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
8th Grade Pre-Ap Math
Help students by solving for x in the following equations, step by step.
![The equation provided is a cubic equation, expressed as:
\[ x^3 - 6x^2 + 5x = 0 \]
**Explanation:**
This is a polynomial equation where the highest degree is 3, indicating that it is a cubic equation. The terms are:
- \( x^3 \): The cubic term
- \(-6x^2\): The quadratic term
- \(5x\): The linear term
**Steps to Solve:**
1. **Factor the Equation:**
You can start by factoring out the greatest common factor, which is \(x\):
\[ x(x^2 - 6x + 5) = 0 \]
2. **Find the Solutions:**
- The solutions to \(x(x^2 - 6x + 5) = 0\) occur when \(x = 0\) or the quadratic equation \(x^2 - 6x + 5 = 0\) is solved.
3. **Solve the Quadratic:**
- Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -6\), and \(c = 5\).
This process allows you to find the roots of the equation, which are the values of \(x\) that satisfy the equation \(x^3 - 6x^2 + 5x = 0\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5ae18fd2-7331-48d4-90a1-bc335103e09b%2F85e6175c-2e93-4270-88d9-72a488c226a0%2Fi470wio_processed.png&w=3840&q=75)
Transcribed Image Text:The equation provided is a cubic equation, expressed as:
\[ x^3 - 6x^2 + 5x = 0 \]
**Explanation:**
This is a polynomial equation where the highest degree is 3, indicating that it is a cubic equation. The terms are:
- \( x^3 \): The cubic term
- \(-6x^2\): The quadratic term
- \(5x\): The linear term
**Steps to Solve:**
1. **Factor the Equation:**
You can start by factoring out the greatest common factor, which is \(x\):
\[ x(x^2 - 6x + 5) = 0 \]
2. **Find the Solutions:**
- The solutions to \(x(x^2 - 6x + 5) = 0\) occur when \(x = 0\) or the quadratic equation \(x^2 - 6x + 5 = 0\) is solved.
3. **Solve the Quadratic:**
- Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -6\), and \(c = 5\).
This process allows you to find the roots of the equation, which are the values of \(x\) that satisfy the equation \(x^3 - 6x^2 + 5x = 0\).
![The equation shown is:
\[
(x + 4)^2 = 36
\]
This equation is a quadratic equation in the form of a perfect square trinomial set equal to a constant. To solve for \( x \), take the square root of both sides, resulting in two potential solutions where:
\[
x + 4 = 6 \quad \text{or} \quad x + 4 = -6
\]
Solving these equations gives:
1. \( x + 4 = 6 \) leads to \( x = 2 \)
2. \( x + 4 = -6 \) leads to \( x = -10 \)
Therefore, the two solutions for \( x \) are \( x = 2 \) and \( x = -10 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5ae18fd2-7331-48d4-90a1-bc335103e09b%2F85e6175c-2e93-4270-88d9-72a488c226a0%2F8isicr_processed.png&w=3840&q=75)
Transcribed Image Text:The equation shown is:
\[
(x + 4)^2 = 36
\]
This equation is a quadratic equation in the form of a perfect square trinomial set equal to a constant. To solve for \( x \), take the square root of both sides, resulting in two potential solutions where:
\[
x + 4 = 6 \quad \text{or} \quad x + 4 = -6
\]
Solving these equations gives:
1. \( x + 4 = 6 \) leads to \( x = 2 \)
2. \( x + 4 = -6 \) leads to \( x = -10 \)
Therefore, the two solutions for \( x \) are \( x = 2 \) and \( x = -10 \).
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