Problem 37 - Algebra 2

Question

Solve each equation by factoring. \(x^{2}-16=0\)

Step-by-Step Solution

Answer

The solutions are \( x = 4 \) and \( x = -4 \).

Step 1: Recognize the Form

Notice that the equation \( x^2 - 16 = 0 \) is a difference of squares. A difference of squares has the form \( a^2 - b^2 = (a - b)(a + b) \).

Step 2: Identify the Perfect Squares

Identify the perfect squares in the expression. Here, \( x^2 \) is the square of \( x \) and \( 16 \) is the square of \( 4 \). Therefore, the equation can be expressed as \((x)^2 - (4)^2 = 0 \).

Step 3: Apply the Difference of Squares Formula

Using the difference of squares formula, factor \( x^2 - 16 \) as \((x - 4)(x + 4) = 0 \).

Step 4: Solve the Factored Equation

Set each factor equal to zero to find the solutions: \( x - 4 = 0 \) or \( x + 4 = 0 \). Solving these gives \( x = 4 \) and \( x = -4 \).

Key Concepts

Difference of Squares

The concept of the difference of squares is a fundamental algebraic identity. It allows you to factor expressions of the form \(a^2 - b^2\). This specific identity is given as \[(a - b)(a + b)\]It's important because it provides a straightforward way to simplify and solve certain types of equations. You apply this when you recognize both terms as perfect squares.

• For example, in the equation \(x^2 - 16 = 0\), \(x^2\) is the square of \(x\), and \(16\) is the square of \(4\).
• The expression transforms as \((x)^2 - (4)^2\).
• Using the difference of squares formula, it factors into \((x - 4)(x + 4)\).

By understanding this identity, you can quickly factor expressions and find solutions to simple quadratic equations.

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). In simple terms, they are polynomials of degree two. Solving quadratic equations is a significant part of algebra because these equations frequently appear in various mathematical and real-world applications.

• The standard form is helpful when you need to quickly identify how to factor or apply special methods to solve it.
• In the example \(x^2 - 16 = 0\), our equation is already in a simplified quadratic form with \(a = 1\), \(b = 0\), and \(c = -16\).

Sometimes, quadratic equations are complex and require formulas like the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), but when special forms like difference of squares are identified, they provide shortcuts to solutions.

Solving Equations

Solving equations involves finding the values of the variables that make the equation true. With quadratic equations, once you factor an expression, solving becomes a process of setting each factor equal to zero.

• In \((x - 4)(x + 4) = 0\), you solve by setting each part to zero:
• \(x - 4 = 0\) gives \(x = 4\).
• \(x + 4 = 0\) gives \(x = -4\).

This process relies on the zero-product property, which states that if a product equals zero, at least one of the factors must be zero. Therefore, solving quadratic equations by factoring is efficient when expressions can be factored outright or recognized as a difference of squares.