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
![### Question 10
**Solve by completing the square, show all work:**
\[2x^2 - 8x - 5 = 0\]
#### Solution Steps:
1. **Divide the entire equation by the coefficient of \(x^2\):**
\[\frac{2x^2 - 8x - 5}{2} = \frac{0}{2}\]
\[x^2 - 4x - \frac{5}{2} = 0\]
2. **Isolate the constant term:**
\[x^2 - 4x = \frac{5}{2}\]
3. **Complete the square:**
- Take the coefficient of \(x\) (which is \(-4\)), divide it by 2, and square it:
\[\left(\frac{-4}{2}\right)^2 = (-2)^2 = 4\]
- Add and subtract this value on the left-hand side of the equation:
\[x^2 - 4x + 4 - 4 = \frac{5}{2}\]
\[x^2 - 4x + 4 = \frac{5}{2} + 4\]
\[x^2 - 4x + 4 = \frac{5}{2} + \frac{8}{2}\]
\[x^2 - 4x + 4 = \frac{13}{2}\]
- Now, express the left-hand side as a perfect square:
\[(x-2)^2 = \frac{13}{2}\]
4. **Solve the resulting equation:**
- Take the square root of both sides:
\[x - 2 = \pm \sqrt{\frac{13}{2}}\]
- Simplify:
\[x - 2 = \pm \frac{\sqrt{26}}{2}\]
- Solve for \(x\):
\[x = 2 \pm \frac{\sqrt{26}}{2}\]
5. **Write the final answer:**
\[x = 2 + \frac{\sqrt{26}}{2}\]
\[x = 2 - \frac{\sqrt{26}}{2}\]
Thus, the solutions to the equation \(2x^2 -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5fe3654-2655-4e19-bbf9-3baa2703a47c%2Fac40d3e8-e675-4943-8399-790ce3fc6409%2Fah8a19i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 10
**Solve by completing the square, show all work:**
\[2x^2 - 8x - 5 = 0\]
#### Solution Steps:
1. **Divide the entire equation by the coefficient of \(x^2\):**
\[\frac{2x^2 - 8x - 5}{2} = \frac{0}{2}\]
\[x^2 - 4x - \frac{5}{2} = 0\]
2. **Isolate the constant term:**
\[x^2 - 4x = \frac{5}{2}\]
3. **Complete the square:**
- Take the coefficient of \(x\) (which is \(-4\)), divide it by 2, and square it:
\[\left(\frac{-4}{2}\right)^2 = (-2)^2 = 4\]
- Add and subtract this value on the left-hand side of the equation:
\[x^2 - 4x + 4 - 4 = \frac{5}{2}\]
\[x^2 - 4x + 4 = \frac{5}{2} + 4\]
\[x^2 - 4x + 4 = \frac{5}{2} + \frac{8}{2}\]
\[x^2 - 4x + 4 = \frac{13}{2}\]
- Now, express the left-hand side as a perfect square:
\[(x-2)^2 = \frac{13}{2}\]
4. **Solve the resulting equation:**
- Take the square root of both sides:
\[x - 2 = \pm \sqrt{\frac{13}{2}}\]
- Simplify:
\[x - 2 = \pm \frac{\sqrt{26}}{2}\]
- Solve for \(x\):
\[x = 2 \pm \frac{\sqrt{26}}{2}\]
5. **Write the final answer:**
\[x = 2 + \frac{\sqrt{26}}{2}\]
\[x = 2 - \frac{\sqrt{26}}{2}\]
Thus, the solutions to the equation \(2x^2 -
![### Question 9
**Problem Statement:**
Solve by completing the square, show all work:
\[ x^2 - 6x = 13 \]
**Solution:**
To solve the equation \( x^2 - 6x = 13 \) by completing the square, follow these steps:
1. **Move the constant term to the right side of the equation:**
\[ x^2 - 6x + \_ = 13 + \_ \]
2. **Complete the square on the left side of the equation:**
To complete the square, add the square of half the coefficient of \( x \). The coefficient of \( x \) is \(-6\), and half of that is \(-3\). The square of \(-3\) is \(9\).
