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1Copyright © 2014, 2010, and 2006 Pearson Education, Inc.
Absolute-Value Equations and Inequalities
Equations with Absolute Value
Inequalities with Absolute Value
9.3
2Copyright © 2014, 2010, and 2006 Pearson Education, Inc.
Warm Up
– Solve and graph each compound inequality.
– 1. -3x – 4 < -16 and 2x - 5 < 15– 2. 4x + 2 > 10 or -2x + 6 > 16– 3. 5 < -3x+2 < 8– 4. 2x < -2 and x + 3 > 6– 5. 5x > 15 or 6x - 4 < 32
3Copyright © 2014, 2010, and 2006 Pearson Education, Inc.
Absolute Value
The absolute value of a number is its distance from 0. Since absolute value is distance it is never negative.
, if 0,
, if 0.
x xx
x x
Definition of absolute value.
4Copyright © 2014, 2010, and 2006 Pearson Education, Inc.
Find each absolute value.
|-3| |0| |4|
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Solution
Example
a) |x| = 6; b) |x| = 0; c) |x| = –2
a) |x| = 6 means that the distance from x to 0 is 6. Only two numbers meet that requirement. Thus the solution set is {–6, 6}.
b) |x| = 0 means that the distance from x to 0 is 0. The only number that satisfies this is zero itself. Thus the solution set is {0}.
c) Since distance is always nonnegative, |x| = –2 has no solution. Thus the solution set is .
6Copyright © 2014, 2010, and 2006 Pearson Education, Inc.
This brings us to…
The Absolute-Value Principlefor Equations If |x| = p, then x = p or x = -p.
Note: The equation |x| = 0 is equivalent to the equation x = 0 and the equation |x| = –p has no solution.
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Example: a) |2x +1| = 5; b) |3 – 4x| = –10
The solution set is {–3, 2}. The check is left for the student.
x = –3 or x = 2 2x = –6 or 2x = 4
Substituting
Solutiona) We use the absolute-value principle, knowing
that 2x + 1 must be either 5 or –5:
|2x +1| = 5
|x| = p
2x +1 = –5 or 2x +1 = 5
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Solution(continued)
The absolute-value principle reminds us that absolute value is always nonnegative. The equation |3 – 4x| = –10 has no solution. The solution set is
To apply the absolute value principle we must make sure the absolute value expression is ISOLATED.
.
b) |3 – 4x| = –10
9Copyright © 2014, 2010, and 2006 Pearson Education, Inc.
Solution Since we
Given that f (x) = 3|x + 5| – 4, find all x for which f (x) = 11
are looking for f(x) = 11, we substitute:
Replacing f (x) with 3|x + 5| − 4
x + 5 = –5 or x + 5 = 5 x = –10 or x = 0
The solution set is {–10, 0}. The check is left for the student.
3|x+5| – 4 = 11
f(x) = 11
3|x + 5| = 15
|x + 5| = 5
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Sometimes an equation has two absolute-value expressions like |x+1| = |2x|. This means that x+1 and 2x are the same distance from zero.
Since they are the same distance from zero, they are the same number or they are opposites.
So solve x+1 = 2x and x+1 = -2x separately to get your answers. As always check by substitution.
When more than one absolute value expression appears…
11Copyright © 2014, 2010, and 2006 Pearson Education, Inc.
Solve |3x – 5| = |8 + 4x|.
You are not done until you solve each equation.
3x – 5 = 8 + 4x
This equation sets both sides the same.
This equation sets them opposite. Notice parenthesis.
3x – 5 = –(8 + 4x) or
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3 5 8 4x x 3 5 (8 4 )x x
5 8 x 13 x
3 5 8 4x x 7 5 8x
7 3x 3
7x
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The Absolute-Value Inequalities Principle
For any positive number p and any expression x:
a) |x| < p is equivalent to –p < x < p. (conjunction)
b) |x| > p is equivalent to x < -p or x > p. (disjunction)
c) Similar rules apply for less than or equal, greater than or equal.
14Copyright © 2014, 2010, and 2006 Pearson Education, Inc.
So the solutions of |X| < p are those numbers that satisfy –p < X < p.
And the solutions of |X| > p are those numbers that satisfy X < –p or p < X.
–p p
–p p( )
()
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Solution
Solve |x| < 3. Then graph.
The solutions of |x| < 3 are all numbers whose distance from zero is less than 3. By substituting we see that numbers like –2, –1, –1/2, 0, 1/3, 1, and 2 are all solutions. The solution set is {x| –3 < x < 3}. In interval notation, the solution set is (–3, 3). The graph is as follows:
-3 0 3( )
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Solution
The solutions of are all numbers whose distance from zero is at least 3 units. The solution set is In interval notation, the solution set is
The graph is as follows:
Solve 3. Then graph.x
3x
{ | 3 3}. x x or x
( , 3] [3, ).
[-3 3]
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SolutionThe number 3x + 7 must be less than 8 units from 0.
|3x + 7| < 8
|X| < pReplacing X with 3x + 7 and p with 8
–8 < 3x + 7 < 8
The solution set is {x|–5 < x < 1/3}. The graph is as follows:
Solve |3x + 7| < 8. Then graph.
−5 < x < 1/3
–15 < 3x < 1
–5 1/3( )
18Copyright © 2014, 2010, and 2006 Pearson Education, Inc.
Solution
The number 3x + 7 must be greater than 8 units from 0.
|3x + 7| < 8
|X| < pReplacing X with 3x + 7 and p with 8
–8 < 3x + 7 < 8
The solution set is {x|–5 < x < 1/3}. The graph is as follows:
Solve |3x + 7| > 8. Then graph.
−5 < x < 1/3
–15 < 3x < 1
–5 1/3( )