Problem 78 - Elementary and Intermediate Algebra

Question

Solve each equation. $$ z^{2}-25=0 $$

Step-by-Step Solution

Answer

z = 5 or z = -5

Step 1: Identify the Equation Type

Notice that the equation \( z^2 - 25 = 0 \) is a difference of squares. This type of equation can be factored using the identity \( a^2 - b^2 = (a - b)(a + b) \). In this equation, \( a = z \) and \( b = 5 \) since \( 5^2 = 25 \).

Step 2: Factor the Equation

Apply the difference of squares formula to factor the expression: \( z^2 - 25 = (z - 5)(z + 5) = 0 \). This gives us two separate factors that can each be set to zero

Key Concepts

Difference of Squares

The concept of "difference of squares" is an essential tool when dealing with certain quadratic equations. It's based on a simple algebraic identity:

• \[ a^2 - b^2 = (a - b)(a + b) \]

This states that the difference between two squared terms can be expressed as the product of a sum and a difference. Let's break it down. If you have two perfect squares, like \( z^2 \) and \( 25 \), you can express them in a subtraction form as \( z^2 - 25 \). Here, \( z^2 \equiv a^2 \) and \( 25 \equiv b^2 \), where \( a = z \) and \( b = 5 \) because \( 25 = 5^2 \).
Applying the formula, the equation becomes \((z - 5)(z + 5)\). Understanding this principle is pivotal as it simplifies solving more complicated equations through factoring.

Factoring Quadratic Equations

Factoring quadratic equations involves rewriting the equation in a form that equates the expression to zero using its factors. Given an equation such as \( z^2 - 25 = 0 \), the goal is to express it in terms of its linear factors:

• Identify the equation's type. Recognize special patterns like difference of squares that allow easy factoring.
• Use appropriate identities or methods to rewrite the original equation into its factors.

In our specific problem \( z^2 - 25 = 0 \), recognizing it as a difference of squares immediately guides us to the solution. Thus, we rewrite it as \((z - 5)(z + 5) = 0\).
Factoring reveals the roots of the equation effectively, making this technique very efficient.

Solving Quadratic Equations

Solving quadratic equations typically follows factoring the equation. Once factored, you can find the solutions by setting each factor to zero, as a product equals zero only if one or more of its factors are zero themselves.

• Start with a factored equation like \((z - 5)(z + 5) = 0\).
• Set each factor to zero: \( z - 5 = 0 \) and \( z + 5 = 0 \).

Solve each simple equation:

• For \( z - 5 = 0 \), add 5 to both sides to get \( z = 5 \).
• For \( z + 5 = 0 \), subtract 5 from both sides to get \( z = -5 \).

Thus, the solutions to the equation \( z^2 - 25 = 0 \) are \( z = 5 \) and \( z = -5 \). This demonstrates how factoring can lead directly to finding solutions, providing a straightforward path through which to resolve quadratic equations.