# Solving Other Types of Equations

### Learning Objectives

- Solve polynomial equations
- Solve absolute value equations

### A General Note: Polynomial Equations

A polynomial of degree*n*is an expression of the type

[latex]{a}_{n}{x}^{n}+{a}_{n - 1}{x}^{n - 1}+\cdot \cdot \cdot +{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}[/latex]

where *n*is a positive integer and [latex]{a}_{n},\dots ,{a}_{0}[/latex] are real numbers and [latex]{a}_{n}\ne 0[/latex]. Setting the polynomial equal to zero gives a

**polynomial equation**. The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent

*n*.

## Solve an Absolute Value Equation

Next, we will learn how to slve an**absolute value equation**. To solve an equation such as [latex]|2x - 6|=8[/latex], we notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is [latex]8[/latex] or [latex]-8[/latex]. This leads to two different equations we can solve independently.

[latex]\begin{array}{lll}2x - 6=8\hfill & \text{ or }\hfill & 2x - 6=-8\hfill \\ 2x=14\hfill & \hfill & 2x=-2\hfill \\ x=7\hfill & \hfill & x=-1\hfill \end{array}[/latex]

Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.
### A General Note: Absolute Value Equations

The absolute value of*x*is written as [latex]|x|[/latex]. It has the following properties:

[latex]\begin{array}{l}\text{If } x\ge 0,\text{ then }|x|=x.\hfill \\ \text{If }x<0,\text{ then }|x|=-x.\hfill \end{array}[/latex]

For real numbers [latex]A[/latex] and [latex]B[/latex], an equation of the form [latex]|A|=B[/latex], with [latex]B\ge 0[/latex], will have solutions when [latex]A=B[/latex] or [latex]A=-B[/latex]. If [latex]B<0[/latex], the equation [latex]|A|=B[/latex] has no solution.
An **absolute value equation**in the form [latex]|ax+b|=c[/latex] has the following properties:

[latex]\begin{array}{l}\text{If }c<0,|ax+b|=c\text{ has no solution}.\hfill \\ \text{If }c=0,|ax+b|=c\text{ has one solution}.\hfill \\ \text{If }c>0,|ax+b|=c\text{ has two solutions}.\hfill \end{array}[/latex]

### How To: Given an absolute value equation, solve it.

- Isolate the absolute value expression on one side of the equal sign.
- If [latex]c>0[/latex], write and solve two equations: [latex]ax+b=c[/latex] and [latex]ax+b=-c[/latex].

### Example: Solving Absolute Value Equations

Solve the following absolute value equations:- [latex]|6x+4|=8[/latex]
- [latex]|3x+4|=-9[/latex]
- [latex]|3x - 5|-4=6[/latex]
- [latex]|-5x+10|=0[/latex]

Answer: a. [latex]|6x+4|=8[/latex] Write two equations and solve each:

[latex]\begin{array}{ll}6x+4\hfill&=8\hfill& 6x+4\hfill&=-8\hfill \\ 6x\hfill&=4\hfill& 6x\hfill&=-12\hfill \\ x\hfill&=\frac{2}{3}\hfill& x\hfill&=-2\hfill \end{array}[/latex]

The two solutions are [latex]x=\frac{2}{3}[/latex], [latex]x=-2[/latex]. b. [latex]|3x+4|=-9[/latex] There is no solution as an absolute value cannot be negative. c. [latex]|3x - 5|-4=6[/latex] Isolate the absolute value expression and then write two equations.[latex]\begin{array}{lll}\hfill & |3x - 5|-4=6\hfill & \hfill \\ \hfill & |3x - 5|=10\hfill & \hfill \\ \hfill & \hfill & \hfill \\ 3x - 5=10\hfill & \hfill & 3x - 5=-10\hfill \\ 3x=15\hfill & \hfill & 3x=-5\hfill \\ x=5\hfill & \hfill & x=-\frac{5}{3}\hfill \end{array}[/latex]

There are two solutions: [latex]x=5[/latex], [latex]x=-\frac{5}{3}[/latex].
d. [latex]|-5x+10|=0[/latex]
The equation is set equal to zero, so we have to write only one equation.
[latex]\begin{array}{l}-5x+10\hfill&=0\hfill \\ -5x\hfill&=-10\hfill \\ x\hfill&=2\hfill \end{array}[/latex]

There is one solution: [latex]x=2[/latex].
### Try It

Solve the absolute value equation: [latex]|1 - 4x|+8=13[/latex].Answer: [latex]x=-1[/latex], [latex]x=\frac{3}{2}[/latex]

## Solving Rational Equations Resulting in a Quadratic

Earlier, we solved rational equations. Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.### Example: Solving a Rational Equation Leading to a Quadratic

Solve the following rational equation: [latex]\frac{-4x}{x - 1}+\frac{4}{x+1}=\frac{-8}{{x}^{2}-1}[/latex].Answer: We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, [latex]{x}^{2}-1=\left(x+1\right)\left(x - 1\right)[/latex]. Then, the LCD is [latex]\left(x+1\right)\left(x - 1\right)[/latex]. Next, we multiply the whole equation by the LCD.

[latex]\begin{array}{l}\left(x+1\right)\left(x - 1\right)\left[\frac{-4x}{x - 1}+\frac{4}{x+1}\right]\hfill&=\left[\frac{-8}{\left(x+1\right)\left(x - 1\right)}\right]\left(x+1\right)\left(x - 1\right)\hfill \\ -4x\left(x+1\right)+4\left(x - 1\right)\hfill&=-8\hfill \\ -4{x}^{2}-4x+4x - 4\hfill&=-8\hfill \\ -4{x}^{2}+4\hfill&=0\hfill \\ -4\left({x}^{2}-1\right)\hfill&=0\hfill \\ -4\left(x+1\right)\left(x - 1\right)\hfill&=0\hfill \\ x\hfill&=-1\hfill \\ x\hfill&=1\hfill \end{array}[/latex]

In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.
### Try It

Solve [latex]\frac{3x+2}{x - 2}+\frac{1}{x}=\frac{-2}{{x}^{2}-2x}[/latex].Answer: [latex]x=-1[/latex], [latex]x=0[/latex] is not a solution.

## Licenses & Attributions

### CC licensed content, Original

- Revision and Adaptation.
**Provided by:**Lumen Learning**License:**CC BY: Attribution.

### CC licensed content, Shared previously

- College Algebra.
**Provided by:**OpenStax**Authored by:**Abramson, Jay et al..**License:**CC BY: Attribution.**License terms:**Download for free at http://cnx.org/contents/[email protected]. - Question ID 34186.
**Authored by:**Jim Smart.**License:**CC BY: Attribution.**License terms:**IMathAS Community License CC- BY + GPL. - Question ID 60839.
**Authored by:**Alyson Day.**License:**CC BY: Attribution.**License terms:**IMathAS Community License CC- BY + GPL. - Question ID 1883.
**Authored by:**Barbara Goldner.**License:**CC BY: Attribution.**License terms:**IMathAS Community License CC- BY + GPL. - Question ID 3496.
**Authored by:**Shawn Triplett.**License:**CC BY: Attribution.**License terms:**IMathAS Community License CC- BY + GPL.

### CC licensed content, Specific attribution

- College Algebra.
**Provided by:**OpenStax**Authored by:**OpenStax College Algebra.**Located at:**https://cnx.org/contents/[email protected]:1/Preface.**License:**CC BY: Attribution.