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,...
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![**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:
\[ \boxed{x = 1} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70fe4f99-9e69-4bef-afe6-80d54e8903ee%2F1ad39039-aa47-4f69-8eea-5eac890cafaf%2F39ju6tn_processed.png&w=3840&q=75)
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:
\[ \boxed{x = 1} \]
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