how to solve quadratic equations with fraction exponents

Fractional Exponents

If an exponent of a number is a fraction, it is called a fractional exponent . Exponents show the number of times a number is replicated in multiplication. For example, 4 2 = 4×4 = 16. Here, exponent 2 is a whole number. In the number, say x 1/y , x is the base and 1/y is the fractional exponent.

In this article, we will discuss the concept of fractional exponents, and their rules, and learn how to solve them. We shall also explore negative fractional exponents and solve various examples for a better understanding of the concept. 

What are Fractional Exponents?

Fractional exponents are ways to represent powers and roots together. In any general exponential expression of the form a b , a is the base and b is the exponent . When b is given in the fractional form, it is known as a fractional exponent. A few examples of fractional exponents are 2 1/2 , 3 2/3 , etc. The general form of a fractional exponent is x m/n , where x is the base and m/n is the exponent.

Look at the figure given below to understand how fractional exponents are represented.

Some examples of fractional exponents that are widely used are given below:

Fractional Exponents Rules

There are certain rules to be followed that help us to multiply or divide numbers with fractional exponents easily. Many people are familiar with whole-number exponents, but when it comes to fractional exponents, they end up doing mistakes that can be avoided if we follow these rules of fractional exponents .

• Rule 1: a 1/m × a 1/n = a (1/m + 1/n)
• Rule 2: a 1/m ÷ a 1/n = a (1/m - 1/n)
• Rule 3: a 1/m × b 1/m = (ab) 1/m
• Rule 4: a 1/m ÷ b 1/m = (a÷b) 1/m
• Rule 5: a -m/n = (1/a) m/n

These rules are very helpful while simplifying fractional exponents. Let us now learn how to simplify fractional exponents.

Simplifying Fractional Exponents

Simplifying fractional exponents can be understood in two ways which are multiplication and division. It involves reducing the expression or the exponent to a reduced form that is easy to understand. For example, 9 1/2 can be reduced to 3. Let us understand the simplification of fractional exponents with the help of some examples.

1) Solve 3 √8 = 8 1/3

We know that 8 can be expressed as a cube of 2 which is given as, 8 = 2 3 . Substituting the value of 8 in the given example we get, (2 3 ) 1/3 = 2 since the product of the exponents gives 3×1/3=1. ∴ 3 √8=8 1/3 =2.

2) Simplify (64/125) 2/3

In this example, both the base and the exponent are in fractional form. 64 can be expressed as a cube of 4 and 125 can be expressed as a cube of 5. They are given as, 64=4 3 and 125=5 3 . Substituting their values in the given example we get, (4 3 /5 3 ) 2/3 . 3 is a common power for both the numbers, hence (4 3 /5 3 ) 2/3 can be written as ((4/5) 3 ) 2/3 , which is equal to (4/5) 2 as 3×2/3=2. Now, we have (4/5) 2 , which is equal to 16/25. Therefore, (64/125) 2/3 = 16/25.

Multiply Fractional Exponents With the Same Base

To multiply fractional exponents with the same base, we have to add the exponents and write the sum on the common base. The general rule for multiplying exponents with the same base is a 1/m × a 1/n = a (1/m + 1/n) . For example, to multiply 2 2/3 and 2 3/4 , we have to add the exponents first. So, 2/3 + 3/4 = 17/12. Therefore, 2 2/3 × 2 3/4 = 2 17/12 .

How to Divide Fractional Exponents?

The division of fractional exponents can be classified into two types.

• Division of fractional exponents with different powers but the same bases
• Division of fractional exponents with the same powers but different bases

When we divide fractional exponents with different powers but the same bases, we express it as a 1/m ÷ a 1/n = a (1/m - 1/n) . Here, we have to subtract the powers and write the difference on the common base. For example, 5 3/4 ÷ 5 1/2 = 5 (3/4-1/2) , which is equal to 5 1/4 .

