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: $$|u|=a$$ $$u=a$$ $$or$$ $$u=-a$$ Let's take a look at a quick example: $$|2x + 9|=3$$ $$2x + 9=3$$ $$or$$ $$2x + 9=-3$$ 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.

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: $$2x^2 + 9x=5$$ $$or$$ $$2x^2 + 9x=-5$$ 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 this expression, 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: $$ x=\frac{1}{2}, -5, \frac{-9 \pm \sqrt{41}}{4}$$

If u is an algebraic expression and a is a positive real number, then: $$|u|=a$$ $$u=a$$ $$or$$ $$u=-a$$ Let's take a look at a quick example: $$|2x + 9|=3$$ $$2x + 9=3$$ $$or$$ $$2x + 9=-3$$ 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: $$2x^2 + 9x=5$$ $$or$$ $$2x^2 + 9x=-5$$ 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 this expression, 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. $$\frac{x - 3}{x - 5}=7$$ $$or$$ $$\frac{x - 3}{x - 5}=-7$$ 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: $$x=\frac{16}{3}, \frac{19}{4}$$#### Skills Check:

Example #1

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

Please choose the best answer.

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

Please choose the best answer.

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. $$|x^2 - 2x - 3|=3$$

Please choose the best answer.

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

$$x=-1, 3, 4 \pm \sqrt{13}$$

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