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2.5 Other Types of Equations

Cynthia Singleton; Ginny Bradley; Karen Perilloux; and Prakash Ghimire

Learning Objectives

In this section, you will:

  • Solve equations involving rational exponents.
  • Solve equations using factoring.
  • Solve radical equations.
  • Solve absolute value equations.
  • Solve other types of 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 rational exponents, polynomial equations, radical equations, 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 Equations Involving Rational Exponents

Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, [latex]{16}^{\frac{1}{2}}[/latex] is another way of writing [latex]\sqrt{16}[/latex], and [latex]{8}^{\frac{1}{3}}[/latex] is another way of writing [latex]\sqrt[3]{8}.[/latex]  The ability to work with rational exponents is a useful skill, as it is highly applicable in calculus.

We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals [latex]1[/latex]. For example, [latex]\frac{2}{3}\left(\frac{3}{2}\right)=1[/latex], and [latex]3 \cdot \frac{1}{3}=1[/latex], and so on.

Rational Exponents

A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:

[latex]{a}^{\frac{m}{n}}={\left({a}^{\frac{1}{n}}\right)}^{m}={\left({a}^{m}\right)}^{\frac{1}{n}}=\sqrt[n]{{a}^{m}}={\left(\sqrt[n]{a}\right)}^{m}[/latex]

Example

Evaluate [latex]{8}^{\frac{2}{3}}.[/latex]

Show Solution

Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite [latex]{8}^{\frac{2}{3}}[/latex] as [latex]{\left({8}^{\frac{1}{3}}\right)}^{2}.[/latex]

[latex]\begin{array}{ccc}\hfill {\left({8}^{\frac{1}{3}}\right)}^{2}& =\hfill & {\left(2\right)}^{2}\hfill \\ & =& 4\hfill \end{array}[/latex]

Try It

Evaluate [latex]{64}^{-\frac{1}{3}}.[/latex]

Show Solution

[latex]\frac{1}{4}[/latex]

Example

Solve the equation in which a variable is raised to a rational exponent: [latex]{x}^{\frac{5}{4}}=32.[/latex]

Show Solution

The way to remove the exponent on [latex]x[/latex] is by raising both sides of the equation to a power that is the reciprocal of [latex]\frac{5}{4}[/latex] which is [latex]\frac{4}{5}.[/latex]

[latex]\begin{array}{cccc}\hfill {x}^{\frac{5}{4}}& =& 32\hfill & \\ \hfill {\left({x}^{\frac{5}{4}}\right)}^{\frac{4}{5}}& =& {\left(32\right)}^{\frac{4}{5}}\hfill & \\ \hfill x& =& {\left(2\right)}^{4}\hfill & \phantom{\rule{2em}{0ex}}\text{The fifth root of 32 is 2.}\hfill \\ & =& 16\hfill & \end{array}[/latex]

Try It

Solve the equation [latex]x^{\frac{3}{2}}=125.[/latex]

Show Solution

[latex]25[/latex]

Example

Solve [latex]3{x}^{\frac{3}{4}}={x}^{\frac{1}{2}}.[/latex]

Show Solution

This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero.

[latex]\begin{array}{ccc}\hfill 3{x}^{\frac{3}{4}}-\left({x}^{\frac{1}{2}}\right)& =& {x}^{\frac{1}{2}}-\left({x}^{\frac{1}{2}}\right)\hfill \\ \hfill 3{x}^{\frac{3}{4}}-{x}^{\frac{1}{2}}& =& 0\hfill \end{array}[/latex]

Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the lowest exponent. Rewrite [latex]{x}^{\frac{1}{2}}[/latex] as [latex]{x}^{\frac{2}{4}}.[/latex]  Then, factor out [latex]{x}^{\frac{2}{4}}[/latex] from both terms on the left.