When we divide fractional exponents with the same powers but different bases, we express it as a 1/m ÷ b 1/m = (a÷b) 1/m . Here, we are dividing the bases in the given sequence and writing the common power on it. For example, 9 5/6 ÷ 3 5/6 = (9/3) 5/6 , which is equal to 3 5/6 .

Negative Fractional Exponents

Negative fractional exponents are the same as rational exponents. In this case, along with a fractional exponent, there is a negative sign attached to the power. For example, 2 -1/2 . To solve negative exponents , we have to apply exponents rules that say a -m = 1/a m . It means before simplifying an expression further, the first step is to take the reciprocal of the base to the given power without the negative sign. The general rule for negative fractional exponents is a -m/n = (1/a) m/n .

For example, let us simplify 343 -1/3 . Here the base is 343 and the power is -1/3. The first step is to take the reciprocal of the base, which is 1/343, and remove the negative sign from the power. Now, we have (1/343) 1/3 . As we know that 343 is the third power of 7 as 7 3 = 343, we can re-write the expression as 1/(7 3 ) 1/3 . Since 3 and 1/3 cancel each other, the final answer is 1/7.

Related Articles

• Non-Integer Rational Exponents
• Irrational Exponents
• Exponential Terms
• Negative Exponents

Fractional Exponents Examples

Example 1: Evaluate 18 1/2 ÷ 2 1/2 .

Solution: In this question, fractional exponents are given. The powers are the same but the bases are different. Hence, we can solve this problem as, 18 1/2 ÷ 2 1/2 = (18/2) 1/2 = 9 1/2 = 3. Therefore, 3 is the required answer.

Example 2: Solve the given expression involving the multiplication of terms with fractional exponents.

2 1/2 × 4 1/4 × 8 1/8

Solution: 4 can be expressed as a square of 2, i.e. 4 = 2 2 . So, 4 1/4 can be written as (2 2 ) 1/4 . It is equal to 2 1/2 . Now, 8 can be expressed as a cube of 2, i.e. 8 = 2 3 . So, 8 1/8 can be written as (2 3 ) 1/8 . It is equal to 2 3/8 . Therefore, the given expression can be re-written as,

2 1/2 × 2 1/2 × 2 3/8

Multiplication of fractional exponents with the same base is done by adding the powers and writing the sum on the common base.

⇒ 2 (1/2 + 1/2 + 3/8)

Therefore, 2 1/2 × 4 1/4 × 8 1/8 = 2 11/8 .

Example 3: Evaluate 3 2/3 ÷ 9 1/2

Solution: To solve this, we will reduce 9 1/2 to the simplest form. So, we have

9 1/2 = (3 2 ) 1/2

3 2/3 ÷ 9 1/2 = 3 2/3 ÷ 3 1

= 3 2/3 - 1

go to slide go to slide go to slide

Book a Free Trial Class

Fractional Exponents Questions

go to slide go to slide

FAQs on Fractional Exponents

What do fractional exponents mean.

Fractional exponents mean the power of a number is in terms of fraction rather than an integer . For example, in a m/n the base is 'a' and the power is m/n which is a fraction .

What is the Rule for Fractional Exponents?

In the case of fractional exponents, the numerator is the power and the denominator is the root. This is the general rule of fractional exponents. We can write x m/n as n √(x m ).

What To Do With Negative Fractional Exponents?

If the exponent is given in negative, it means we have to take the reciprocal of the base and remove the negative sign from the power. For example, 2 -1/2 = (1/2) 1/2 .

How To Solve Fractional Exponents?

To solve fractional exponents, we use the laws of exponents or the exponent rules . The fractional exponents' rules are stated below:

How To Add Fractional Exponents?

There is no rule for the addition of fractional exponents. We can add them only by simplifying the powers, if possible. For example, 9 1/2 + 125 1/3 = 3 + 5 = 8.

How To Divide Fractional Exponents?

