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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![Title: Solving Polynomial Inequality
---
**Problem:**
What is the solution set of the inequality \( x^2 + 3x - 10 > 8 \)?
**Options:**
1. \(\{ x \mid -6 < x < 3 \}\)
2. \(\{ x \mid x < -6 \text{ or } x > 3 \}\)
3. \(\{ x \mid -3 < x < 6 \}\)
4. \(\{ x \mid x < -3 \text{ or } x > 6 \}\)
---
**Explanation:**
To solve the inequality \( x^2 + 3x - 10 > 8 \), follow these steps:
1. **Rewrite the inequality:**
\[ x^2 + 3x - 10 > 8 \]
\[ x^2 + 3x - 18 > 0 \]
2. **Factor the quadratic expression \( x^2 + 3x - 18 \):**
\[ x^2 + 3x - 18 = (x + 6)(x - 3) \]
Thus, the inequality is:
\[ (x + 6)(x - 3) > 0 \]
3. **Find the critical points:**
Set \( x + 6 = 0 \) and \( x - 3 = 0 \):
\[ x + 6 = 0 \implies x = -6 \]
\[ x - 3 = 0 \implies x = 3 \]
4. **Determine the intervals to test:**
The critical points divide the number line into three intervals:
- \( x < -6 \)
- \( -6 < x < 3 \)
- \( x > 3 \)
5. **Test a point from each interval to determine where the inequality holds:**
- For \( x < -6 \), choose \( x = -7 \):
\[ (x + 6)(x - 3) \text{ at } x = -7 \Rightarrow (-7 + 6)(-7 - 3) = (-1)(-10) = 10 > 0 \]
- For \( -6 < x < 3 \), choose \( x =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9c16bde4-41e9-48fa-bdb9-6b51f3db4228%2F8c9a0086-b007-44b7-927c-7ea4c4ec2128%2Fhc8k9ie_processed.png&w=3840&q=75)
Transcribed Image Text:Title: Solving Polynomial Inequality
---
**Problem:**
What is the solution set of the inequality \( x^2 + 3x - 10 > 8 \)?
**Options:**
1. \(\{ x \mid -6 < x < 3 \}\)
2. \(\{ x \mid x < -6 \text{ or } x > 3 \}\)
3. \(\{ x \mid -3 < x < 6 \}\)
4. \(\{ x \mid x < -3 \text{ or } x > 6 \}\)
---
**Explanation:**
To solve the inequality \( x^2 + 3x - 10 > 8 \), follow these steps:
1. **Rewrite the inequality:**
\[ x^2 + 3x - 10 > 8 \]
\[ x^2 + 3x - 18 > 0 \]
2. **Factor the quadratic expression \( x^2 + 3x - 18 \):**
\[ x^2 + 3x - 18 = (x + 6)(x - 3) \]
Thus, the inequality is:
\[ (x + 6)(x - 3) > 0 \]
3. **Find the critical points:**
Set \( x + 6 = 0 \) and \( x - 3 = 0 \):
\[ x + 6 = 0 \implies x = -6 \]
\[ x - 3 = 0 \implies x = 3 \]
4. **Determine the intervals to test:**
The critical points divide the number line into three intervals:
- \( x < -6 \)
- \( -6 < x < 3 \)
- \( x > 3 \)
5. **Test a point from each interval to determine where the inequality holds:**
- For \( x < -6 \), choose \( x = -7 \):
\[ (x + 6)(x - 3) \text{ at } x = -7 \Rightarrow (-7 + 6)(-7 - 3) = (-1)(-10) = 10 > 0 \]
- For \( -6 < x < 3 \), choose \( x =
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