Problem 6 - College Algebra

Question

Solve the quadratic equation by factoring. $$ x^{2}+4 x-21=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = -7\) and \(x = 3\).

Step 1: Identify the Quadratic Equation

The given equation is a quadratic equation of the form \( ax^2 + bx + c = 0 \). Here, the coefficients are: \( a = 1 \), \( b = 4 \), \( c = -21 \).

Step 2: Find Two Numbers that Multiply to \( c \) and Add to \( b \)

To factor the quadratic, we need two numbers that multiply to \(-21\) (the constant term) and add to \(4\) (the coefficient of \(x\)). After exploring possibilities, the numbers \(7\) and \(-3\) satisfy these conditions because \(7 \times (-3) = -21\) and \(7 + (-3) = 4\).

Step 3: Write Middle Term as Two Separate Terms

Split the middle term \(4x\) into \(7x\) and \(-3x\) so that the equation is expressed as: \[ x^2 + 7x - 3x - 21 = 0. \]

Step 4: Group Terms and Factor by Grouping

Group the terms to factor by grouping: \( (x^2 + 7x) + (-3x - 21) = 0 \). Factor out the common factors from each group. The expression becomes \( x(x + 7) - 3(x + 7) = 0 \).

Step 5: Factor the Common Binomial

Notice that \((x + 7)\) is a common factor. Factor it out: \[ (x + 7)(x - 3) = 0. \]

Step 6: Solve for \(x\)

Set each factor equal to zero and solve for \(x\):1. \( x + 7 = 0 \rightarrow x = -7 \).2. \( x - 3 = 0 \rightarrow x = 3 \).

Step 7: Verify the Solutions

Substitute \(x = -7\) and \(x = 3\) back into the original equation to verify both solutions are correct. The solutions satisfy the equation \( x^2 + 4x - 21 = 0 \).

Key Concepts

Factoring Method

The factoring method is a powerful tool in solving quadratic equations. It involves rewriting the quadratic expression as a product of two binomial expressions. For a quadratic equation of the form:

• \( ax^2 + bx + c = 0 \)

we find two numbers that multiply to the constant term \( c \) and add to the linear coefficient \( b \). This step helps in splitting the middle term and rewriting the quadratic as a factored expression.

In our example given by the equation \( x^2 + 4x - 21 = 0 \), we need numbers that multiply to -21 and add to 4. The numbers 7 and -3 work because:

• \( 7 \times (-3) = -21 \)
• \( 7 + (-3) = 4 \)

This identification allows the middle term \( 4x \) to be split into \( 7x - 3x \), so that the quadratic can be rewritten in a way that reveals its factorable components.

Solving Quadratic Equations

Quadratic equations can be solved via several methods, one of which is factoring, as demonstrated in this exercise. Solving involves finding the values of \( x \) that make the equation equal zero. After factoring the quadratic equation, the next step:

• Express the quadratic in the form \((x - p)(x - q) = 0\)

The solutions to the equation are

• \( x = p \)
• \( x = q \)

For the equation \( (x + 7)(x - 3) = 0 \), it leads to:

• \( x + 7 = 0 \Rightarrow x = -7 \)
• \( x - 3 = 0 \Rightarrow x = 3 \)

These solutions are then verified by substituting back into the original equation to ensure they satisfy the quadratic.

Algebraic Techniques

To efficiently tackle quadratic equations, certain algebraic techniques are essential. Understanding the concept of factoring, grouping, and solving is key.

Let's take a closer look at the original equation:

• Group terms for ease of factorization, i.e., \( (x^2 + 7x) + (-3x - 21) \).
• Extract common factors from each group, resulting in \( x(x + 7) - 3(x + 7) \).
• Notice the repeated binomial \( (x + 7) \), which is a common factor.

By factoring the common binomial, the expression simplifies, leading us to the solutions.

This use of algebraic techniques not only aids in solving the equation but also builds foundational skills for more complex algebraic manipulations.