Problem 2 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Solve the equation by factoring, if required: $$ (y-3)(y-4)=0 $$

Step-by-Step Solution

Answer

The equation is already in factored form: \((y-3)(y-4)=0\). Using the Zero Product Property, we solve for \(y\) in each factor: \(y-3=0\) gives \(y=3\), and \(y-4=0\) gives \(y=4\). Therefore, the solutions are \(y=3\) and \(y=4\).

Step 1: Identify the factors

The given equation is already in the factored form: \((y-3)(y-4)=0\), where factors are \(y-3\) and \(y-4\).

Step 2: Apply the Zero Product Property

According to the Zero Product Property, if \(a \times b = 0\), then either \(a=0\) or \(b=0\). So, we need to find the values of \(y\) that make the factors equal to zero. That is, we will check for which values of \(y\), either \(y-3 = 0\) or \(y-4 = 0\).

Step 3: Solve for y in each factor

To find the values of \(y\) that satisfy the equation, we will solve each factor for \(y\): Factor 1: \(y-3 = 0\) $$ y = 3 $$ Factor 2: \(y-4 = 0\) $$ y = 4 $$

Step 4: Write the solution

The solutions to the equation \((y-3)(y-4)=0\) are \(y=3\) and \(y=4\).

Key Concepts

Zero Product Property

The Zero Product Property is a fundamental concept in algebra that's essential for solving quadratic equations by factoring. Whenever you multiply two numbers or expressions and the result is zero, at least one of the numbers or expressions must be zero. In mathematical terms, if \(a \times b = 0\), then either \(a = 0\) or \(b = 0\) or both.
This property is particularly useful when dealing with equations that have been factored into a product of two or more expressions. It helps us determine the values of the variable that solve the equation by setting each factor equal to zero and solving them individually.
In the example given, \((y-3)(y-4) = 0\), the Zero Product Property indicates that either \(y-3 = 0\) or \(y-4 = 0\). Solving these equations gives us the solutions to the problem. So this property helps eliminate possibilities and narrows down solutions simply by equating the factors to zero.

Solving Quadratic Equations

Solving quadratic equations involves finding the value(s) of the variable that make the equation true. There are various methods to achieve this, with factoring being one of the most straightforward when applicable.
When you have a quadratic equation in its expanded form, such as \(ax^2 + bx + c = 0\), the first step is to transform it into a format that allows factoring. If it's already in a product of factors, like the exercise provided \((y-3)(y-4) = 0\), we are one step ahead.

• Identify the factors of the equation.
• Apply the Zero Product Property by setting each factor to zero.
• Solve each equation to find the possible values of the variable.

Solving these resulting linear equations leads directly to the solutions of the original quadratic equation. Practice with multiple examples enhances understanding of this method and its efficiency in finding roots of a quadratic without needing to use more complex techniques.

Algebraic Solutions

Algebraic solutions are the expressions or numbers that solve a given algebraic equation. In our exercise, finding the algebraic solutions entails solving each linear equation generated from the factors of the original quadratic equation.
Once the Zero Product Property is applied, each factor is individually set to zero forming simple linear equations, such as \(y - 3 = 0\) and \(y - 4 = 0\). Solving these is straightforward and involves basic algebraic manipulation:

• For \(y - 3 = 0\), add 3 to both sides to solve for \(y\), resulting in \(y = 3\).
• For \(y - 4 = 0\), add 4 to both sides for \(y = 4\).

The solutions \(y = 3\) and \(y = 4\) are the algebraic solutions to the original equation. These solutions correspond to the points where the function intersects the y-axis, verifying that the zero-product property and factoring method are reliable techniques for solving quadratic equations efficiently.