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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![### Question 7
**Solve by factoring, show all work:**
\[ 4x^2 - 12x + 8 = 0 \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5fe3654-2655-4e19-bbf9-3baa2703a47c%2Ff4c996f3-cd8b-40e7-856c-0fe073ad418a%2Fwq0ii4h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 7
**Solve by factoring, show all work:**
\[ 4x^2 - 12x + 8 = 0 \]
![### Question 8
Solve by factoring, show all work:
\[2x^2 + 9x - 5 = 0\]
---
**Step-by-Step Solution:**
1. **Identify the quadratic equation** in the standard form \(ax^2 + bx + c = 0\). In this case, \(a = 2\), \(b = 9\), and \(c = -5\).
2. **Multiply** \(a\) and \(c\):
\[ a \cdot c = 2 \cdot (-5) = -10 \]
3. **Find two numbers** that multiply to \(-10\) (the result from step 2) and add up to \(9\) (the coefficient \(b\)):
\[ x_1 = 10 \]
\[ x_2 = -1 \]
4. **Rewrite the middle term** (\(9x\)) using the two numbers found in step 3:
\[ 2x^2 + 10x - x - 5 = 0 \]
5. **Group the terms** to factor by grouping:
\[ (2x^2 + 10x) + (-x - 5) = 0 \]
6. **Factor out the greatest common factor** (GCF) from each group:
\[ 2x(x + 5) - 1(x + 5) = 0 \]
7. **Factor out the common binomial** (\(x + 5\)):
\[ (2x - 1)(x + 5) = 0 \]
8. **Apply the Zero Product Property** by setting each factor equal to zero:
\[ 2x - 1 = 0 \quad \text{or} \quad x + 5 = 0 \]
9. **Solve each equation** for \(x\):
\[ 2x - 1 = 0 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2} \]
\[ x + 5 = 0 \Rightarrow x = -5 \]
There are **two solutions** to the equation:
\[ x = \frac{1}{2} \quad \text{and} \quad x = -5 \]
---
This solution involves](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5fe3654-2655-4e19-bbf9-3baa2703a47c%2Ff4c996f3-cd8b-40e7-856c-0fe073ad418a%2Fnry1kq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 8
Solve by factoring, show all work:
\[2x^2 + 9x - 5 = 0\]
---
**Step-by-Step Solution:**
1. **Identify the quadratic equation** in the standard form \(ax^2 + bx + c = 0\). In this case, \(a = 2\), \(b = 9\), and \(c = -5\).
2. **Multiply** \(a\) and \(c\):
\[ a \cdot c = 2 \cdot (-5) = -10 \]
3. **Find two numbers** that multiply to \(-10\) (the result from step 2) and add up to \(9\) (the coefficient \(b\)):
\[ x_1 = 10 \]
\[ x_2 = -1 \]
4. **Rewrite the middle term** (\(9x\)) using the two numbers found in step 3:
\[ 2x^2 + 10x - x - 5 = 0 \]
5. **Group the terms** to factor by grouping:
\[ (2x^2 + 10x) + (-x - 5) = 0 \]
6. **Factor out the greatest common factor** (GCF) from each group:
\[ 2x(x + 5) - 1(x + 5) = 0 \]
7. **Factor out the common binomial** (\(x + 5\)):
\[ (2x - 1)(x + 5) = 0 \]
8. **Apply the Zero Product Property** by setting each factor equal to zero:
\[ 2x - 1 = 0 \quad \text{or} \quad x + 5 = 0 \]
9. **Solve each equation** for \(x\):
\[ 2x - 1 = 0 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2} \]
\[ x + 5 = 0 \Rightarrow x = -5 \]
There are **two solutions** to the equation:
\[ x = \frac{1}{2} \quad \text{and} \quad x = -5 \]
---
This solution involves
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