Division of fractional exponents with the same base and different powers is done by subtracting the powers, and the division with different bases and same powers is done by dividing the bases first and writing the common power on the answer.

Simplify Fraction Exponents

Formula and examples of how to simplify Fraction exponents

Examples of Rewriting Fractional Exponents

So how does this work.

We can use one of the laws of exponents to explain how fractional exponents work.

$ \\ 9^{\frac 1 2 } \cdot 9^{\frac 1 2 } = 9^{\frac 1 2 + \frac 1 2 } \\ = \boxed{ 9 ^1 } $

We can do the same thing with $$ \sqrt[3] 8 \cdot \sqrt[3] 8 \cdot \sqrt[3] 8 = 8 $$

$ \\ 8^{\frac 1 3} \cdot 8^{\frac 1 3 } \cdot 8^{\frac 1 3 } = 8^{\frac 1 3 + \frac 1 3+ \frac 1 3 } \\ = \boxed{ 8 ^1 } $

General Formula

With fractional exponents whose numerator is 1.

Below is the general formula for a fractional exponent with a numerator of 1 .

$ \sqrt[n] x = x ^ {\frac 1 n} $

$$ \frac 1 n $$ is another way of asking: What number can you multiply by itself n times to get x?

When the numerator is not 1

Below is a specific example illustrating the formula for fraction exponents when the numerator is not one. There are two ways to simplify a fraction exponent such $$ \frac 2 3$$ . You can either apply the numerator first or the denominator. See the example below.

Practice Problems

Simplify $$ 125^{\frac 1 3 }$$

Simplify $$ 125^{\frac 2 3 }$$

Simplify $$ 64^{\frac 2 3 }$$

Simplify $$ 81^{\frac 3 4 }$$

• Formula for Fraction Exponents
• Formula Fraction Exponent: Numerator Not One

Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!

Popular pages @ mathwarehouse.com.

• PRO Courses Guides New Tech Help Pro Expert Videos About wikiHow Pro Upgrade Sign In
• EDIT Edit this Article
• EXPLORE Tech Help Pro About Us Random Article Quizzes Request a New Article Community Dashboard This Or That Game Popular Categories Arts and Entertainment Artwork Books Movies Computers and Electronics Computers Phone Skills Technology Hacks Health Men's Health Mental Health Women's Health Relationships Dating Love Relationship Issues Hobbies and Crafts Crafts Drawing Games Education & Communication Communication Skills Personal Development Studying Personal Care and Style Fashion Hair Care Personal Hygiene Youth Personal Care School Stuff Dating All Categories Arts and Entertainment Finance and Business Home and Garden Relationship Quizzes Cars & Other Vehicles Food and Entertaining Personal Care and Style Sports and Fitness Computers and Electronics Health Pets and Animals Travel Education & Communication Hobbies and Crafts Philosophy and Religion Work World Family Life Holidays and Traditions Relationships Youth
• Browse Articles
• Learn Something New
• Quizzes Hot
• This Or That Game
• Train Your Brain
• Explore More
• Support wikiHow
• About wikiHow
• Log in / Sign up
• Education and Communications
• Mathematics

How to Solve a Quadratic Equation: A Step-by-Step Guide

Last Updated: May 3, 2024 Fact Checked

Factoring the Equation

Using the quadratic formula, completing the square, practice problems and answers, expert q&a.

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 9 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,414,569 times.

A quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2. [1] X Research source There are three main ways to solve quadratic equations: 1) to factor the quadratic equation if you can do so, 2) to use the quadratic formula, or 3) to complete the square. If you want to know how to master these three methods, just follow these steps.

Quadradic Formula for Solving Equations

• Then, use the process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4. [3] X Research source

• 3x = -1 ..... by subtracting
• 3x/3 = -1/3 ..... by dividing
• x = -1/3 ..... simplified
• x = 4 ..... by subtracting
• x = (-1/3, 4) ..... by making a set of possible, separate solutions, meaning x = -1/3, or x = 4 seem good.

