(a%2B1)%3D3.jpeg "(a-1)(a+1)=3")
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.