## Solving Other Types of Equations

### Learning Outcomes

• Solve polynomial equations.
• Solve absolute value equations.

We have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring.

### A General Note: Polynomial Equations

A polynomial of degree n is an expression of the type

${a}_{n}{x}^{n}+{a}_{n - 1}{x}^{n - 1}+\cdot \cdot \cdot +{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$

where n is a positive integer and ${a}_{n},\dots ,{a}_{0}$ are real numbers and ${a}_{n}\ne 0$.

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.

## Solving an Absolute Value Equation

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

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

Knowing how to solve problems involving absolute value is useful. For example, we may need to identify numbers or points on a line that are a specified distance from a given reference point.

### A General Note: Absolute Value Equations

The absolute value of x is written as $|x|$. It has the following properties:

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

For real numbers $A$ and $B$, an equation of the form $|A|=B$, with $B\ge 0$, will have solutions when $A=B$ or $A=-B$. If $B<0$, the equation $|A|=B$ has no solution.

An absolute value equation in the form $|ax+b|=c$ has the following properties:

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

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

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

### Example: Solving Absolute Value Equations

Solve the following absolute value equations:

1. $|6x+4|=8$
2. $|3x+4|=-9$
3. $|3x - 5|-4=6$
4. $|-5x+10|=0$

### Try It

Solve the absolute value equation: $|1 - 4x|+8=13$.

## Other Types of Equations

There are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form and rational equations that result in a quadratic.

## Solving Equations in Quadratic Form

Equations in quadratic form are equations with three terms. The first term has a power other than 2. The middle term has an exponent that is one-half the exponent of the leading term. The third term is a constant. We can solve equations in this form as if they were quadratic. A few examples of these equations include ${x}^{4}-5{x}^{2}+4=0,{x}^{6}+7{x}^{3}-8=0$, and ${x}^{\frac{2}{3}}+4{x}^{\frac{1}{3}}+2=0$. In each one, doubling the exponent of the middle term equals the exponent on the leading term. We can solve these equations by substituting a variable for the middle term.

### A General Note: Quadratic Form

If the exponent on the middle term is one-half of the exponent on the leading term, we have an equation in quadratic form which we can solve as if it were a quadratic. We substitute a variable for the middle term to solve equations in quadratic form.

### How To: Given an equation quadratic in form, solve it

1. Identify the exponent on the leading term and determine whether it is double the exponent on the middle term.
2. If it is, substitute a variable, such as u, for the variable portion of the middle term.
3. Rewrite the equation so that it takes on the standard form of a quadratic.
4. Solve using one of the usual methods for solving a quadratic.
5. Replace the substitution variable with the original term.
6. Solve the remaining equation.

### Example: Solving a Fourth-Degree Equation in Quadratic Form

Solve this fourth-degree equation: $3{x}^{4}-13{x}^{2}+4=0$.

### Try It

Solve using substitution: ${x}^{4}-6{x}^{2}+5=0$.

### Example: Solving an Equation in Quadratic Form Containing a Binomial

Solve the equation in quadratic form: ${\left(x+2\right)}^{2}+11\left(x+2\right)-12=0$.

### Try It

Solve: ${\left(x - 5\right)}^{2}-4\left(x - 5\right)-21=0$.

## 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.

Solve the following rational equation: $\frac{-4x}{x - 1}+\frac{4}{x+1}=\frac{-8}{{x}^{2}-1}$.
Solve $\frac{3x+2}{x - 2}+\frac{1}{x}=\frac{-2}{{x}^{2}-2x}$.