Lesson Objectives
• Demonstrate an understanding of the concept of absolute value
• Learn how to solve absolute value equations with quadratic expressions
• Learn how to solve absolute value equations with rational expressions

## How to Solve Absolute Value Equations with Quadratic/Rational Expressions

In this lesson, we want to go deeper into the topic of how to solve absolute value equations. Previously in our course, we learned how to solve simple absolute value equations. Now, we are going to look at some additional examples that are slightly more challenging. Let's begin by looking at a general rule for solving absolute value equations.
If u is an algebraic expression and a is a positive real number, then: $$|u|=a$$ $$u=a$$ $$\text{or}$$ $$u=-a$$ Let's take a look at a quick example: $$|2x + 9|=3$$ $$2x + 9=3$$ $$\text{or}$$ $$2x + 9=-3$$ Solving both equations gives us two solutions: $$x=-6, -3$$ If a ends up being zero, we just get one equation. This is because zero doesn’t have an opposite like every other number. The only number with an absolute value of zero is zero.
Let's take a look at a quick example: $$|9x - 7|=0$$ $$9x - 7=0$$ $$x=\frac{7}{9}$$ If a is negative, there is no solution.
Let's take a look at a quick example: $$|3x - 1|=-5$$ No solution since the absolute value operation can’t give us a negative result. There isn’t a number that can be plugged in for x, where we would get a true statement.

### Solving Absolute Value Equations with Quadratic Expressions

Let’s use our rule to look at a problem with a quadratic expression inside of absolute value bars.
Example #1: Solve each equation. $$3|2x^2 + 9x|-1=14$$ The first thing we want to do is get our absolute value operation by itself on one side of the equation:
Add 1 to each side: $$3|2x^2 + 9x|=15$$ Divide each side by 3: $$|2x^2 + 9x|=5$$ Now we can use our rule and set up two equations: $$2x^2 + 9x=5$$ $$\text{or}$$ $$2x^2 + 9x=-5$$ Let's solve the top equation first: $$2x^2 + 9x - 5=0$$ $$(2x - 1)(x + 5)=0$$ $$x=\frac{1}{2}, -5$$ Now let's move on to the bottom equation: $$2x^2 + 9x + 5=0$$ Since we can't factor the left side, we will use the quadratic formula:
$$a=2, b=9, c=5$$ $$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$x=\frac{-9 \pm \sqrt{81 - 4(2)(5)}}{2(2)}$$ $$x=\frac{-9 \pm \sqrt{81 - 40}}{4}$$ $$x=\frac{-9 \pm \sqrt{41}}{4}$$ We have four solutions: $$x=\frac{1}{2}, -5, \frac{-9 \pm \sqrt{41}}{4}$$

### Solving Absolute Value Equations with Rational Expressions

Example #2: Solve each equation. $$\left|\frac{x - 3}{x - 5}\right|=7$$ Again, we use our rule and split this up into two separate equations. $$\frac{x - 3}{x - 5}=7$$ $$\text{or}$$ $$\frac{x - 3}{x - 5}=-7$$ Let's begin with the top equation: $$\frac{x - 3}{x - 5}=7$$ Clear the denominator: $$x - 3=7(x - 5)$$ Simplify: $$x - 3=7x - 35$$ $$-6x=-32$$ Solve: $$x=\frac{32}{6}=\frac{16}{3}$$ For the bottom equation, we repeat the process: $$\frac{x - 3}{x - 5}=-7$$ Clear the denominator: $$x - 3=-7(x - 5)$$ Simplify: $$x - 3=-7x + 35$$ $$8x=38$$ Solve: $$x=\frac{38}{8}=\frac{19}{4}$$ We have two solutions: $$x=\frac{16}{3}, \frac{19}{4}$$

### Solving Absolute Value Equations with Two Absolute Value Expressions

In both of our previous examples, we had only one absolute value expression in our equation. What happens when there are two absolute value expressions? When this happens and we can set one absolute value expression equal to the other, we can use the following method. Note, that the harder case is covered in the next lesson. $$|u|=|w|$$ $$u=w$$ $$\text{or}$$ $$-u=-w$$ $$\text{or}$$ $$-u=w$$ $$\text{or}$$ $$u=-w$$ We don't actually need to solve four equations, only two. The first two lead to the same equation since we can divide both sides of our second equation by -1: $$u=w$$ In the last two, we only have to consider that the two have opposite signs since we can divide both sides of either equation by -1: $$u=-w$$ In summary, we will consider that both are positive (just drop the absolute value bars) and then consider that the two are opposites (drop the absolute value bars and change the expression on the right into its opposite). Let's look at an example.
Example #3: Solve each equation. $$\left|\frac{5}{3x-1}\right|=\left|\frac{3}{x+1}\right|$$ First, we will simply drop the absolute value bars and solve the resulting equation. $$\frac{5}{3x - 1}=\frac{3}{x + 1}$$ Cross Multiply: $$5(x + 1)=3(3x - 1)$$ $$5x + 5=9x - 3$$ $$-4x=-8$$ $$x=2$$ Second, we will drop the absolute value bars and change the expression on the right into its opposite. $$\frac{5}{3x - 1}=-\frac{3}{x + 1}$$ Cross Multiply: $$5(x + 1)=-3(3x - 1)$$ $$5x + 5=-9x + 3$$ $$14x=-2$$ $$x=-\frac{1}{7}$$ We have two solutions: $$x = 2, -\frac{1}{7}$$

#### Skills Check:

Example #1

Solve each equation. $$|15x^2 - 26x|=7$$

A
$$x=\frac{1}{5}, \frac{19}{5}, \frac{13 \pm \sqrt{274}}{15}$$
B
$$x=\frac{1}{5}, \frac{12}{7}, \pm 2i$$
C
$$x=\frac{1}{3}, 5, 7, 9$$
D
$$x=\frac{1}{3}, \frac{7}{5}, \frac{13 \pm \sqrt{274}}{15}$$
E
$$x=-\frac{1}{3}, \frac{7}{5}, \frac{19 \pm \sqrt{13}}{2}$$

Example #2

Solve each equation. $$\left|\frac{3}{x-1}\right|=\left|\frac{2}{x + 1}\right|$$

A
$$x=-10, \frac{13}{5}$$
B
$$x=-5, -\frac{1}{5}$$
C
$$x=-20, 4$$
D
$$x=-\frac{1}{3}, \frac{19}{5}$$
E
$$x=-7, 3$$

Example #3

Solve each equation. $$|x^2 - 2x - 3|=3$$

A
$$x=0, 2, 1 \pm \sqrt{19}$$
B
$$x=0, 2, 1 \pm \sqrt{14}$$
C
$$x=-5, 5, 6 \pm 3i$$
D
$$x=0, 2, 1 \pm \sqrt{7}$$
E
$$x=-1, 3, 4 \pm \sqrt{13}$$         