[latex]\begin{array}{ccc}\hfill 3{x}^{\frac{3}{4}}-{x}^{\frac{2}{4}}& =& 0\hfill \\ \hfill {x}^{\frac{2}{4}}\left(3{x}^{\frac{1}{4}}-1\right)& =& 0\hfill \end{array}[/latex]

Where did [latex]{x}^{\frac{1}{4}}[/latex] come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply [latex]{x}^{\frac{2}{4}}[/latex] back in using the distributive property, we get the expression we had before the factoring, which is what should happen. We need an exponent such that when added to [latex]\frac{2}{4}[/latex] equals [latex]\frac{3}{4}.[/latex]  Thus, the exponent on [latex]x[/latex] in the parentheses is [latex]\frac{1}{4}.[/latex]

Let us continue. Now we have two factors and can use the zero factor theorem.

[latex]\begin{array}{ccc}\hfill {x}^{\frac{2}{4}}\left(3{x}^{\frac{1}{4}}-1\right)& = 0\hfill & \\ \hfill {x}^{\frac{2}{4}}& = 0\hfill & \\ \hfill x& = 0\hfill & \hfill \\ 3{x}^{\frac{1}{4}}-1& = 0\hfill & \\ \hfill 3{x}^{\frac{1}{4}}& = 1\hfill & \hfill \\ {x}^{\frac{1}{4}}& = \frac{1}{3}\hfill & \phantom{\rule{2em}{0ex}}\text{Divide both sides by 3}.\hfill \\ \hfill {\left({x}^{\frac{1}{4}}\right)}^{4}& = {\left(\frac{1}{3}\right)}^{4}\hfill & \phantom{\rule{2em}{0ex}}\text{Raise both sides to the reciprocal of }\frac{1}{4}.\hfill \\ \hfill x& = \frac{1}{81}\hfill & \end{array}[/latex]

The two solutions are [latex]0[/latex] and [latex]\frac{1}{81}[/latex].

Try It

Solve: [latex]\left(x+5\right)^{\frac{3}{2}}=8.[/latex]

Show Solution

[latex]-1[/latex]

https://youtube.com/watch?v=xs-D0MixxLM%26%2391%3B

Example

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

Show 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 [latex]u=x+2.[/latex]  Then rewrite the equation in [latex]u[/latex].

[latex]\begin{array}{ccc}\hfill {u}^{2}+11u-12& =& 0\hfill \\ \hfill \left(u+12\right)\left(u-1\right)& =& 0\hfill \end{array}[/latex]

Solve using the zero-factor property and then replace [latex]u[/latex] with the original expression.

[latex]\begin{array}{ccc}\hfill u+12& =& 0\hfill \\ \hfill u& =& -12\hfill \\ \hfill x+2& =& -12\hfill \\ \hfill x& =& -14\hfill \end{array}[/latex]

The second factor results in

[latex]\begin{array}{ccc}\hfill u-1& =& 0\hfill \\ \hfill u& =& 1\hfill \\ \hfill x+2& =& 1\hfill \\ \hfill x& =& -1\hfill \end{array}[/latex]

We have two solutions: [latex]-14[/latex] and [latex]-1[/latex].

Try It

Solve: [latex]{\left(x-5\right)}^{2}-4\left(x-5\right)-21=0.[/latex]

Show Solution

[latex]x=2, x=12[/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

Solve the following rational equation: [latex]\frac{-4x}{x-1}+\frac{4}{x+1}=\frac{-8}{{x}^{2}-1}.[/latex]

Show Solution

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}{ccc}\hfill \left(x+1\right)\left(x-1\right)\left[\frac{-4x}{x-1}+\frac{4}{x+1}\right]& =& \left[\frac{-8}{\left(x+1\right)\left(x-1\right)}\right]\left(x+1\right)\left(x-1\right)\hfill \\ \hfill -4x\left(x+1\right)+4\left(x-1\right)& =& -8\hfill \\ \hfill -4{x}^{2}-4x+4x-4& =& -8\hfill \\ \hfill -4{x}^{2}+4& =& 0\hfill \\ \hfill -4\left({x}^{2}-1\right)& =& 0\hfill \\ \hfill -4\left(x+1\right)\left(x-1\right)& =& 0\hfill \\ \hfill x& =& -1, 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]

Show Solution

[latex]x=-1[/latex] is a solution, but [latex]x=0[/latex] is not a solution.