\[ x^2 - 6x + 9 = 13 + 9 \]
\[ x^2 - 6x + 9 = 22 \]
3. **Rewrite the left side of the equation as a squared binomial:**
\[ (x - 3)^2 = 22 \]
4. **Solve for \( x \):**
Take the square root of both sides of the equation:
\[ x - 3 = \pm \sqrt{22} \]
5. **Isolate \( x \):**
\[ x = 3 \pm \sqrt{22} \]
Therefore, the solutions to the equation \( x^2 - 6x = 13 \) are:
\[ x = 3 + \sqrt{22} \quad \text{or} \quad x = 3 - \sqrt{22} \]
### Final Answer:
The solutions are \( x = 3 + \sqrt{22} \) and \( x = 3 - \sqrt{22} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5fe3654-2655-4e19-bbf9-3baa2703a47c%2Fac40d3e8-e675-4943-8399-790ce3fc6409%2F8e6na7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 9
**Problem Statement:**
Solve by completing the square, show all work:
\[ x^2 - 6x = 13 \]
**Solution:**
To solve the equation \( x^2 - 6x = 13 \) by completing the square, follow these steps:
1. **Move the constant term to the right side of the equation:**
\[ x^2 - 6x + \_ = 13 + \_ \]
2. **Complete the square on the left side of the equation:**
To complete the square, add the square of half the coefficient of \( x \). The coefficient of \( x \) is \(-6\), and half of that is \(-3\). The square of \(-3\) is \(9\).
\[ x^2 - 6x + 9 = 13 + 9 \]
\[ x^2 - 6x + 9 = 22 \]
3. **Rewrite the left side of the equation as a squared binomial:**
\[ (x - 3)^2 = 22 \]
4. **Solve for \( x \):**
Take the square root of both sides of the equation:
\[ x - 3 = \pm \sqrt{22} \]
5. **Isolate \( x \):**
\[ x = 3 \pm \sqrt{22} \]
Therefore, the solutions to the equation \( x^2 - 6x = 13 \) are:
\[ x = 3 + \sqrt{22} \quad \text{or} \quad x = 3 - \sqrt{22} \]
### Final Answer:
The solutions are \( x = 3 + \sqrt{22} \) and \( x = 3 - \sqrt{22} \).
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 1 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.Recommended textbooks for you
Algebra and Trigonometry (6th Edition)
Algebra
ISBN:
9780134463216
Author:
Robert F. Blitzer
Publisher:
PEARSON
Contemporary Abstract Algebra
Algebra
ISBN:
9781305657960
Author:
Joseph Gallian
Publisher:
Cengage Learning
Linear Algebra: A Modern Introduction
Algebra
ISBN:
9781285463247
Author:
David Poole
Publisher:
Cengage Learning
Algebra and Trigonometry (6th Edition)
Algebra
ISBN:
9780134463216
Author:
Robert F. Blitzer
Publisher:
PEARSON
Contemporary Abstract Algebra
Algebra
ISBN:
9781305657960
Author:
Joseph Gallian
Publisher:
Cengage Learning
Linear Algebra: A Modern Introduction
Algebra
ISBN:
9781285463247
Author:
David Poole
Publisher:
Cengage Learning
Algebra And Trigonometry (11th Edition)
Algebra
ISBN:
9780135163078
Author:
Michael Sullivan
Publisher:
PEARSON
Introduction to Linear Algebra, Fifth Edition
Algebra
ISBN:
9780980232776
Author:
Gilbert Strang
Publisher:
Wellesley-Cambridge Press
College Algebra (Collegiate Math)
Algebra
ISBN:
9780077836344
Author:
Julie Miller, Donna Gerken
Publisher:
McGraw-Hill Education