Problem 20 - Algebra 1: Concepts and Skills

Question

Use the zero-product property to solve the equation. \((y-2)(y+1)=0\)

Step-by-Step Solution

Answer

The solutions to the equation are \(y=2\) and \(y=-1\).

Step 1: Apply the Zero-Product Property

Applying the zero-product property to the equation \((y-2)(y+1)=0\), this equation is true if \(y-2=0\) or \(y+1=0\).

Step 2: Solve for y in each equation

Solving the equation \(y-2=0\), find \(y=2\). Similarly, solving the equation \(y+1=0\), find \(y=-1\)

Key Concepts

Solving Quadratic Equations

Quadratic equations are mathematical expressions that include a variable raised to the power of two. These equations are typically written in the standard form: \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. Quadratic equations are called so because 'quad' means four, indicating the variable squared.
To solve quadratic equations, there are several methods that can be applied, such as:

• Factoring: Writing the quadratic equation as a product of its factors can simplify it into a solvable format.
• Completing the Square: Transforming the quadratic into a perfect square trinomial helps isolate the variable.
• Using the Quadratic Formula: This is a general formula that can solve any quadratic equation.

By choosing the appropriate method, one can find the values of \( x \) that satisfy the equation. In our exercise, we use the zero-product property, which involves the factoring technique.

Zero-Product Property

The zero-product property is a fundamental concept in algebra. It states that if the product of two numbers is zero, then at least one of the numbers must be zero. In equation form, if \( a \cdot b = 0 \), then either \( a = 0 \) or \( b = 0 \).
This property is particularly useful in solving quadratic equations that can be factored into a product of binomials. For instance, in the original exercise with the equation \((y-2)(y+1)=0\), the zero-product property allows us to split the problem into two separate equations:

• First equation: \(y-2=0\)
• Second equation: \(y+1=0\)

By solving each equation separately, we find the solutions to the original quadratic equation. This eliminates the need for more complex methods when factoring is feasible.

Algebraic Equations

Algebraic equations involve a combination of numbers and variables linked by operations such as addition, multiplication, etc. Solving these equations requires finding the value of the variable that makes the equation true.
There are various types of algebraic equations, and each has a distinct method of solving:

• Linear Equations: These take the form \( ax + b = 0 \) and can be solved by isolating the variable.
• Quadratic Equations: As discussed, they have a squared variable and require specific techniques like factoring.
• Polynomial Equations: These include variables raised to more than two powers and require factoring or synthetic division.

In our specific exercise, we dealt with a quadratic equation that was already factored, simplifying the process of applying the zero-product property to find the solutions. A strong grasp of algebraic equations and properties allows for effective and efficient problem-solving in any mathematical context.