Solving Absolute Value Equations
Lesson 3 of 3
HS Algebra
4060 minutes
Description
Solving absolute value equations graphically and algebraically.
Materials
Introduce
 Begin with the Dynagraph Applet set to Relationship 2 and k = 5 (this would be x → x  5).
 Ask students how the Dynagraph can show the solutions to the equation x  5 = 2.
 In the discussion, the idea that the output point needs to coincide with the desired answer should be addressed. The output point tells us what the input xvalue should be. In this case it is x = 7. Therefore,
x = 7 is a solution.
 The next example should begin with the Dynagraph Applet set to Relationship 3 and k = 5
(this would be x → x  5).
 Ask students to consider the solutions to the equation x5=2.
 Recalling the previous example, students should suggest that the output point needs to coincide with the desired output of 2.
 Use the Dynagraph to show the input values that give output values of 2. In this case there are two input values, x = 3 and x = 7.
 This opening activity sets the stage for the two connections that will be made to solving absolute value equations in this lesson.
Explore
Our goal today is to take what we have seen using the Dynagraphs and look at solving absolute value problems from both a graphical approach on the coordinate plane and algebraically.
To solve x5=2 graphically, we can think of it as a system of equations where Y1=x5 and Y2=2. How is a solution to a system of equations represented graphically? Students should recall that the solution is where the two functions intersect on the graph. Additionally students should be encouraged to connect how this approach connects to getting solutions from the Dynagraph.
Have students graph each function by hand and identify where the solutions are located on the graph. [x=3 and x=7.]
How can we find the solutions algebraically?
If we think back to Lesson 2, we can write the absolute value function as a piecewise function consisting of two lines:
x5 for the domain values where the output y=x5 is above the xaxis
and the opposite of x5 for the domain values where the output y=x5 is below the xaxis.
We are then figuring out where each of these pieces of the function are equal to 2 to solve the equation.
x  5 = 2
x = 7

(x  5) = 2
x  5 = 2
x =
3 
There is a difference in the way we are setting up our equations for the piecewise function and for solving the absolute value problem. Did you notice we did not take the function x5 and set it equal to 2 and 2? Instead we took the function equal to 2 and the opposite of the function equal to 2.
We are thinking about (x5) as being a reflection over the xaxis for values in the domain that had a negative output value. So, instead of thinking of where that portion of the graph is equal to 2 we reflect the function (or take the opposite of it) and figure out where it is equal to 2. We are finding values for the function x5 that are a distance of 2 away from the xaxis.
Have students complete the Solving Absolute Value Equations AS. [SMP 5, SMP 7]
Discuss student work from Solving Absolute Value Equations activity sheet. Make sure to address why x  6=4 has no solution.
Teacher Note:
If time permits, consider the following extension.
After discussing the first example, have students consider another absolute value equation: x^29x+14=3. Ask students, "What is a solution? What is a nonsolution? Explain."
Students will most likely struggle to find a solution to this equation because the solutions turn out to be irrational. This might be a good time to discuss the pros and cons of solving absolute value equations by graphing as compared to algebraically.
Have students graph the system: Y1=x^29x+14 and Y2=3. (Recall, from lesson 2, that you can graph 〖x〗^29x+14 by graphing the quadratic x^29x+14 as long as you reflect the portion that has negative output values over the xaxis.)
Ask students to rewrite the absolute value equation using a piecewise function. The portion of the parabola that is below the xaxis is reflected over the xaxis since it is an absolute value; that is why we can write that part of the function as the "opposite" or (x^2+2x3).
So, solving this system algebraically, students can set each of these pieces equal to 3 and solve for x.
Finally, have students consider the following question: Why is there no solution to the following problem: x〖^2〗4=2? [SMP 7]
[The students may say that there is no value for x that makes the equation true. If needed, help students also notice that the system of equations to be graphed will not have an intersection.]
Synthesize
Summarize or explain how to rewrite an absolute value function as a piecewise function.
How does solving absolute value equations graphically as a system of equations support the process used to solve absolute value equations algebraically? When is one method more appropriate than the other?
Teacher Reflection
 What connections did students make between the graphical and algebraic approaches to solving absolute value equations?
 How did students respond to solving absolute value equations that had no solution?
 How did the use of graphing help students understand the procedures involved in algebraically solving absolute value equations?
Leave your thoughts in the comments below.