## Solving the Equation: (a-1)(a+1) = 3

This equation is a great example of how to use algebraic manipulation and the zero product property to find solutions. Let's break it down step by step:

### 1. Expanding the Equation

We can begin by expanding the left side of the equation using the **difference of squares pattern**: (a-b)(a+b) = a² - b²

Applying this to our equation: (a-1)(a+1) = a² - 1² = a² - 1

Now we have: a² - 1 = 3

### 2. Rearranging the Equation

To solve for 'a', we need to get all the terms on one side and set it equal to zero: a² - 1 - 3 = 0 a² - 4 = 0

### 3. Factoring the Equation

Now we can factor the quadratic expression: (a - 2)(a + 2) = 0

### 4. Applying the Zero Product Property

The **zero product property** states that if the product of two or more factors is zero, then at least one of the factors must be zero.

Therefore, we have two possibilities:

**a - 2 = 0**Solving for 'a', we get: a = 2**a + 2 = 0**Solving for 'a', we get: a = -2

### 5. Solution

Therefore, the solutions to the equation (a-1)(a+1) = 3 are **a = 2** and **a = -2**.