Solve the equation by using a u-substitution: x 5 + x5-2=0

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
icon
Related questions
Question
**Solving the Equation Using u-Substitution**

Solve the equation by using a u-substitution:

\[ x^{\frac{2}{5}} + x^{\frac{1}{5}} - 2 = 0 \]

#### Steps for Solving:

1. **Identify the Substitution:**
   Let \( u = x^{\frac{1}{5}} \). Therefore, \( u^2 = x^{\frac{2}{5}} \).

2. **Substitute in the Original Equation:**
   Replace \( x^{\frac{2}{5}} \) and \( x^{\frac{1}{5}} \) with \( u \) and \( u^2 \):

   \[ u^2 + u - 2 = 0 \]

3. **Solve the Quadratic Equation:**
   Factor the quadratic equation.

   \[ u^2 + u - 2 = (u + 2)(u - 1) = 0 \]

   This gives us two solutions for \( u \):

   \[ u + 2 = 0 \Rightarrow u = -2 \]
   \[ u - 1 = 0 \Rightarrow u = 1 \]

4. **Back-Substitute to Find \( x \):** 
   Recall that \( u = x^{\frac{1}{5}} \).

   For \( u = -2 \):

   \[ -2 = x^{\frac{1}{5}} \]
   Since \( x^{\frac{1}{5}} \) must be a real number, \( x \) cannot be negative, and thus \( u = -2 \) is not a valid solution in the real number system.

   For \( u = 1 \):

   \[ 1 = x^{\frac{1}{5}} \]
   Raising both sides to the power of 5:

   \[ x = 1^5 \]
   Hence, \( x = 1 \).

Our valid solution is:

\[ \boxed{x = 1} \]
Transcribed Image Text:**Solving the Equation Using u-Substitution** Solve the equation by using a u-substitution: \[ x^{\frac{2}{5}} + x^{\frac{1}{5}} - 2 = 0 \] #### Steps for Solving: 1. **Identify the Substitution:** Let \( u = x^{\frac{1}{5}} \). Therefore, \( u^2 = x^{\frac{2}{5}} \). 2. **Substitute in the Original Equation:** Replace \( x^{\frac{2}{5}} \) and \( x^{\frac{1}{5}} \) with \( u \) and \( u^2 \): \[ u^2 + u - 2 = 0 \] 3. **Solve the Quadratic Equation:** Factor the quadratic equation. \[ u^2 + u - 2 = (u + 2)(u - 1) = 0 \] This gives us two solutions for \( u \): \[ u + 2 = 0 \Rightarrow u = -2 \] \[ u - 1 = 0 \Rightarrow u = 1 \] 4. **Back-Substitute to Find \( x \):** Recall that \( u = x^{\frac{1}{5}} \). For \( u = -2 \): \[ -2 = x^{\frac{1}{5}} \] Since \( x^{\frac{1}{5}} \) must be a real number, \( x \) cannot be negative, and thus \( u = -2 \) is not a valid solution in the real number system. For \( u = 1 \): \[ 1 = x^{\frac{1}{5}} \] Raising both sides to the power of 5: \[ x = 1^5 \] Hence, \( x = 1 \). Our valid solution is: \[ \boxed{x = 1} \]
Expert Solution
steps

Step by step

Solved in 2 steps

Blurred answer
Recommended textbooks for you
Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
Algebra
ISBN:
9780134463216
Author:
Robert F. Blitzer
Publisher:
PEARSON
Contemporary Abstract Algebra
Contemporary Abstract Algebra
Algebra
ISBN:
9781305657960
Author:
Joseph Gallian
Publisher:
Cengage Learning
Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
Algebra
ISBN:
9781285463247
Author:
David Poole
Publisher:
Cengage Learning
Algebra And Trigonometry (11th Edition)
Algebra And Trigonometry (11th Edition)
Algebra
ISBN:
9780135163078
Author:
Michael Sullivan
Publisher:
PEARSON
Introduction to Linear Algebra, Fifth Edition
Introduction to Linear Algebra, Fifth Edition
Algebra
ISBN:
9780980232776
Author:
Gilbert Strang
Publisher:
Wellesley-Cambridge Press
College Algebra (Collegiate Math)
College Algebra (Collegiate Math)
Algebra
ISBN:
9780077836344
Author:
Julie Miller, Donna Gerken
Publisher:
McGraw-Hill Education