Lesson Objectives

- Demonstrate an understanding of absolute value
- Learn how to solve nested absolute value equations
- Learn how to solve advanced absolute value equations

## How to Solve Advanced Absolute Value Equations

In this lesson, we want to discuss how to solve advanced absolute value equations. Up to this point, we have seen absolute value equations that can be solved pretty easily. In most cases, we just needed to isolate the absolute value operation and set up a compound equation with "or". We then solve the two resulting equations and have our answer. Now, we will push a little further and look at some examples that are much more tedious. Let’s begin by thinking about how to solve an absolute value equation where one absolute value operation is nested inside of another.

Example #2: Solve each equation. $$|x - 4| + |x - 1|=12$$ First, we need to find the values that make the expressions inside of each absolute value operation equal to zero. This is where a sign change would occur, from positive to negative or from negative to positive. This can only happen at a value of zero. $$x - 4=0$$ $$x=4$$ $$x - 1=0$$ $$x=1$$ Next, we think about the value of the absolute value expression in each interval. If it is positive in the interval, we will simply drop the absolute value bars. If it is negative in the interval, we will also drop the absolute value bars, but we must first change the expression inside of the absolute value operation into its opposite. We can do this by wrapping our expression with parentheses and then placing a negative outside.

This information is usually easiest to organize using a table or number line.

Example #3: Solve each equation. $$x^2 + 3|x| - 10=0$$ First, let's realize that squaring x leads to a non-negative answer. $$x^2 ≥ 0$$ This leads us to the following: $$x^2=|x|^2$$ Let's replace this in our problem. $$|x|^2 + 3|x| - 10=0$$ From here, we will make a simple substitution: $$\text{let}\hspace{.2em}u=|x|$$ $$u^2 + 3u - 10=0$$ $$(u - 2)(u + 5)=0$$ $$u=2, -5$$ Since u is |x| we will substitute again: $$|x|=2$$ $$x=2, -2$$ $$|x|=-5$$ This second scenario leads to no solution. The result of the absolute value operation is non-negative. We can't find a number whose absolute value is -5. Our final answer: $$x=2, -2$$

### How to solve Nested Absolute Value Equations

Example #1: Solve each equation. $$||2x - 1| + 5|=14$$ To solve this problem, let's first consider the larger absolute value expression: $$|2x - 1| + 5=14$$ $$\text{or}$$ $$|2x - 1| + 5=-14$$ Now, we can just solve each as a normal absolute value equation. $$|2x - 1| + 5=14$$ $$|2x - 1|=9$$ $$2x - 1=9$$ $$2x=10$$ $$x=5$$ $$\text{or}$$ $$2x - 1=-9$$ $$2x=-8$$ $$x=-4$$ So far, our solutions are: $$x=-4, 5$$ Now, let's look at the other part: $$|2x - 1| + 5=-14$$ $$|2x - 1|=-19$$ This part doesn't have a solution. We can conclude that we just have two solutions: $$x=-4, 5$$### Solving Absolute Value Equations with Two Absolute Value Operations

Here, we want to look at an example where we have two absolute value operations and a loose number. The loose number stops us from setting the two absolute value operations equal to each other.Example #2: Solve each equation. $$|x - 4| + |x - 1|=12$$ First, we need to find the values that make the expressions inside of each absolute value operation equal to zero. This is where a sign change would occur, from positive to negative or from negative to positive. This can only happen at a value of zero. $$x - 4=0$$ $$x=4$$ $$x - 1=0$$ $$x=1$$ Next, we think about the value of the absolute value expression in each interval. If it is positive in the interval, we will simply drop the absolute value bars. If it is negative in the interval, we will also drop the absolute value bars, but we must first change the expression inside of the absolute value operation into its opposite. We can do this by wrapping our expression with parentheses and then placing a negative outside.

