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,...
Related questions
Question
![**Problem 4: Solving the Inequality**
Solve the inequality:
\[
\frac{x^2 - 9}{x + 1} > 0
\]
Express your answer in interval form.
**Approach:**
1. **Factor the numerator:**
\[
x^2 - 9 = (x - 3)(x + 3)
\]
2. **Rewrite the inequality:**
\[
\frac{(x - 3)(x + 3)}{x + 1} > 0
\]
3. **Determine the critical points:**
- From \(x - 3 = 0\), we get \(x = 3\).
- From \(x + 3 = 0\), we get \(x = -3\).
- From \(x + 1 = 0\), we get \(x = -1\).
4. **Test the intervals:**
\[
(-\infty, -3), (-3, -1), (-1, 3), (3, \infty)
\]
5. **Consider the sign in each interval:**
- **For \(x < -3\):** All factors are negative, so the expression is positive.
- **For \(-3 < x < -1\):** The factor \(x + 3\) is positive, while the others are negative, making the expression negative.
- **For \(-1 < x < 3\):** \(x - 3\) is negative, and others positive, making the expression negative.
- **For \(x > 3\):** All factors are positive, so the expression is positive.
6. **Intervals satisfying the inequality:**
\[
(-\infty, -3) \cup (3, \infty)
\]
*Note: Exclude \(x = -1\) and \(x = 3\) as they make the denominator zero or do not satisfy the strict inequality.*
**Conclusion:**
The solution in interval form is:
\[
(-\infty, -3) \cup (3, \infty)
\]
This means the inequality holds true for values of \(x\) in these intervals.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6071ed29-7bf0-4a7a-9fbf-55dd6a8bed61%2Fb03c735c-82ad-44dc-9a43-fb04b109e559%2Fy577fk7_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 4: Solving the Inequality**
Solve the inequality:
\[
\frac{x^2 - 9}{x + 1} > 0
\]
Express your answer in interval form.
**Approach:**
1. **Factor the numerator:**
\[
x^2 - 9 = (x - 3)(x + 3)
\]
2. **Rewrite the inequality:**
\[
\frac{(x - 3)(x + 3)}{x + 1} > 0
\]
3. **Determine the critical points:**
- From \(x - 3 = 0\), we get \(x = 3\).
- From \(x + 3 = 0\), we get \(x = -3\).
- From \(x + 1 = 0\), we get \(x = -1\).
4. **Test the intervals:**
\[
(-\infty, -3), (-3, -1), (-1, 3), (3, \infty)
\]
5. **Consider the sign in each interval:**
- **For \(x < -3\):** All factors are negative, so the expression is positive.
- **For \(-3 < x < -1\):** The factor \(x + 3\) is positive, while the others are negative, making the expression negative.
- **For \(-1 < x < 3\):** \(x - 3\) is negative, and others positive, making the expression negative.
- **For \(x > 3\):** All factors are positive, so the expression is positive.
6. **Intervals satisfying the inequality:**
\[
(-\infty, -3) \cup (3, \infty)
\]
*Note: Exclude \(x = -1\) and \(x = 3\) as they make the denominator zero or do not satisfy the strict inequality.*
**Conclusion:**
The solution in interval form is:
\[
(-\infty, -3) \cup (3, \infty)
\]
This means the inequality holds true for values of \(x\) in these intervals.
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