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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![### Solving Quadratic Equations: Example 7
In this example, we are given a quadratic equation that we need to solve for the value of \( x \). The equation is as follows:
\[ x^2 + 14x - 15 = 0 \]
To solve this quadratic equation, we need to find the values of \( x \) that satisfy the equation. Quadratic equations of the form \( ax^2 + bx + c = 0 \) can be solved using various methods including factoring, completing the square, and using the quadratic formula.
In this specific case, let's explore the detailed steps to solve it using the factoring method:
1. **Identifying the coefficients:** Here, \( a = 1 \), \( b = 14 \), and \( c = -15 \).
2. **Factoring the quadratic expression:** We need to find two numbers that multiply to give \( a \cdot c = 1 \cdot (-15) = -15 \) and add to give \( b = 14 \).
After examining the pairs of factors of -15, we find that:
- \( -1 \) and \( 15 \) multiply to give -15, and add to give 14.
3. **Rewriting the middle term:** Use these numbers to rewrite the middle term, \( 14x \):
\[ x^2 + 15x - x - 15 = 0 \]
4. **Grouping and factoring:** Group the terms to factor by grouping:
\[ (x^2 + 15x) - (x + 15) = 0 \]
\[ x(x + 15) - 1(x + 15) = 0 \]
5. **Factoring out the common binomial factor:**
\[ (x - 1)(x + 15) = 0 \]
6. **Solving for \( x \):** Set each factor equal to zero:
\[ x - 1 = 0 \quad \text{or} \quad x + 15 = 0 \]
Therefore, the solutions are:
\[ x = 1 \quad \text{or} \quad x = -15 \]
By graphing these solutions, you would find the points where the parabola \( y = x^2 + 14x - 15 \) intersects the x-axis. The points of intersection are \( (1,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe93494c9-6043-4390-b772-e35af6d121f7%2F2a5d764a-dba0-404a-804f-15c616eeed2c%2Fj0fx056_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Solving Quadratic Equations: Example 7
In this example, we are given a quadratic equation that we need to solve for the value of \( x \). The equation is as follows:
\[ x^2 + 14x - 15 = 0 \]
To solve this quadratic equation, we need to find the values of \( x \) that satisfy the equation. Quadratic equations of the form \( ax^2 + bx + c = 0 \) can be solved using various methods including factoring, completing the square, and using the quadratic formula.
In this specific case, let's explore the detailed steps to solve it using the factoring method:
1. **Identifying the coefficients:** Here, \( a = 1 \), \( b = 14 \), and \( c = -15 \).
2. **Factoring the quadratic expression:** We need to find two numbers that multiply to give \( a \cdot c = 1 \cdot (-15) = -15 \) and add to give \( b = 14 \).
After examining the pairs of factors of -15, we find that:
- \( -1 \) and \( 15 \) multiply to give -15, and add to give 14.
3. **Rewriting the middle term:** Use these numbers to rewrite the middle term, \( 14x \):
\[ x^2 + 15x - x - 15 = 0 \]
4. **Grouping and factoring:** Group the terms to factor by grouping:
\[ (x^2 + 15x) - (x + 15) = 0 \]
\[ x(x + 15) - 1(x + 15) = 0 \]
5. **Factoring out the common binomial factor:**
\[ (x - 1)(x + 15) = 0 \]
6. **Solving for \( x \):** Set each factor equal to zero:
\[ x - 1 = 0 \quad \text{or} \quad x + 15 = 0 \]
Therefore, the solutions are:
\[ x = 1 \quad \text{or} \quad x = -15 \]
By graphing these solutions, you would find the points where the parabola \( y = x^2 + 14x - 15 \) intersects the x-axis. The points of intersection are \( (1,
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