2.6: Other Types of Equations

Learning Objectives

• Solve absolute value equations.
• Solve equations in quadratic form.
• Solve rational equations that result in quadratic equations.

We have solved linear equations, rational equations, and quadratic equations using several methods. However, there are many other types of equations, and we will investigate a few more types in this section. We will look at equations involving absolute value equations, equations in quadratic form, and some rational equations that can be transformed into quadratics. Solving any equation, however, employs the same basic algebraic rules. We will learn some new techniques as they apply to certain equations, but the algebra never changes.

Solving an Absolute Value Equation

First, we will learn how to solve an absolute value equation. To solve an equation such as \(|2x−6|=8\), we 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.

\[\begin{align*} 2x-6&= 8\\ 2x&= 14\\ x&= 7 \end{align*}\]

OR

\[\begin{align*} 2x-6&= -8\\ 2x&= -2\\ x&= -1 \end{align*}\]

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.

ABSOLUTE VALUE EQUATIONS

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

If \(x≥0\), then \(|x|=x\).If \(x<0\), then \(x=−x\).

For real numbers \(A\) and \(B\), an equation of the form \(|A|=B\), with \(B≥0\), 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:

• If \(c<0\),\(|ax+b|=c\) has no solution.
• If \(c=0\),\(|ax+b|=c\) has one solution.
• If \(c>0\),\(|ax+b|=c\) has two solutions.

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 \(\PageIndex{8}\): Solving Absolute Value Equations

Solve the following absolute value equations:

1. \(|6x+4|=8\)
2. \(|3x+4|=−9\)
3. \(|3x−5|−4=6\)
4. \(|−5x+10|=0\)

Solution

1. \(|6x+4|=8\)

Write two equations and solve each:

\[\begin{align*} 6x+4&= 8\\ 6x&= 4\\ x&= \dfrac{2}{3} \end{align*}\]

OR

\[\begin{align*} 6x+4&= -8\\ 6x&= -12\\ x&= -2 \end{align*}\]

The two solutions are \(\dfrac{2}{3}\) and \(−2\).

2. \(|3x+4|=−9\)

There is no solution as an absolute value cannot be negative.

3. \(|3x−5|−4=6\)

Isolate the absolute value expression and then write two equations.

\[\begin{align*} |3x-5|-4&= 6\\ |3x-5|&= 10\\ 3x-5&= 10\\ 3x&= 15\\ x&= 5 \end{align*}\]

OR

\[\begin{align*} 3x-5&= -10\\ 3x=-5\\ x=\dfrac{5}{3} \end{align*}\]

There are two solutions: \(5\), and \(-\dfrac{5}{3}\).

4. \(|−5x+10|=0\)

The equation is set equal to zero, so we have to write only one equation.

\[\begin{align*} -5x+10&= 0\\ -5x&= -10\\ x&= 2 \end{align*}\]

There is one solution: \(2\).

Exercise \(\PageIndex{7}\)

Solve the absolute value equation: \(|1−4x|+8=13\).

Answer

\(x=−1, x=\dfrac{3}{2}\)

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 \(x^4−5x^2+4=0\),\(x^6+7x^3−8=0\), and \(x^{\tfrac{2}{3}} +4x^{\tfrac{1}{3}}+2=0\). 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.

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.

Howto: 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 \(\PageIndex{9}\): Solving a Fourth-degree Equation in Quadratic Form

Solve this fourth-degree equation: \(3x^4−2x^2−1=0\).

Solution

This equation fits the main criteria, that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let \(u =x^2\). Rewrite the equation in \(u\).

\[3u^2−2u−1=0 \nonumber\]

Now solve the quadratic.

\[\begin{align*} 3u^2-2u-1&= 0\\ (3u+1)(u-1)&= 0 \end{align*}\]

Solve each factor and replace the original term for \(u\).

\[\begin{align*} 3u+1&= 0\\ 3u&= -1\\ u&= -\dfrac{1}{3}\\ x^2&= -\dfrac{1}{3}\\ x&= \pm i\sqrt{\dfrac{1}{3}}\\ u-1&= 0\\ u&= 1\\ x^2&= 1\\ x&= \pm 1 \end{align*}\]

The solutions are \(x=±i\sqrt{\dfrac{1}{3}}\) and \(x=±1\)

Exercise \(\PageIndex{8}\)

Solve using substitution: \(x^4−8x^2−9=0\).

Answer

\(x=−3,3,−i,i\)

Example \(\PageIndex{10}\): Solving an Equation in Quadratic Form Containing a Binomial

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

Solution

This equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution, letting \(u =x+2\). Then rewrite the equation in \(u\).

\[\begin{align*} u^2+11u-12&= 0\\ (u+12)(u-1)&= 0 \end{align*}\]

Solve using the zero-factor property and then replace \(u\) with the original expression.

\[\begin{align*} u+12&= 0\\ u&= -12\\ x+2&= -12\\ x&= -14 \end{align*}\]

The second factor results in

\[\begin{align*} u-1&= 0\\ u&= 1\\ x+2&= 1\\ x&= -1 \end{align*}\]

We have two solutions: \(−14\), and \(−1\).

Exercise \(\PageIndex{9}\)

Solve: \({(x−5)}^2−4(x−5)−21=0\).

Answer

\(x=2,x=12\)

Solving Rational Equations Resulting in a Quadratic

In Chapter 1, 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 \(\PageIndex{11}\): Solving a Rational Equation Leading to a Quadratic

Solve the following rational equation: \(\dfrac{-4x}{x-1}+\dfrac{4}{x+1}=\dfrac{-8}{x^2-1}\)

Solution

We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, \(x^2−1=(x+1)(x−1)\). Then, the LCD is \((x+1)(x−1)\). Next, we multiply the whole equation by the LCD.

\[\begin{align*} (x+1)(x-1)\left (\dfrac{-4x}{x-1}+\dfrac{4}{x+1} \right )&= \left (\dfrac{-8}{x^2-1} \right )(x+1)(x-1)\\ -4x(x+1)+4(x-1)&= -8\\ -4x^2-4x+4x-4&= -8\\ -4x^2+4&= 0\\
-4(x^2-1)&= 0\\ -4(x+1)(x-1)&= 0\\ x&= -1\\ x&= 1 \end{align*}\]

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

Exercise \(\PageIndex{10}\)

Solve \(\dfrac{3x+2}{x-2}+\dfrac{1}{x}=\dfrac{-2}{x^2-2x}\)

Answer

\(x=−1, x= 0\) is not a solution.

Key Concepts

• To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value. See Example.
• Equations in quadratic form are easy to spot, as the exponent on the first term is double the exponent on the second term and the third term is a constant. We may also see a binomial in place of the single variable. We use substitution to solve. See Example and Example.
• Solving a rational equation may also lead to a quadratic equation or an equation in quadratic form. See Example.