Problem 60 - Algebra for College Students

Question

Set up an equation and solve each problem. Suppose that the length of a certain rectangle is two centimeters more than three times its width. If the area of the rectangle is 56 square centimeters, find its length and width.

Step-by-Step Solution

Answer

Width = 4 cm, Length = 14 cm.

Step 1: Define Variables

Let the width of the rectangle be denoted by the variable \( w \) (in centimeters). Then, according to the problem, the length of the rectangle can be defined as \( 3w + 2 \) (in centimeters).

Step 2: Write the Equation for Area

The area of a rectangle is given by the formula \( \, \text{Area} = \text{length} \times \text{width} \, \). Substituting the expressions for length and width, we have:\[ (3w + 2) \times w = 56 \].

Step 3: Expand and Simplify the Equation

Expand the equation:\[ 3w^2 + 2w = 56 \].Rearrange the equation to form a standard quadratic equation:\[ 3w^2 + 2w - 56 = 0 \].

Step 4: Solve the Quadratic Equation

We will use the quadratic formula, \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 2 \), and \( c = -56 \).Calculate the discriminant: \[ b^2 - 4ac = 2^2 - 4 \times 3 \times (-56) = 4 + 672 = 676 \].Find the square root of the discriminant: \( \sqrt{676} = 26 \).

Step 5: Find the Values of Width

Substitute these values into the quadratic formula:\[ w = \frac{-2 \pm 26}{6} \].Calculate the two possible values:1. \( w = \frac{-2 + 26}{6} = \frac{24}{6} = 4 \).2. \( w = \frac{-2 - 26}{6} = \frac{-28}{6} = -\frac{14}{3} \).

Step 6: Select the Valid Width

Since a negative width does not make sense in this context, we choose \( w = 4 \) cm.

Step 7: Calculate the Length

Substitute \( w = 4 \) cm back into the expression for length:\[ 3w + 2 = 3(4) + 2 = 12 + 2 = 14 \].Thus, the length is \( 14 \) cm.

Key Concepts

Area of a Rectangle

Understanding how to calculate the area of a rectangle is essential in this problem-solving process. The area of a rectangle is determined by multiplying its length by its width. This is expressed in the formula:

• Area = Length \( \times \) Width

This formula helps determine how much space is contained within the rectangle's boundaries. In this specific problem, you know the area is 56 square centimeters, providing a starting point to find the missing length and width. Recognizing this relationship between length, width, and area is key to setting up the correct equation.

Defining Variables

When solving algebraic problems, like the area of a rectangle, defining variables is a crucial first step. Variables represent the unknowns you need to find. In this exercise:

• "\( w \)" is the variable chosen to represent the width of the rectangle.
• The problem states the length is "two centimeters more than three times the width," which translates to the expression "\( 3w + 2 \)" for the length.

Defining these variables correctly allows you to translate the word problem into a mathematical equation. This translation forms the foundation for solving the problem using algebraic methods.

Quadratic Formula

The quadratic formula is a powerful tool for solving equations of the form \( ax^2 + bx + c = 0 \). It is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In this exercise, after creating the equation \( 3w^2 + 2w - 56 = 0 \), you apply the quadratic formula to find the variable \( w \). Here, you have:

• \( a = 3 \)
• \( b = 2 \)
• \( c = -56 \)

Calculate the discriminant: \( 2^2 - 4 \times 3 \times (-56) = 676 \), and then find its square root, \( 26 \). Substituting into the formula gives potential solutions for \( w \), helping you determine the rectangle's width.

Problem-Solving in Algebra

Solving algebraic problems often involves several methodical steps. Here’s a breakdown using this rectangle problem:

• **Read and understand the problem:** Identify what is being asked and the relationship between variables.
• **Define the variables:** Set up the variables to represent unknown quantities (like width and length).
• **Formulate an equation:** Use known mathematical formulas and relationships to set up an equation. In this case, the formula for the area of a rectangle.
• **Solve the equation:** Utilize algebraic techniques such as expanding, simplifying, and applying the quadratic formula to find solutions.
• **Interpret the solution:** Evaluate the solutions to make sure they are sensible in the given context (e.g., dismissing negative widths).

Following these steps helps solve not only this problem but lays down a foundation for tackling other algebraic problems efficiently.