This information is usually easiest to organize using a table or number line.

$$(-\infty, 1)$$ | $$(1, 4)$$ | $$(4, \infty)$$ |
---|---|---|

$$-(x-1)$$ | $$(x-1)$$ | $$(x-1)$$ |

$$-(x-4)$$ | $$-(x-4)$$ | $$(x-4)$$ |

Note: You may include 1 in the second interval [1, 4) and include 4 in the third interval [4, ∞). This is a matter of personal preference and will only cause concern if you obtain 1 or 4 (known as the breakpoints or where an expression inside of the absolute value bars is zero) as a solution. Most students find it easier to set the problem up this way. You can always check a breakpoint in the original problem if it comes up as a solution and then accept it if it works (gives a true statement) or reject it otherwise (gives a false statement). You can take a look at #3a and #4a on the practice test where you will get an opportunity to think about the different ways of setting up a problem with breakpoint answers. If you are still lost, the full step-by-step solutions are given in video form. As for this tutorial, we will continue with (1, 4), and (4, ∞) as our second and third intervals.

We will set up three different equations based on the expressions in the interval. From there, we will accept the solution if it lies in the interval and reject the solution if it's outside of the interval. Let's begin with: $$(-\infty, 1)$$ $$|x - 4| + |x - 1|=12$$ Both expressions are negative in this interval: $$-(x - 4) - (x - 1)=12$$ $$-x + 4 - x + 1=12$$ $$-2x + 5=12$$ $$-2x=7$$ $$x=-\frac{7}{2}$$ Since -7/2 or -3.5 is in our interval, we can accept this as part of our solution set. Let's move on to the next interval: $$(1,4)$$ $$|x - 4| + |x - 1|=12$$ One expression is positive and the other is negative in this interval: $$-(x - 4) + (x - 1)=12$$ $$-x + 4 + x - 1=12$$ $$3=12$$ This leads to a false statement, so there won't be a solution in this interval. Let's move on to the final interval: $$(4, \infty)$$ Both expressions are positive, so just drop the absolute value bars: $$|x - 4| + |x - 1|=12$$ $$x - 4 + x - 1=12$$ $$2x - 5=12$$ $$2x=17$$ $$x=\frac{17}{2}$$ Since 17/2 or 8.5 is in our interval, we can accept this as part of our solution set. We can state our solution as: $$x=-\frac{7}{2}, \frac{17}{2}$$ ### Solving an Absolute Value Equation that is Quadratic in Form

For our last example, we will learn how to solve an absolute value equation that is quadratic in form. As we have previously seen, equations that are quadratic in form are often solved using a simple substitution technique.Example #3: Solve each equation. $$x^2 + 3|x| - 10=0$$ First, let's realize that squaring x leads to a non-negative answer. $$x^2 ≥ 0$$ This leads us to the following: $$x^2=|x|^2$$ Let's replace this in our problem. $$|x|^2 + 3|x| - 10=0$$ From here, we will make a simple substitution: $$\text{let}\hspace{.2em}u=|x|$$ $$u^2 + 3u - 10=0$$ $$(u - 2)(u + 5)=0$$ $$u=2, -5$$ Since u is |x| we will substitute again: $$|x|=2$$ $$x=2, -2$$ $$|x|=-5$$ This second scenario leads to no solution. The result of the absolute value operation is non-negative. We can't find a number whose absolute value is -5. Our final answer: $$x=2, -2$$

#### Skills Check:

Example #1

Solve each equation. $$||2x + 7| - 3|=20$$

Please choose the best answer.

A

$$x=-7, -5, 3, 1$$

B

$$x=13, \frac{1}{2}$$

C

$$x=-15, 8$$

D

$$x=15, -8$$

E

$$x=-13, \frac{1}{2}$$

Example #2

Solve each equation. $$|2x - 3| + |x - 7|=25$$

Please choose the best answer.

A

$$x=-\frac{5}{3}, -1$$

B

$$x=-1, \frac{3}{2}$$

C

$$x=-5, \frac{35}{3}$$

D

$$x=-1, 3, 7, 9$$

E

$$x=-4, -2$$

Example #3

Solve each equation. $$|9x - 1| + |x + 2|=8$$

Please choose the best answer.

A

$$x=-\frac{5}{8}, \frac{7}{10}$$

B

$$x=-4, \frac{1}{2}$$

C

$$x=-7, 10$$

D

$$x=-19, \frac{3}{5}$$

E

$$x=-8, 7$$

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