Problem 13 - College Algebra

Question

Solve the quadratic equation by factoring. $$ 4 x^{2}=9 $$

Step-by-Step Solution

Answer

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

Step 1: Move Terms to One Side

To solve the quadratic equation by factoring, we first need to write it in the standard form. The given equation is: \ \[4x^2 = 9\] \ Subtract 9 from both sides to set the equation to zero: \ \[4x^2 - 9 = 0\]

Step 2: Recognize the Difference of Squares

The equation \(4x^2 - 9 = 0\) is a difference of squares. Recall that \(a^2 - b^2 = (a - b)(a + b)\). \ In this equation, \((2x)^2 - 3^2\), so we can apply the difference of squares: \ \[4x^2 - 9 = (2x - 3)(2x + 3) = 0\]

Step 3: Apply the Zero Product Property

To solve \((2x - 3)(2x + 3) = 0\), use the zero product property, which states if \(ab = 0\), then either \(a = 0\) or \(b = 0\). \ Set each factor to zero: \ \(2x - 3 = 0\) or \(2x + 3 = 0\).

Step 4: Solve Each Equation

Solve \(2x - 3 = 0\): \ Add 3 to both sides: \(2x = 3\) \ Divide by 2: \(x = \frac{3}{2}\) \ Now solve \(2x + 3 = 0\): \ Subtract 3 from both sides: \(2x = -3\) \ Divide by 2: \(x = -\frac{3}{2}\)

Step 5: Verify the Solutions

Substitute \(x = \frac{3}{2}\) and \(x = -\frac{3}{2}\) back into the original equation \(4x^2 = 9\) to verify both solutions work. \ For \(x = \frac{3}{2}\): \ \[4\left(\frac{3}{2}\right)^2 = 4 \times \frac{9}{4} = 9\] \ For \(x = -\frac{3}{2}\): \ \[4\left(-\frac{3}{2}\right)^2 = 4 \times \frac{9}{4} = 9\] \ Both solutions satisfy the original equation.

Key Concepts

Factoring

Factoring is a powerful method to solve quadratic equations. It's about breaking down the quadratic expression into simpler multiplicative components. Here's how you can think about it: imagine the equation is like a puzzle. By factoring, you're taking the puzzle apart into pieces that are easier to understand. For example, consider the equation \(4x^2 - 9 = 0\). Here, the goal is to express this as a product of two expressions.

• We recognize it can be rewritten using specific identities, such as the difference of squares.
• Factoring makes it easier to apply further methods, like the zero product property.

By converting complex expressions into a factored form, you simplify the process of solving the equation.

Difference of Squares

The difference of squares is a specific algebraic pattern that allows a quadratic expression to be broken into two binomials. This pattern is evident in expressions like \(a^2 - b^2\). The formula for the difference of squares is \(a^2 - b^2 = (a - b)(a + b)\).In the example \(4x^2 - 9 = 0\), the terms \((2x)^2 - 3^2\) indicate a difference of squares:

• This identifies that \(4x^2\) is the square of \(2x\), and \(9\) is the square of \(3\).
• When factored, this becomes \((2x - 3)(2x + 3) = 0\).

Recognizing and applying the difference of squares pattern simplifies factoring, making it easier to solve quadratic equations.

Zero Product Property

The Zero Product Property states that if the product of two quantities is zero, then at least one of the quantities must also be zero. Mathematically, if \(ab = 0\), then \(a = 0\) or \(b = 0\).Applying this property to the factored expression \((2x - 3)(2x + 3) = 0\), we establish that:

• Either \(2x - 3 = 0\) or \(2x + 3 = 0\).

This simplifies the process of finding the values of \(x\) that satisfy the original equation. By resolving each factor separately, we systematically determine the solutions to the quadratic equation.

Solving Equations

Solving equations involves finding unknown values that satisfy the expression. In our case, we solve each portion of the equation separately:

• Starting with \(2x - 3 = 0\), we add 3 to both sides to give \(2x = 3\), and then divide by 2 to find \(x = \frac{3}{2}\).
• Then, for \(2x + 3 = 0\), subtracting 3 from both sides results in \(2x = -3\). Dividing by 2 yields \(x = -\frac{3}{2}\).

Checking these solutions by substituting back into the original equation \(4x^2 = 9\) verifies their correctness. Both solutions satisfy the equation, confirming the method used is accurate and effective for determining solutions to quadratic equations.