• So, both solutions do "check" separately, and both are verified as working and correct for two different solutions.

• 4x 2 - 5x - 13 = x 2 -5
• 4x 2 - x 2 - 5x - 13 +5 = 0
• 3x 2 - 5x - 8 = 0

• {-b +/-√ (b 2 - 4ac)}/2
• {-(-5) +/-√ ((-5) 2 - 4(3)(-8))}/2(3) =
• {-(-5) +/-√ ((-5) 2 - (-96))}/2(3)

• {-(-5) +/-√ ((-5) 2 - (-96))}/2(3) =
• {5 +/-√(25 + 96)}/6
• {5 +/-√(121)}/6

• (5 + 11)/6 = 16/6
• (5-11)/6 = -6/6

• x = (-1, 8/3)

• 2x 2 - 9 = 12x =
• In this equation, the a term is 2, the b term is -12, and the c term is -9.

• 2x 2 - 12x - 9 = 0
• 2x 2 - 12x = 9

• 2x 2 /2 - 12x/2 = 9/2 =
• x 2 - 6x = 9/2

• -6/2 = -3 =
• (-3) 2 = 9 =
• x 2 - 6x + 9 = 9/2 + 9

• x = 3 + 3(√6)/2
• x = 3 - 3(√6)/2)

• If the number under the square root is not a perfect square, then the last few steps run a little differently. Here is an example: [14] X Research source Thanks Helpful 0 Not Helpful 0
• If the "b" is an even number, the formula is : {-(b/2) +/- √(b/2)-ac}/a. Thanks Helpful 3 Not Helpful 0
• As you can see, the radical sign did not disappear completely. Therefore, the terms in the numerator cannot be combined (because they are not like terms). There is no purpose, then, to splitting up the plus-or-minus. Instead, we divide out any common factors --- but ONLY if the factor is common to both of the constants AND the radical's coefficient. Thanks Helpful 1 Not Helpful 0

You Might Also Like

• ↑ https://www.mathsisfun.com/definitions/quadratic-equation.html
• ↑ http://www.mathsisfun.com/algebra/factoring-quadratics.html
• ↑ https://www.mathportal.org/algebra/solving-system-of-linear-equations/elimination-method.php
• ↑ https://www.cuemath.com/algebra/quadratic-equations/
• ↑ https://www.purplemath.com/modules/solvquad4.htm
• ↑ http://www.purplemath.com/modules/quadform.htm
• ↑ https://uniskills.library.curtin.edu.au/numeracy/algebra/quadratic-equations/
• ↑ http://www.mathsisfun.com/algebra/completing-square.html
• ↑ http://www.umsl.edu/~defreeseca/intalg/ch7extra/quadmeth.htm

About This Article

To solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. Then, factor the expression, and set each set of parentheses equal to 0 as separate equations. Finally, solve each equation separately to find the 2 possible values for x. To learn how to solve quadratic equations using the quadratic formula, scroll down! Did this summary help you? Yes No

• Send fan mail to authors

Reader Success Stories

Sep 24, 2022

Did this article help you?

Kalo Morris

Mar 12, 2017

Matthew Mathers

Mar 25, 2017

Kapela Davis

Oct 10, 2017

Jan 29, 2018

Featured Articles

Trending Articles

Solver Title

Generating PDF...

• Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Number Line Expanded Form Mean, Median & Mode
• Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi (Product) Notation Induction Logical Sets Word Problems
• Pre Calculus Equations Inequalities Scientific Calculator Scientific Notation Arithmetics Complex Numbers Polar/Cartesian Simultaneous Equations System of Inequalities Polynomials Rationales Functions Arithmetic & Comp. Coordinate Geometry Plane Geometry Solid Geometry Conic Sections Trigonometry
• Calculus Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Fourier Transform
• Functions Line Equations Functions Arithmetic & Comp. Conic Sections Transformation
• Linear Algebra Matrices Vectors
• Trigonometry Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify
• Statistics Mean Geometric Mean Quadratic Mean Average Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge Standard Normal Distribution
• Physics Mechanics
• Chemistry Chemical Reactions Chemical Properties
• Finance Simple Interest Compound Interest Present Value Future Value
• Economics Point of Diminishing Return
• Conversions Roman Numerals Radical to Exponent Exponent to Radical To Fraction To Decimal To Mixed Number To Improper Fraction Radians to Degrees Degrees to Radians Hexadecimal Scientific Notation Distance Weight Time Volume
• Pre Algebra
• One-Step Addition
• One-Step Subtraction
• One-Step Multiplication
• One-Step Division
• One-Step Decimals
• Two-Step Integers
• Two-Step Add/Subtract
• Two-Step Multiply/Divide
• Two-Step Fractions
• Two-Step Decimals
• Multi-Step Integers
• Multi-Step with Parentheses
• Multi-Step Rational
• Multi-Step Fractions
• Multi-Step Decimals
• Solve by Factoring
• Completing the Square
• Quadratic Formula
• Biquadratic
• Logarithmic
• Exponential
• Rational Roots
• Floor/Ceiling
• Equation Given Roots
• Newton Raphson
• Substitution
• Elimination
• Cramer's Rule
• Gaussian Elimination
• System of Inequalities
• Perfect Squares
• Difference of Squares
• Difference of Cubes
• Sum of Cubes
• Polynomials
• Distributive Property
• FOIL method
• Perfect Cubes
• Binomial Expansion
• Negative Rule
• Product Rule
• Quotient Rule
• Expand Power Rule
• Fraction Exponent
• Exponent Rules
• Exponential Form
• Logarithmic Form
• Absolute Value
• Rational Number
• Powers of i
• Complex Form
• Partial Fractions
• Is Polynomial
• Leading Coefficient
• Leading Term
• Standard Form
• Complete the Square
• Synthetic Division
• Linear Factors
• Rationalize Denominator
• Rationalize Numerator
• Identify Type
• Convergence
• Interval Notation
• Pi (Product) Notation
• Boolean Algebra
• Truth Table
• Mutual Exclusive
• Cardinality
• Caretesian Product
• Age Problems
• Distance Problems
• Cost Problems
• Investment Problems
• Number Problems
• Percent Problems
• Addition/Subtraction
• Multiplication/Division
• Dice Problems
• Coin Problems
• Card Problems
• Pre Calculus
• Linear Algebra
• Trigonometry
• Conversions

Most Used Actions

Number line.

• ax^2+bx+c=0
• x^2+2x+1=3x-10
• 2x^2+4x-6=0
• How do you calculate a quadratic equation?
• To solve a quadratic equation, use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
• What is the quadratic formula?
• The quadratic formula gives solutions to the quadratic equation ax^2+bx+c=0 and is written in the form of x = (-b ± √(b^2 - 4ac)) / (2a)
• Does any quadratic equation have two solutions?
• There can be 0, 1 or 2 solutions to a quadratic equation. If the discriminant is positive there are two solutions, if negative there is no solution, if equlas 0 there is 1 solution.
• What is quadratic equation in math?
• In math, a quadratic equation is a second-order polynomial equation in a single variable. It is written in the form: ax^2 + bx + c = 0 where x is the variable, and a, b, and c are constants, a ≠ 0.
• How do you know if a quadratic equation has two solutions?
• A quadratic equation has two solutions if the discriminant b^2 - 4ac is positive.

quadratic-equation-calculator

• High School Math Solutions – Quadratic Equations Calculator, Part 1 A quadratic equation is a second degree polynomial having the general form ax^2 + bx + c = 0, where a, b, and c...

Please add a message.

Message received. Thanks for the feedback.

