Problem 14 - Basic Technical Mathematics with Calculus

Question

$$\text { Solve the given quadratic equations by factoring.}$$ $$x^{2}+x-6=0$$

Step-by-Step Solution

Answer

The solutions are \(x = 2\) and \(x = -3\).

Step 1: Identify a Quadratic Equation

Recognize that the given equation \(x^2 + x - 6 = 0\) is a quadratic equation where the standard form is \(ax^2 + bx + c = 0\). Here, \(a = 1, b = 1, c = -6\).

Step 2: Find Factor Pairs of the Constant Term

List the pairs of integers that multiply to give the constant term \(-6\). Possible pairs are \((1, -6), (-1, 6), (2, -3), (-2, 3)\).

Step 3: Choose the Correct Pair to Factor

Choose the pair \((-2, 3)\) because their sum is \(1\) (the value of coefficient \(b\) in the equation).

Step 4: Write Equation in Factored Form

Rewrite the equation \(x^2 + x - 6 = 0\) using the factors to get \((x - 2)(x + 3) = 0\).

Step 5: Apply the Zero Product Property

Set each factor equal to zero: \(x - 2 = 0\) and \(x + 3 = 0\).

Step 6: Solve Each Equation

Solve for \(x\) in each of the equations: \(x - 2 = 0\) gives \(x = 2\), and \(x + 3 = 0\) gives \(x = -3\).

Step 7: State the Solution Set

The solutions to the quadratic equation are \(x = 2\) and \(x = -3\).

Key Concepts

Factoring

When working with quadratic equations, one of the most useful techniques for finding solutions is factoring. Factoring is essentially rewriting a quadratic equation as a product of two simpler binomials.
This process involves looking for two numbers that multiply to give the constant term at the end of the equation and add up to the middle coefficient, which is sometimes a challenging step.

• Example: In the equation \(x^2 + x - 6 = 0\), you need a pair of numbers that multiply to \(-6\) and add up to \(1\) (the coefficient of \(x\)).
• Possible pairs: \((1, -6), (-1, 6), (2, -3), (-2, 3)\).
• The pair \((-2, 3)\) works because \(-2 + 3 = 1\).

Rewriting the original equation using these factors gives us \((x - 2)(x + 3) = 0\).
This is the factored form.

Zero Product Property

The zero product property is a fundamental concept when solving quadratic equations by factoring. It states that if the product of two numbers is zero, then at least one of the numbers must be zero.
This principle allows us to break down an equation into simpler parts and then find solutions.

• Once in factored form, for example \((x - 2)(x + 3) = 0\), each factor can be set to zero individually: \(x - 2 = 0\) and \(x + 3 = 0\).
• By solving these linear equations, you determine the possible values for \(x\).
• This ultimately provides the solutions to the quadratic equation based on these individual roots.

The property is crucial because it simplifies the process of solving quadratic equations into more manageable steps.

Standard Form of Quadratic Equation

Every quadratic equation needs to be initially identified in its standard form: \(ax^2 + bx + c = 0\).
Recognizing this arrangement is essential for proceeding with methods like factoring, the quadratic formula, or completing the square.

• In the problem \(x^2 + x - 6 = 0\), the coefficients are \(a = 1\), \(b = 1\), and \(c = -6\).
• This standard form clearly outlines the terms of the equation and helps us see the structure required for factoring.
• Understanding this format ensures that we can consistently approach various problems, applying the same strategies effectively.

By consistently transforming equations into this form, solving them becomes a structured process.

Solving Quadratic Equations

Solving quadratic equations involves finding the value of \(x\) that makes the equation true, i.e., it equals zero. There are several methods available, but factoring followed by implementing the zero product property is often a simple and direct approach.
Here's a streamlined look at the process:

• Verify the equation is in the standard form: \(ax^2 + bx + c = 0\).
• If possible, rewrite the quadratic in factored form, such as \((x - 2)(x + 3) = 0\).
• Apply the zero product property to set each binomial factor to zero: \(x - 2 = 0\) or \(x + 3 = 0\).
• Solve these simpler equations to find \(x\) values that satisfy the original equation.
• For the example equation \(x = 2\) and \(x = -3\) are the solutions.

Once mastered, these steps facilitate solving a wide range of quadratic equations efficiently.