Problem 9 - Basic Technical Mathematics with Calculus

Question

$$\text { Solve the given quadratic equations by factoring.}$$ $$x^{2}-4=0$$

Step-by-Step Solution

Answer

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

Step 1: Move Equation to Standard Form

To solve the given quadratic equation by factoring, we must first write it in standard form which is \(ax^2 + bx + c = 0\). The given equation is \(x^2 - 4 = 0\). This equation is already in standard form.

Step 2: Identify Terms for Factoring

Identify terms for factoring. We have \(x^2 - 4\) which is a difference of squares. Recall the formula for the difference of squares: \(a^2 - b^2 = (a - b)(a + b)\).

Step 3: Apply Difference of Squares Formula

Apply the difference of squares formula to factor the expression. Here, \(a^2 = x^2\) and \(b^2 = 4\), so \(a = x\) and \(b = 2\). Therefore, the equation factors to \((x - 2)(x + 2) = 0\).

Step 4: Solve for Roots

To find the roots, set each factor equal to zero. This gives us the equations \(x - 2 = 0\) and \(x + 2 = 0\). Solving these gives \(x = 2\) and \(x = -2\), respectively. These are the solutions to the quadratic equation.

Key Concepts

Factoring Quadratics

When we talk about factoring quadratics, what we mean is breaking down a quadratic expression into simpler factors that, when multiplied, give us the original quadratic. A quadratic equation is anything in the form of \( ax^2 + bx + c = 0 \). The goal with factoring is to transform this into a product of linear equations, like \( (x - r_1)(x - r_2) = 0 \), where \( r_1 \) and \( r_2 \) are the solutions to the equation. The power of factoring lies in its simplicity.

• It reduces a complex equation into simpler, more manageable parts.
• This makes finding the roots or zeros of the equation straightforward.

Factoring is useful when the quadratic can easily be rewritten as a product of polynomials. When the factors are straightforward, it's both fast and efficient. It's also one of the most satisfying methods because it feels like solving a puzzle.

Difference of Squares

In our exercise, the concept of difference of squares comes into play significantly. The difference of squares is a property used in algebra to simplify expressions that involve the subtraction of two perfect squares. The formula is \( a^2 - b^2 = (a-b)(a+b) \).

• This property holds true because when you multiply \( (a-b)(a+b) \), the middle terms cancel out, leaving you with only \( a^2 - b^2 \).
• Recognizing this pattern allows you to factor expressions quickly, as is done in the equation \( x^2 - 4 \).

For our quadratic \( x^2 - 4 = 0 \), we identify \( x^2 \) and \( 2^2 \) as our \( a^2 \) and \( b^2 \), giving us the factors \( (x - 2)(x + 2) \). Thus, this method is powerful when dealing with quadratic equations that can be rewritten as the subtraction of two perfect squares.

Standard Form

The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). This is the starting point for many algebra techniques used to solve these problems. In this equation, \( a, b, \) and \( c \) are constants, and \( x \) is the variable.

• The term \( ax^2 \) represents the quadratic part, and \( a \) is the coefficient of \( x^2 \).
• The term \( bx \) is the linear part, and \( b \) is the coefficient of \( x \).
• Finally, \( c \) is the constant term without any variables attached.

Our given equation \( x^2 - 4 = 0 \) is already in this form, making it easy to proceed with factoring. By ensuring the equation is in standard form first, you simplify the process of executing other methods like factoring, solving graphically, or using the quadratic formula. Each part of the standard form equation plays a role in identifying the solutions to the quadratic.