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6.5: Absolute Value Equations

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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 120\;\mathrm{units} 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, | a - b | = | b - a |.
  • Have students rethink the simple absolute value equations |x| = 3 and |x| = 10 in Example 3 as |?| = 3 and |?| = 10, respectively. (Which numbers are 3\;\mathrm{units} from the origin? 10\;\mathrm{units} from the origin?)
  • Have students interpret absolute value equations out loud. |x - 2| = 7 means “those numbers on the number line 7\;\mathrm{units} 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 |x - 2| = 0 and |x - 2| = 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 3\;\mathrm{units} away from -1. Therefore, the solution set is \left \{2, -4\right \}.

b. Solve the equation and interpret the answer.

|3x - 6| = 8

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

\left |\frac{\cancel{3}x} {\cancel{3}} - \frac{6} {3}\right | & = \frac{8} {3}\\|x - 2| & = \frac{8} {3}

This last equation can now be interpreted as those numbers 8/3\;\mathrm{units} away from 2.

Treat the absolute value as a grouping symbol when appropriate.

  • The distributive law holds in expressions such as 3|x - 4| = |3x - 12| and \frac{1} {3} |3x - 6| = \left |\frac{\cancel{3}x} {\cancel{3}} - \frac{6} {3}\right | = |x - 2|.
  • The distributive law does not hold in an expression such as -2 |x - 3| \neq | -2x + 6 |
  • In general, distribute into absolute value a \cdot |b| = |a \cdot b| when a = |a|; i.e., for positive numbers a.
  • These steps are based on the property |a| \cdot |b| = |a \cdot b|.

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|x - 3| \neq | -2x + 6 |.

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

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