Problem 43 - Elementary and Intermediate Algebra

Question

Quadratic Equation Model Problems A student was asked to solve the following problem: The length of a rectangular room is 2 feet more than twice the width. If the area of the room is 60 square feet, find its dimensions. Here is the student's solution: since \(10 \cdot 6=60,\) the length of the room is 10 feet and the width is 6 feet. Explain why his solution is incorrect.

Step-by-Step Solution

Answer

The student's length and width don't satisfy the condition \(l = 2 + 2w\). Correct room dimensions: width = 5 feet, length = 12 feet.

Step 1: Understand the problem

The problem states that the length of the room is 2 feet more than twice the width, and the area of the room is 60 square feet. We need to find the correct dimensions of the room.

Step 2: Define variables

Let's set "w" as the width of the room in feet and "l" as the length of the room in feet. According to the problem statement, the length can be expressed in terms of the width: \( l = 2 + 2w \).

Step 3: Set up the equation for the area

Since the area of a rectangle is given by the formula \( ext{Area} = ext{length} imes ext{width} \), we can write the equation: \( w imes (2 + 2w) = 60 \).

Step 4: Simplify the equation

Simplify the equation from Step 3: \[ w(2+2w) = 60 \]Which expands to:\[ 2w + 2w^2 = 60\]

Step 5: Write the quadratic equation

Rearrange the equation to form a standard quadratic equation:\[ 2w^2 + 2w - 60 = 0\]

Step 6: Simplify the quadratic equation

Divide the entire equation by 2 to simplify it:\[ w^2 + w - 30 = 0\]

Step 7: Solve the quadratic equation

To solve for the width, factor the quadratic equation:\[ (w + 6)(w - 5) = 0 \]Setting each factor equal to zero gives us the possible solutions for "w":- \(w + 6 = 0\) leads to \(w = -6\) which isn't possible since width can't be negative.- \(w - 5 = 0\) leads to \(w = 5\).

Step 8: Find the length using width

Using \(w = 5\), calculate the length: \[ l = 2 + 2w = 2 + 2 \times 5 = 12 \]Thus, the correct dimensions are width = 5 feet and length = 12 feet.

Key Concepts

Rectangular Dimensions

The dimensions of a rectangle are crucial to understanding problems involving length, width, and area. In our exercise, we are dealing with a rectangular room where the relationship between the length and width is explicitly stated. The length is described to be "2 feet more than twice the width." This relationship is expressed algebraically as: - **Length (l)** = 2 + 2 w - **Width (w)** = w, where w represents the width of the room. Understanding how to convert these relationships into algebraic expressions is key when tackling such problems. Rectangular dimensions play a vital role in calculating the area, leading us to solve equations for unknown values effectively.

Problem Solving

Problem solving in mathematics often requires a systematic approach. For the given problem, the challenge is to correctly find the dimensions of a room based on the area and a relationship between length and width. The initial step is always to **understand the problem fully**. Here, it involves:

• Recognizing that the problem gives a specific relationship between the room's length and width.
• Understanding that the total area is provided, which will help us form an equation to solve.

Once we understand the problem, setting up the right equations using the given relationships allows us to find solutions logically. Correctly substituting values and simplifying provide insights that lead to accurate results. In this exercise, the oversight by the student reinforced the importance of using algebra to verify assumptions.

Algebraic Expressions

Algebraic expressions form the backbone of solving problems like these. They provide a means to describe relationships mathematically. Initially, expressions help us translate the words of a problem into symbolic representations. For example, the statement "the length is 2 feet more than twice the width" is translated to the expression: - **Expression:** \( l = 2 + 2w \) Using algebraic expressions, we then construct equations to find unknowns. In our exercise, the equation for the area offers an opportunity to create a quadratic equation: - **Equation:** \( w(2 + 2w) = 60 \) Simplifying and rearranging the equation result in a quadratic form, which we solve to find the possible values for width (w). Such expressions and equations are crucial in not only solving but also in verifying the solution to problems in algebra.