Chapter 1: Problem 5
Solve and check each of the equations. \(x^{2}-5 x-6=0\)
Short Answer
Expert verified
The solutions are \(x = 6\) and \(x = -1\).
Step by step solution
01
Identify the Quadratic Equation
Let's start with the equation: \[x^2 - 5x - 6 = 0\]This is a quadratic equation in standard form \(ax^2 + bx + c = 0\) where \(a = 1\), \(b = -5\), and \(c = -6\).
02
Factor the Quadratic Equation
To solve the equation by factoring, we need to express it in the form \[(x - p)(x - q) = 0\]where the product of \(p\) and \(q\) is equal to \(c\) and their sum is equal to \(b\).We look for two numbers that multiply to \(-6\) and add up to \(-5\). These numbers are \(-6\) and \(1\). Therefore, we can factor the equation as:\[(x - 6)(x + 1) = 0\]
03
Set Each Factor to Zero
Next, we set each factor to zero and solve for \(x\):1. \(x - 6 = 0\) - Add 6 to both sides to solve for \(x\): \[x = 6\]2. \(x + 1 = 0\) - Subtract 1 from both sides to solve for \(x\): \[x = -1\]
04
Verify the Solutions
We will substitute our solutions \(x = 6\) and \(x = -1\) back into the original equation to verify.1. For \(x = 6\): - Substitute 6 into the equation: \[6^2 - 5(6) - 6 = 36 - 30 - 6 = 0\] - Since the equation holds, \(x = 6\) is a solution.2. For \(x = -1\): - Substitute -1 into the equation: \[(-1)^2 - 5(-1) - 6 = 1 + 5 - 6 = 0\] - Since the equation holds, \(x = -1\) is a solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factoring Quadratic Equations
Factoring a quadratic equation is a method to find the solutions of the equation by expressing it as the product of its factors. The standard form of a quadratic equation is given by \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients. To factor the quadratic equation \(x^2 - 5x - 6 = 0\), we look for two numbers whose product is \(c = -6\) and whose sum is \(b = -5\).
The factors \(x - 6\) and \(x + 1\) help us solve for the roots of the equation by setting each factor equal to zero. This results in two solutions for \(x\): \(x = 6\) and \(x = -1\).
Factoring is a powerful and efficient method to solve quadratic equations, especially when the numbers are easy to recognize and factor.
- These numbers are \(-6\) and \(1\).
- They satisfy the condition \(p \times q = -6\) and \(p + q = -5\).
The factors \(x - 6\) and \(x + 1\) help us solve for the roots of the equation by setting each factor equal to zero. This results in two solutions for \(x\): \(x = 6\) and \(x = -1\).
Factoring is a powerful and efficient method to solve quadratic equations, especially when the numbers are easy to recognize and factor.
Verifying Solutions
Once we have found the possible solutions \(x = 6\) and \(x = -1\) by factoring, it is crucial to verify them. Verification ensures that both solutions truly satisfy the original quadratic equation. We do this by substituting each solution back into the equation. Let's examine both solutions:
- Substitute \(x = 6\):
\[6^2 - 5(6) - 6 = 36 - 30 - 6 = 0\] - Substitute \(x = -1\):
\[(-1)^2 - 5(-1) - 6 = 1 + 5 - 6 = 0\]
Standard Form of a Quadratic Equation
Understanding the standard form of a quadratic equation is fundamental in solving and manipulating these types of equations. The standard form is written as \(ax^2 + bx + c = 0\), where:
- \(a\) is the coefficient of \(x^2\) and determines the direction and width of the parabola.
- \(b\) is the coefficient of \(x\), influencing the axis of symmetry and the vertex.
- \(c\) is the constant term, indicating the equation's value when \(x = 0\).