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons

Margin Size

• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

1.5: Equations with Rational Exponents

• Last updated
• Save as PDF
• Page ID 38268

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

We have solved linear equations, rational equations, radical equations, and quadratic equations using several methods. However, there are many other types of equations, such as equations involving rational exponents, polynomial 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. 

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, \({16}^{\tfrac{1}{2}}\) is another way of writing \(\sqrt{16}\); \(8^{\tfrac{1}{3}}\) is another way of writing \(\sqrt[3]{8}\). The ability to work with rational exponents is a useful skill, as it is highly applicable in calculus.

Equations in which a variable expression is raised to a rational exponent can be solved by raising both sides of the equation to the reciprocal of the exponent. The reason the expression is raised to the reciprocal of its exponent is because the product of a number and its reciprocal is one. Therefore the exponent on the variable expression becomes one and is thus eliminated. 

Definition: 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:

\[a^{\tfrac{m}{n}}={\left (a^{\tfrac{1}{n}} \right )}^m={(a^m)}^{\tfrac{1}{n}}=\sqrt[n]{a^m}={(\sqrt[n]{a})}^m \nonumber\]

Example \(\PageIndex{1}\): Evaluate a Number Raised to a Rational Exponent

Evaluate \(8^{\tfrac{2}{3}}\)

Solution.  It does not matter whether the root  or the power is done first  because \(8^{\tfrac{2}{3}} = (8^2)^{\tfrac{1}{3}}= (8^{\tfrac{1}{3}})^2\). Since the cube root of \(8\) is easy to find, \(8^{\tfrac{2}{3}}\) can be evaluated as \({\left (8^{\tfrac{1}{3}} \right )}^2= {(2)}^2= 4\).

Evaluate \({64}^{-\tfrac{1}{3}}\)

\(\dfrac{1}{4}\)

• Isolate the expression with the rational exponent
• If the numerator of the reciprocal power is an even number, the solution must be checked because the solution involves the squaring process which can introduce extraneous roots.
• If the denominator of the reciprocal power is an even number, this is equivalent to taking an even root so +/- must be included.

Example \(\PageIndex{2}\): Solve an Equation Containing a Variable Raised to a Rational Exponent

Solve the equation in which a variable is raised to a rational exponent: \(x^{\tfrac{3}{4}} = 8\).

Solution   The  exponent on \(x\) is removed by raising both sides of the equation to a power that is the reciprocal of \(\dfrac{3}{4}.\)  The reciprocal of \(\dfrac{3}{4}\)  is \(\dfrac{4}{3}\). The numerator of this exponent we are applying is an even number, which means that both sides are being raised to an even power.

\[\begin{align*} x^{\tfrac{3}{4}}&= 8\\ {\left(x^{\tfrac{3}{4}}\right)}^{\tfrac{4}{3}}&= {\left(8\right)}^{\tfrac{4}{3}}\\ x&= (8^{1/3})^4\\ &= (2)^4\\ &= 16 \end{align*}\]

It is necessary to check our result because the solution involved raising both sides of the equation to an even power. Raising both sides of an equation to an even power can introduce "extraneous" roots.  Therefore our answer must be checked:  \( 16 ^ {\tfrac{3}{4}} = (16^ \tfrac{1}{4})^3= 2^3=8 \). \(\color{Cerulean}{✓} \)  The solution set is \( \{ 16 \} \).

Example \(\PageIndex{3}\)

Solve \(x^\tfrac{5}{4}+36=4\).

Solution  

\(x^\frac{5}{4}=-32\) \(  (x^\frac{5}{4})^\frac{4}{5} =(-32)^\frac{4}{5}  \) \(x = (\sqrt[5]{-32}) ^ 4 \) \(x = (-2) ^ 4 \) \(x = 16 \)

It is necessary in this case to check our result because the solution involved raising both sides of the equation to an even power. Raising both sides of an equation to an even power can introduce "extraneous" roots.     \(  16^\tfrac{5}{4}+36= (\sqrt[4]{16}) ^ 5 +36= 2^5  +36 = 32 + 36 =68 \ne 4 \). Therefore, the solution \(x=16\) must be rejected. Therefore this problem has no solution. The solution set is \( \{ \quad \} \).

