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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![**Problem Statement:**
Find the solution set of the inequality:
\[x^2 + 3x - 10 > 8\]
**Options:**
1. \(\{x | -6 < x < 3\}\)
2. \(\{x | x < -6 \text{ or } x > 3\}\)
3. \(\{x | -3 < x < 6\}\)
4. \(\{x | x < -3 \text{ or } x > 6\}\)
Explanation:
To solve the inequality \(x^2 + 3x - 10 > 8\), we need to first rearrange it to a standard form:
\[x^2 + 3x - 10 - 8 > 0\]
Simplifying, we get:
\[x^2 + 3x - 18 > 0\]
Next, we find the roots of the equation \(x^2 + 3x - 18 = 0\). To find the roots, we can factorize the quadratic equation.
\[x^2 + 3x - 18 = (x + 6)(x - 3)\]
The roots of the equation are \(x = -6\) and \(x = 3\).
Now we need to determine the intervals where the inequality \(x^2 + 3x - 18 > 0\) holds true.
To do this, we test the values in the regions determined by the roots:
1. \(x < -6\)
2. \(-6 < x < 3\)
3. \(x > 3\)
By testing values in these intervals:
- For \(x < -6\), let's take \(x = -7\):
\[(-7 + 6)(-7 - 3) = (-1)(-10) = 10 > 0\]
- For \(-6 < x < 3\), let's take \(x = 0\):
\[(0 + 6)(0 - 3) = (6)(-3) = -18 < 0\]
- For \(x > 3\), let's take \(x = 4\):
\[(4 + 6)(4 - 3) = (10)(1) = 10 > 0\]
From](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F679cf050-cce2-40a1-ad84-31514e2ece1c%2F57c518b1-da39-4124-89a1-eda525930871%2F1ftvth8_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the solution set of the inequality:
\[x^2 + 3x - 10 > 8\]
**Options:**
1. \(\{x | -6 < x < 3\}\)
2. \(\{x | x < -6 \text{ or } x > 3\}\)
3. \(\{x | -3 < x < 6\}\)
4. \(\{x | x < -3 \text{ or } x > 6\}\)
Explanation:
To solve the inequality \(x^2 + 3x - 10 > 8\), we need to first rearrange it to a standard form:
\[x^2 + 3x - 10 - 8 > 0\]
Simplifying, we get:
\[x^2 + 3x - 18 > 0\]
Next, we find the roots of the equation \(x^2 + 3x - 18 = 0\). To find the roots, we can factorize the quadratic equation.
\[x^2 + 3x - 18 = (x + 6)(x - 3)\]
The roots of the equation are \(x = -6\) and \(x = 3\).
Now we need to determine the intervals where the inequality \(x^2 + 3x - 18 > 0\) holds true.
To do this, we test the values in the regions determined by the roots:
1. \(x < -6\)
2. \(-6 < x < 3\)
3. \(x > 3\)
By testing values in these intervals:
- For \(x < -6\), let's take \(x = -7\):
\[(-7 + 6)(-7 - 3) = (-1)(-10) = 10 > 0\]
- For \(-6 < x < 3\), let's take \(x = 0\):
\[(0 + 6)(0 - 3) = (6)(-3) = -18 < 0\]
- For \(x > 3\), let's take \(x = 4\):
\[(4 + 6)(4 - 3) = (10)(1) = 10 > 0\]
From
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