Problem 10 - College Algebra with Corequisite Support

Question

For the following exercises, solve the quadratic equation by factoring. $$ 4 x^{2}-12 x+8=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = 1 \) and \( x = 2 \).

Step 1: Identify the Common Factor

First, observe the equation \( 4x^2 - 12x + 8 = 0 \). Notice that all terms have a common factor of 4. We can factor this out to simplify the equation. Factoring out 4 gives us: \[ 4(x^2 - 3x + 2) = 0 \]. Since the factor of 4 does not affect the solutions, we can focus on solving \( x^2 - 3x + 2 = 0 \).

Step 2: Set Up the Factored Form

We need to express \( x^2 - 3x + 2 \) as a product of two binomials. We write it in this form: \( (x - p)(x - q) = 0 \). Our goal is to find numbers \( p \) and \( q \) such that \( pq = 2 \) (the constant term) and \( p + q = 3 \) (the coefficient of the middle term).

Step 3: Determine the Numbers p and q

We look for two numbers that multiply to 2 and add to 3. These numbers are 1 and 2 because \( 1 \times 2 = 2 \) and \( 1 + 2 = 3 \).

Step 4: Write the Factored Equation

Using the values of \( p \) and \( q \), we can now write the equation as \( (x - 1)(x - 2) = 0 \).

Step 5: Solve for the Roots

To find the solutions to \( (x - 1)(x - 2) = 0 \), we set each factor equal to zero: \( x - 1 = 0 \) or \( x - 2 = 0 \). Solving these gives us \( x = 1 \) or \( x = 2 \).

Key Concepts

Factoring

When solving quadratic equations, factoring is a powerful method. It involves writing the quadratic as a product of simpler expressions, often binomials, which can be easier to solve. Let's break down the process of factoring:

• First, identify any common factors in all the terms of the quadratic equation. By factoring out the greatest common factor, you simplify the problem significantly.
• For example, in the equation \( 4x^2 - 12x + 8 = 0 \), you notice that 4 is a common factor for all terms. Factoring it out, you get \( 4(x^2 - 3x + 2) = 0 \).
• Once the equation is simplified by removing the common factor, focus on the remaining expression inside the parentheses.

After simplifying, the next step is to express the quadratic as a product of binomials. This is typically written as \((x - p)(x - q)\) where \( p \) and \( q \) are numbers that add and multiply to give you the coefficients of the middle and constant terms, respectively.

Polynomials

Polynomials are expressions consisting of variables and coefficients, combined using addition, subtraction, multiplication, and non-negative integer exponents. In the context of quadratic equations, polynomials usually appear as a second-degree expression, meaning the highest power of the variable is 2. Quadratics come in the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.

• The equation \( 4x^2 - 12x + 8 = 0 \) is a typical quadratic polynomial. Here, the variable \( x \) is adjusted through degrees and coefficients to model various situations in algebra.
• By restructuring these terms, we simplify analysis and solutions. This makes the polynomial easier to handle.

Understanding polynomials' structure is crucial, as it allows for easier manipulation when we set the goal of isolating the variable to find solutions.

Solving Equations

Solving equations involves finding the value(s) of the variable(s) that satisfy the equation. For quadratic equations, solving is particularly about finding the roots of the equation—values which make the equation true when substituted back into it. To solve the equation \( (x - 1)(x - 2) = 0 \), you use the property of zero products which states that if a product of factors equals zero, at least one factor must also equal zero. To find the roots:

• Set each factor equal to zero: \( x - 1 = 0 \) and \( x - 2 = 0 \).
• Solve for \( x \). When \( x - 1 = 0 \), then \( x = 1 \). Similarly, \( x - 2 = 0 \) gives \( x = 2 \).

Hence, the solutions or the roots of the quadratic equation are 1 and 2. By substituting these values back, you can verify that they do indeed satisfy the original equation.