Problem 62 - Algebra 1: Concepts and Skills

Question

Solve the equation. $$ (x+12)(x+7)=0 $$

Step-by-Step Solution

Answer

The solutions to the given equation \((x+12)(x+7)=0\) are \(x = -12\) and \(x = -7\).

Step 1: Break Down the Equation

We are given the equation \((x+12)(x+7)=0\). We can see that this equation is a product of two binomials. This suggests using the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. For any real numbers a and b, if \(a \cdot b = 0\), then either \(a=0\), \(b=0\), or both \(a\) and \(b\) are 0.

Step 2: Set Each Factor Equal to Zero

We need to set each factor equal to zero and solve the resulting equations. This means we solve \(x+12=0\) and \(x+7=0\)

Step 3: Solve the Equations

For \(x+12=0\), subtract 12 from each side of the equation to solve for x. This gives \(x = -12\). For the equation \(x+7=0\), subtract 7 from each side of the equation to solve for x. This gives \(x = -7\).

Key Concepts

Zero Product Property

The zero product property is a fundamental aspect of algebra that simplifies solving certain types of equations. In essence, this property states that if the product of two elements is zero, at least one of the elements must be zero itself. This is mathematically expressed as: if \(a \cdot b = 0\), then \(a = 0\), \(b = 0\), or both \(a\) and \(b\) are zero.

In the context of solving quadratic equations, this becomes extremely useful. Quadratic equations, such as \((x+12)(x+7)=0\), are often presented this way to encourage the use of the zero product property. By setting each binomial factor (i.e., \(x+12\) and \(x+7\)) equal to zero, we can quickly find the possible solutions for \(x\).

Here's why this works:

• If \(x+12 = 0\), solving for \(x\) gives us \(x = -12\).
• If \(x+7 = 0\), solving for \(x\) gives us \(x = -7\).

By breaking down the quadratic into its components using this method, even complex polynomials can be tackled seamlessly.

Solving Equations

Solving equations is the process of finding the values that satisfy the given mathematical statement. This is a crucial skill in algebra as it forms the basis for solving more complex problems. Here, solving starts with understanding the structure of the equation given, which usually involves variables, numbers, and operations. Once the structure is understood, you can begin to manipulate the equation to isolate the variable and find its value.

For the problem \((x+12)(x+7)=0\), we use the zero product property to set each factor equal to zero:

• Solve \(x+12=0\): Subtract 12 from both sides to find \(x = -12\).
• Solve \(x+7=0\): Subtract 7 from both sides to find \(x = -7\).

The steps involve simple arithmetic—adding, subtracting, multiplying, or dividing both sides of the equation to maintain balance and reach the solution. In this example, solving these linear equations individually after using the zero product property is direct and straightforward.

Algebraic Expressions

An algebraic expression is a combination of numbers, variables, and arithmetic operations. Understanding these expressions is foundational to tackling equations, particularly when dealing with objects like binomials, monomials, and polynomials. Each term in an algebraic expression represents part of the problem's structure and possesses specific numerical or variable components.

In a quadratic equation such as \((x+12)(x+7)=0\), each factor (\(x+12\) and \(x+7\)) is itself an algebraic expression known as a binomial. These binomials can be considered as building blocks, and knowing how to manipulate them is crucial to solving the equation. By rewriting the quadratic in factored form, the equation becomes easier to manage using properties like the zero product property.

Key components of an algebraic expression include:

• Variables (e.g., \(x\)), which represent unknown quantities.
• Constants (e.g., 12 and 7), which are known values.
• Operations (e.g., addition and multiplication) connecting the terms.

Grasping how these elements interact allows for deeper understanding and effective application in solving a variety of equations.