Problem 12 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations by factoring. $$2 x^{2}=0.32$$

Step-by-Step Solution

Answer

The solutions are \( x = 0.4 \) and \( x = -0.4 \).

Step 1: Express in Standard Form

First, let's express the equation in standard quadratic form, which is \( ax^2 + bx + c = 0 \). Start by rewriting the given equation: \( 2x^2 = 0.32 \). Move all terms to one side to set the equation to zero: \( 2x^2 - 0.32 = 0 \).

Step 2: Factor Out the Coefficient

Identify and factor out the greatest common factor from the equation. In this case, both terms have a factor of 2. Factor out 2: \( 2(x^2 - 0.16) = 0 \).

Step 3: Focus on the Quadratic Expression

Since there is a factor of 2 outside the parentheses, our focus is on solving the equation inside: \( x^2 - 0.16 = 0 \). To factor this quadratic expression, recognize it as a difference of squares: \( (x)^2 - (0.4)^2 = 0 \).

Step 4: Factor the Quadratic

Apply the difference of squares formula, \( a^2 - b^2 = (a - b)(a + b) \). With \( a = x \) and \( b = 0.4 \), we factor \( x^2 - 0.16 \) as: \( (x - 0.4)(x + 0.4) = 0 \).

Step 5: Solve for Roots

Set each factor equal to zero and solve for \( x \). First, solve \( x - 0.4 = 0 \), giving \( x = 0.4 \). Next, solve \( x + 0.4 = 0 \), giving \( x = -0.4 \).

Step 6: Verify the Solutions

To ensure accuracy, substitute the roots back into the original equation to verify: For \( x = 0.4 \), substitute into \( 2x^2 = 0.32 \), confirming it equals 0.32. Similarly, for \( x = -0.4 \), substituting also results in 0.32. Thus, both are verified as correct solutions.

Key Concepts

Factoring

Factoring is a method used to solve quadratic equations. It involves expressing the equation as a product of simpler expressions. When we factor, we are essentially reversing the process of expanding two binomials into a quadratic expression. In our specific exercise, after moving all terms to one side to get a zero on the other, the focus is on the expression inside the parentheses: 2(x^2 - 0.16) = 0.

• The first step involves identifying any common factors. In this case, both terms share 2, which we factor out.
• Next, focus on the quadratic expression inside the parentheses. This can often be decomposed into two binomial factors by recognizing patterns like the Difference of Squares.

The goal is to rewrite the quadratic equation so each factor can be individually set to zero, allowing us to find the roots easily.

Standard Form

A quadratic equation is in standard form when it is expressed as \( ax^2 + bx + c = 0 \).This form is essential for recognizing and applying various mathematical techniques to solve the equation, including factoring.

• The exercise starts by expressing the equation \( 2x^2 = 0.32 \) in standard form. \( 2x^2 - 0.32 = 0 \) is the resulting expression.
• By rearranging the equation to set it against zero, you create an equation open to solutions using factoring or other methods.

Being in standard form lays the groundwork for factoring or using alternative methods like the quadratic formula or completing the square.

Difference of Squares

The difference of squares is a specific factoring technique used for expressions that fit the pattern of \( a^2 - b^2 \). This can be rewritten as the product \( (a-b)(a+b) \). Identifying this pattern allows for quick and straightforward factoring.

• In our problem, the quadratic expression \( x^2 - 0.16 \) becomes a difference of squares: \( (x)^2 - (0.4)^2 \).
• We then rewrite it using the formula for the difference of squares to factor it into \( (x - 0.4)(x + 0.4) = 0 \).

Recognizing the difference of squares simplifies factoring because it breaks the quadratic into two binomial expressions, each derived from the square roots of the terms involved.

Roots Verification

Finding the roots of a quadratic equation provides the solutions to the equation when set to zero. The roots are the values of the variable that make the equation true. After factoring and solving each factor separately, the roots are initially hypothetical solutions.

• After determining the roots of the equation \( x^2 - 0.16 =0 \) are \( x = 0.4 \) and \( x = -0.4 \), we verify these solutions by substituting them back into the original equation.
• For example, substituting \( x = 0.4 \) and \( x = -0.4 \) back into \( 2x^2 = 0.32 \) confirms the equation equals 0.32 for both values, thereby validating them as true solutions.

Verification ensures accuracy and confirms that no arithmetic errors were made during calculations. It is a final check in the problem-solving process.