Problem 18 - Algebra 1: Concepts and Skills

Question

Use the zero-product property to solve the equation. \((b-9)(b+8)=0\)

Step-by-Step Solution

Answer

The solutions to the equation \((b-9)(b+8) = 0\) are \(b=9\) and \(b=-8\)

Step 1: Set the first factor equal to zero

Start by setting the first factor equal to zero, that is \(b-9 = 0\). To solve this for \(b\), just add 9 to both sides of the equation which would give \(b=9\)

Step 2: Set the second factor equal to zero

Next, set the second factor equal to zero, that is \(b+8 = 0\). To solve this for \(b\), subtract 8 from both sides of the equation which results in \(b=-8\)

Step 3: Combine the solutions

The zero-product property gives two solutions for the equation, which are independent of each other. Therefore, the solutions for the given equation are \(b=9\) and \(b=-8\)

Key Concepts

Solving Quadratic Equations

When you encounter a quadratic equation, you're dealing with an expression that is usually set to zero and has the highest exponent of 2. The form of a basic quadratic equation is \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable you need to solve for. One effective method to solve these equations is by factoring the quadratic expression into a product of two binomials.

In the exercise provided, \(b^2 - b - 72 = 0\), the equation has already been factored for you as \(b-9)(b+8)=0\). To find the values of \(b\), we use the zero-product property, which states if a product of two factors is zero, at least one of the factors must be zero. Setting each factor equal to zero gives us the potential solutions for \(b\). In simpler terms, you get two smaller and more manageable equations to solve, leading to the final answers.

Factoring Polynomials

Factoring is the process of breaking down a complex polynomial into more straightforward terms or factors that, when multiplied together, give back the original polynomial. It's similar to splitting a number into its multiplication factors. For example, the number 15 can be factored into 3 and 5 because 3 times 5 equals 15.

When factoring quadratic polynomials, the goal is to find two numbers that add up to the middle term and multiply to the constant term. Let's take \(x^2 + 5x + 6\) as an example. We're looking for two numbers that add up to 5 and multiply to 6. The numbers 2 and 3 fit this description, so the polynomial factors to \(x+2)(x+3)\).

Identifying Factors

With the equation from our initial exercise, \(b^2 - b - 72\), the two numbers we would need are -9 and 8, because they multiply to -72 and add to -1. This technique of finding the right combination is crucial when factoring polynomials to solve algebraic equations.

Algebraic Equations

Algebraic equations are fundamental to algebra and serve as recipes for finding unknown values. These equations can range from simple, like \(x + 2 = 5\), to complex quadratic equations such as \(x^2 - 4x + 4 = 0\). Solving them is like unraveling a mystery; you perform a series of operations to isolate the variable and discover its value.

There are various strategies for solving algebraic equations, including factoring, using square roots, completing the square, or applying the quadratic formula. The beauty of algebra lies in its flexibility; often, multiple methods can lead to the same solution. For instance, in our textbook exercise, we isolated the variables by using the zero-product property after the polynomial was factored. This approach is particularly powerful because it simplifies the problem into components we can solve with basic arithmetic.