology on the Important Links webpage.
1
171S3.5 Solving Equations and Inequalities with Absolute Value
3.5 Solving Equations and Inequalities with Absolute Value
• Solve equations with absolute value.
• Solve inequalities with absolute value.
February 29, 2012
Example
Solve:
Solution:
Equations with Absolute Value
For a > 0 and an algebraic expression X:
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 is equivalent to
X = ­a or X = a.
Example
Solve:
Solution: First, add one to both sides to get the expression in the form | X | = a.
Example (continued)
The possible solutions are –2 and 8.
Check x = 8:
Check x = –2:
TRUE
Let’s check the possible solutions –2 and 8.
TRUE
The solutions are –2 and 8.
2
171S3.5 Solving Equations and Inequalities with Absolute Value
February 29, 2012
Inequalities with Absolute Value
More About Absolute Value Equations
Inequalities sometimes contain absolute­value notation. The following properties are used to solve them.
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
For example,
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.
Example
Solve: Solve and graph the solution set:
Solution:
| x | < 3 is equivalent to ­3 < x < 3
| y | ≥ 1 is equivalent to y ≤ ­1 or y ≥ 1
| 2x + 3 | ≤ 4 is equivalent to ­4 < 2x + 3 < 4
3
171S3.5 Solving Equations and Inequalities with Absolute Value
February 29, 2012
3.5 Solving Equations and Inequalities with Absolute Value
Example
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 = 4.5.
3.5 Solving Equations and Inequalities with Absolute Value
287/14. |x ­ 7| = 5
3.5 Solving Equations and Inequalities with Absolute Value
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
Since Y1 and Y2 do intersect, there is a common solution. 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
http://www.sascurriculumpathways.com/portal/resources/mathematics/grap
herseries/medalg/tool.jsf?resourceID=1424
4
171S3.5 Solving Equations and Inequalities with Absolute Value
February 29, 2012
3.5 Solving Equations and Inequalities with Absolute Value
287/28. |5x + 4| + 2 = 5
3.5 Solving Equations and Inequalities with Absolute Value
287/20. |(1/3)x ­ 4| = 13
SOLULTION:
(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
SOLULTION:
|5x + 4| + 2 = 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.
Since Y1 and Y2 do intersect, there is a common solution. Thus, |(1/3)x ­ 4| = 13 has the solution x = ­27 or x = 51.
Y1 = |5x + 4| + 2
Y2 = 5
Since Y1 and Y2 do intersect, there is a common solution. Thus, |5x + 4| +2 = 5 has the solution x = ­1.4 or x = ­0.2.
3.5 Solving Equations and Inequalities with Absolute Value
Solve and write interval notation for the solution set. Then graph the solution set.
288/42. |5x| < 4
3.5 Solving Equations and Inequalities with Absolute Value
288/32. 5 ­ |4x + 3| = 2
Y1 = |5x|
Y2 = 4
Since Y1 and Y2 do intersect, there is a common solution. 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.
Since Y1 and Y2 do intersect, there is a common solution. Thus, 5 ­ |4x + 3| = 2 has the solution x = ­1.5 or x = 0.
Instructions for graphing one­variable inequality with TI calculator.
http://cfcc.edu/faculty/cmoore/ti­inequality­1.htm
5
171S3.5 Solving Equations and Inequalities with Absolute Value
February 29, 2012
3.5 Solving Equations and Inequalities with Absolute Value
Solve and write interval notation for the solution set. Then graph the solution set.
288/56. |5 ­ 2x| > 10
SOLULTION:
|5 ­ 2x| > 10
5 ­ 2x < ­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 one­variable inequality with TI calculator.
http://cfcc.edu/faculty/cmoore/ti­inequality­1.htm
6
171S3.5 Solving Equations and Inequalities with Absolute Value
February 29, 2012
3.5 Solving Equations and Inequalities with Absolute Value
Solve and write interval notation for the solution set. Then graph the solution set.
288/60. |(2x ­ 1) / 3| > 5/6
SOLULTION:
|(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 ≥ 1.75.
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 1.75. Thus, we have closed circles at these two values. So, the solution is (­∞, ­0.75] U [1.75, ∞).
Instructions for graphing one­variable inequality with TI calculator.
http://cfcc.edu/faculty/cmoore/ti­inequality­1.htm
7