Problem 5 - Algebra 2 and Trigonometry

Question

Solve and check each of the equations. \(x^{2}-5 x-6=0\)

Step-by-Step Solution

Answer

The solutions are \(x = 6\) and \(x = -1\).

Step 1: 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 \(c = -6\).

Step 2: 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:\[(x - 6)(x + 1) = 0\]

Step 3: 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 \(x\): \[x = -1\]

Step 4: 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.

Key Concepts

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 \(b = -5\).

• These numbers are \(-6\) and \(1\).
• They satisfy the condition \(p \times q = -6\) and \(p + q = -5\).

Thus, the equation can be rewritten as \(x^2 - 5x - 6 = (x - 6)(x + 1) = 0\).
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 \(x = -1\):
  \[(-1)^2 - 5(-1) - 6 = 1 + 5 - 6 = 0\]

In both cases, the results are \(0\), confirming that both \(x = 6\) and \(x = -1\) are valid solutions of the equation. Verification of solutions is an essential step in solving quadratic equations to ensure accuracy and correctness.

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 \(x = 0\).

In the problem \(x^2 - 5x - 6 = 0\), the coefficients are \(a = 1\), \(b = -5\), and \(c = -6\). Recognizing and writing a quadratic equation in its standard form makes it easier to apply methods like factoring and using the quadratic formula. It allows us to quickly identify the roots of the equation by solving for the zeros of the function. The standard form serves as the starting point for any further analysis or calculation related to quadratic equations.