Key Concepts

  • Rational exponents can be rewritten several ways depending on what is most convenient for the problem. To solve, both sides of the equation are raised to a power that will render the exponent on the variable equal to 1.
  • Factoring extends to higher-order polynomials when it involves factoring out the GCF or factoring by grouping.
  • We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index.
  • To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value.
  • 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.
  • Solving a rational equation may also lead to a quadratic equation or an equation in quadratic form.

Section Exercises

Verbal

  1. In a radical equation, what does it mean if a number is an extraneous solution?
Show Solution

This is not a solution to the radical equation; it is a value obtained from squaring both sides and thus changing the signs of an equation, which has caused it not to be a solution in the original equation.

  1. Explain why possible solutions must be checked in radical equations.
  1. Your friend tries to calculate the value [latex]-{9}^{\frac{3}{2}}[/latex] and keeps getting an ERROR message. What mistake is he or she probably making?
Show Solution

He or she is probably trying to enter negative 9, but taking the square root of [latex]-9[/latex] is not a real number. The negative sign is in front of this, so your friend should be taking the square root of 9, cubing it, and then putting the negative sign in front, resulting in [latex]-27.[/latex]

  1. Explain why [latex]|2x+5|=-7[/latex] has no solutions.
  2. Explain how to change a rational exponent into the correct radical expression.
Show Solution

A rational exponent is a fraction: the denominator of the fraction is the root or index number, and the numerator is the power to which it is raised.

Algebraic

For the following exercises, solve the rational exponent equation. Use factoring where necessary.

  1. [latex]{x}^{\frac{2}{3}}=16[/latex]
  1. [latex]{x}^{\frac{3}{4}}=27[/latex]
Show Solution

[latex]x=81[/latex]

  1. [latex]2{x}^{\frac{1}{2}}-{x}^{\frac{1}{4}}=0[/latex]
  1. [latex]{\left(x-1\right)}^{\frac{3}{4}}=8[/latex]
Show Solution

[latex]x=17[/latex]

  1. [latex]{\left(x+1\right)}^{\frac{2}{3}}=4[/latex]
  1. [latex]{x}^{\frac{2}{3}}-5{x}^{\frac{1}{3}}+6=0[/latex]
Show Solution

[latex]x=8, x=27[/latex]

  1. [latex]{x}^{\frac{7}{3}}-3{x}^{\frac{4}{3}}-4{x}^{\frac{1}{3}}=0[/latex]

For the following exercises, solve the following polynomial equations by grouping and factoring.

  1. [latex]{x}^{3}+2{x}^{2}-x-2=0[/latex]
Show Solution

[latex]x=-2,1,-1[/latex]

  1. [latex]3{x}^{3}-6{x}^{2}-27x+54=0[/latex]
  2. [latex]4{y}^{3}-9y=0[/latex]
Show Solution

[latex]y=0, \frac{3}{2}, \frac{-3}{2}[/latex]

  1. [latex]{x}^{3}+3{x}^{2}-25x-75=0[/latex]
  1. [latex]{m}^{3}+{m}^{2}-m-1=0[/latex]
Show Solution

[latex]m=1,-1[/latex]

  1. [latex]2{x}^{5}-14{x}^{3}=0[/latex]
  1. [latex]5{x}^{3}+45x=2{x}^{2}+18[/latex]
Show Solution

[latex]x=\frac{2}{5},±3i[/latex]

For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions.

  1. [latex]\sqrt{3x-1}-2=0[/latex]
  1. [latex]\sqrt{x-7}=5[/latex]
Show Solution

[latex]x=32[/latex]

  1. [latex]\sqrt{x-1}=x-7[/latex]
  1. [latex]\sqrt{3t+5}=7[/latex]
Show Solution

[latex]t=\frac{44}{3}[/latex]

  1. [latex]\sqrt{t+1}+9=7[/latex]
  1. [latex]\sqrt{12-x}=x[/latex]
Show Solution

[latex]x=3[/latex]

  1. [latex]\sqrt{2x+3}-\sqrt{x+2}=2[/latex]
  1. [latex]\sqrt{3x+7}+\sqrt{x+2}=1[/latex]
Show Solution

[latex]x=-2[/latex]

  1. [latex]\sqrt{2x+3}-\sqrt{x+1}=1[/latex]

For the following exercises, solve the equation involving absolute value.