Example \(\PageIndex{4}\)

Solve  \(x^\tfrac{4}{3}=81\)

Solution . The solution involves raising both sides of the equal sign to the power of \( \frac{3}{4} \). Because the denominator is an even number, that means tha we are actually taking the even root of a quantity, which could be either a positive or negative value.

\(  (x^\frac{4}{3})^\frac{3}{4} = { \color{Cerulean}{\pm }}81^\frac{3}{4}  \) \(x = \pm (\sqrt[4]{81}) ^ 3 \) \(x = \pm (3) ^ 3 \) \(x = \pm 27 \)

No checking is required in this example because the process did not involve raising both sides of the equation to an even power. The even number was in the denominator, not the numerator of the reciprocal power. The solution set is \( \{  -27, 27 \}   \).

Example \(\PageIndex{5}\)

Solve \((x+5)^\tfrac{2}{3}= 64\)

Solution.  Notice here that the reciprocal power has an even denominator which represents taking the square root of both sides of the equation. This requires using \( \pm \) in the solution process.

\(  ((x+5)^\frac{2}{3})^\frac{3}{2} = \pm(64^\frac{3}{2})  \)      \(x+5 =\pm  (\sqrt{64}) ^ 3 \) \(x+5 = \pm (8) ^ 3 \) \(x+5 = \pm 512 \) \(x = -5+512 \) and \(x = -5-512 \) \(x = 509 \) and \(x = -517 \)

The solution does not need to be checked!  Solution Set: \( \{ 509, -517 \} \)

Solve the equation

a.  \( \{ 129, -121 \} \qquad \) b. \( \{ -1 \} \qquad \) c. \( \{  \} \)

Example \(\PageIndex{7}\): Solve an Equation involving Rational Exponents and Factoring

Solve \(3x^{\tfrac{3}{4}} = x^{\tfrac{1}{2}}\).

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.

\[\begin{align*} 3x^{\tfrac{3}{4}}-\left(x^{\tfrac{1}{2}}\right)&= x^{\tfrac{1}{2}}-\left(x^{\tfrac{1}{2}}\right)\\ 3x^{\tfrac{3}{4}}-x^{\tfrac{1}{2}}&= 0 \end{align*}\]

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. The factor with the lowest exponent is \(x^{1/2}\), so \(x^{3/4}\) needs to be rewritten as a product involving \(x^{1/2}\). 

\[\begin{align*} 3x^{\tfrac{3}{4}}-x^{\tfrac{1}{2}}&= 0\\ 3x^{(\tfrac{1}{2}+\tfrac{1}{4})}-x^{\tfrac{1}{2}}&= 0\\ 3x^{\tfrac{1}{2}}x^{\tfrac{1}{4}}-x^{\tfrac{1}{2}}&= 0\\ x^{\tfrac{1}{2}}\left (3x^{\tfrac{1}{4}}-1 \right )&= 0 \end{align*}\]

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

\( x^{\tfrac{1}{2}}\left (3x^{\tfrac{1}{4}}-1 \right )= 0 \)

\( \begin{array}{c|rl} x^{\tfrac{1}{2}}= 0 \qquad & 3x^{\tfrac{1}{4}}-1&= 0\\ x= 0 \qquad & 3x^{\tfrac{1}{4}}&= 1\\  & x^{\tfrac{1}{4}}&= \dfrac{1}{3} \\  & {\left (x^{\tfrac{1}{4}} \right )}^4&= {\left (\dfrac{1}{3} \right )}^4 \\  & x&= \dfrac{1}{81} \end{array} \)

The solution set is \(  {\Large\{} 0, \dfrac{1}{81}  {\Large\}}\).

Game Central

Similar problems from web search.