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6.5: Absolute Value Equations

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

At the end of this lesson, students will be able to:

  • Solve an absolute value equation.
  • Analyze solutions to absolute value equations.
  • Graph absolute value functions.
  • Solve real-world problems using absolute value equations.

Vocabulary

Terms introduced in this lesson:

absolute value
absolute value equations
vertex or cusp

Teaching Strategies and Tips

Focus on the interpretation of absolute value as a distance.

  • In Example 1b, |120|=120 because 120 is 120units from the origin.
  • In Example 2, have students explain, using a distance argument, why the order in which the two numbers are subtracted is not important. In general, for any two numbers (or points) a and b, |ab|=|ba|.
  • Have students rethink the simple absolute value equations |x|=3 and |x|=10 in Example 3 as |?|=3 and |?|=10, respectively. (Which numbers are 3units from the origin? 10units from the origin?)
  • Have students interpret absolute value equations out loud. |x2|=7 means “those numbers on the number line 7units away from 2.” See Examples 4-6. Encourage students to draw the number line and mark the possible solutions.
  • Using the distance interpretation, point out that absolute value equations (involving only linear functions) can have no more than 2 solutions. Have students consider absolute value equations with 1 or 0 solutions such as |x2|=0 and |x2|=5, respectively.

Students have trouble reconciling the definition “|x|=x if x is negative” and the fact that “absolute value changes a negative number into its positive inverse.” Offer an example:

  • Let x=5. Then |5|=(5) since 5 is negative (using the definition). This simplifies to |5|=5.
  • |5|=5 since absolute value changes a negative number into its positive inverse.

Use Examples 5 and 6 to show students how to rewrite absolute value equations so that the distance interpretation is clearer.

Additional Examples:

a. Solve the equation and interpret the answer.

|x+1|=3

Solution: As it stands, the equation cannot be interpreted in terms of distance. Rewrite the equation with a minus sign: |x(1)|=3 which can now be interpreted as those numbers 3units away from 1. Therefore, the solution set is {2,4}.

b. Solve the equation and interpret the answer.

|3x6|=8

Hint: As it stands, the equation cannot be interpreted in terms of distance. Rewrite the equation by dividing both sides by 3:

3x363|x2|=83=83

This last equation can now be interpreted as those numbers 8/3units away from 2.

Treat the absolute value as a grouping symbol when appropriate.

  • The distributive law holds in expressions such as 3|x4|=|3x12| and 13|3x6|=3x363=|x2|.
  • The distributive law does not hold in an expression such as 2|x3||2x+6|
  • In general, distribute into absolute value a|b|=|ab| when a=|a|; i.e., for positive numbers a.
  • These steps are based on the property |a||b|=|ab|.

When beginning to graph absolute value functions, encourage students to make a table of values such as those in Examples 7 and 8.

Have students plot and describe in words the basic graph y=|x|. Allow them to observe the essential properties of the absolute value graph:

  • The graph has a V shape, consisting of two rays that meet at a sharp point, called the vertex or cusp.
  • One side of the V has positive slope and other side negative slope.
  • The vertex is located at the point where the expression inside the absolute value is equal to zero.

Error Troubleshooting

General Tip: Remind students not to distribute a negative into the absolute value expression. For example, 2|x3||2x+6|.

General Tip: Students may misinterpret “absolute value is always positive” and commit the error |35|=2. Suggest to students that in such situations |35|=(1)|35|. The multiplication by \begin{align*}-1\end{align*} happens after the absolute value has been performed, and so \begin{align*}-|3 - 5| = (-1)|3 - 5| = (-1) |-2| = (-1) 2 = -2\end{align*}.

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CK.MAT.ENG.TE.1.Algebra-I.6.5
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