  1. [latex]|3x-4|=8[/latex]
Show Solution

[latex]x=4,\frac{-4}{3}[/latex]

  1. [latex]|2x-3|=-2[/latex]
  1. [latex]|1-4x|-1=5[/latex]
Show Solution

[latex]x=\frac{-5}{4},\frac{7}{4}[/latex]

  1. [latex]|4x+1|-3=6[/latex]
  1. [latex]|2x-1|-7=-2[/latex]
Show Solution

[latex]x=3,-2[/latex]

  1. [latex]|2x+1|-2=-3[/latex]
  1. [latex]|x+5|=0[/latex]
Show Solution

[latex]x=-5[/latex]

  1. [latex]-|2x+1|=-3[/latex]

For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring.

  1. [latex]{x}^{4}-10{x}^{2}+9=0[/latex]
Show Solution

[latex]x=1,-1,3,-3[/latex]

  1. [latex]4{\left(t-1\right)}^{2}-9\left(t-1\right)=-2[/latex]
  1. [latex]{\left({x}^{2}-1\right)}^{2}+\left({x}^{2}-1\right)-12=0[/latex]
Show Solution

[latex]x=2,-2[/latex]

  1. [latex]{\left(x+1\right)}^{2}-8\left(x+1\right)-9=0[/latex]
  1. [latex]{\left(x-3\right)}^{2}-4=0[/latex]
Show Solution

[latex]x=1,5[/latex]

Extensions

For the following exercises, solve for the unknown variable.

  1. [latex]{x}^{-2}-{x}^{-1}-12=0[/latex]
  1. [latex]\sqrt{{|x|}^{2}}=x[/latex]
Show Solution

All real numbers

  1. [latex]{t}^{10}-{t}^{5}+1=0[/latex]
  1. [latex]|{x}^{2}+2x-36|=12[/latex]
Show Solution

[latex]x=4,6,-6,-8[/latex]

Real-World Applications

For the following exercises, use the model for the period of a pendulum, [latex]T[/latex], such that [latex]T=2\pi \sqrt{\frac{L}{g}}[/latex] where the length of the pendulum is [latex]L[/latex] and the acceleration due to gravity is[latex]g.[/latex]

  1. If the acceleration due to gravity is 9.8 m/s2 and the period equals 1 s, find the length to the nearest cm (100 cm = 1 m).
  1. If the gravity is 32 ft/s2 and the period equals 1 s, find the length to the nearest in. (12 in. = 1 ft). Round your answer to the nearest in.
Show Solution

10 in.

For the following exercises, use a model for body surface area, BSA, such that [latex]\text{BSA}=\sqrt{\frac{wh}{3600}}[/latex] where [latex]w[/latex] = weight in kg and [latex]h[/latex] = height in cm.

  1. Find the height of a 72-kg female to the nearest cm whose [latex]\text{BSA}=1.8.[/latex]
  1. Find the weight of a 177-cm male to the nearest kg whose [latex]\text{BSA}=2.1.[/latex]
Show Solution

90 kg

Glossary

absolute value equation
an equation in which the variable appears in absolute value bars, typically with two solutions, one accounting for the positive expression and one for the negative expression
equations in quadratic form
equations with a power other than 2 but with a middle term with an exponent that is one-half the exponent of the leading term
extraneous solutions
any solutions obtained that are not valid in the original equation
polynomial equation
an equation containing a string of terms including numerical coefficients and variables raised to whole-number exponents
radical equation
an equation containing at least one radical term where the variable is part of the radicand

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2.5 Other Types of Equations Copyright © by Cynthia Singleton; Ginny Bradley; Karen Perilloux; and Prakash Ghimire is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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