Solving Rectangle Dimensions with Quadratic Equations

Quadratic equations provide a powerful tool to solve problems involving rectangles, especially when the relationship between length and width is given. This article demonstrates how to solve for the dimensions of a rectangle whose area is known, using the quadratic formula. We'll explore different methods to arrive at the correct solution, ensuring a comprehensive understanding of the problem.

Problem Statement

The length of a rectangle is 3 feet longer than its width, and the area of the rectangle is 54 square feet. We need to find the measure of the width and the length of the rectangle.

Solving the Problem with Variables and Equations

Let's define the variables:

x represents the width of the rectangle in feet.

The length can be expressed as x 3 (since it is 3 feet longer than the width).

The area of a rectangle is given by the formula:

Area length times; width

According to the problem:

The area is 54 square feet.

The length is x 3.

The width is x.

Therefore, we can set up the equation:

x(x 3) 54

Expanding and rearranging to standard quadratic form:

x^2 3x - 54 0

We can solve this quadratic equation using the quadratic formula:

x u00BD(-b u00B1 u221A(b^2 - 4ac))

Here, a 1, b 3, and c -54. Plugging in these values:

ud835udc53 u00BD(-3 u00B1 u221A3^2 - 4 cdot 1 cdot -54)

ud835udc53 u00BD(-3 u00B1 u221A(9 216))

ud835udc53 u00BD(-3 u00B1 u221A225)

ud835udc53 u00BD(-3 u00B1 15)

This gives us two possible solutions:

ud835udc53 u00BD(12) 6 and ud835udc53 u00BD(-18) -9

Since the width cannot be negative, we take the positive solution:

x 6

The length of the rectangle is:

6 3 9

Summary:

The width of the rectangle is 6 feet.

The length of the rectangle is 9 feet.

Alternative Methods to Solve the Problem

Let's explore another method to solve the same problem, this time using factoring:

Factorize 54 to find the possible pairs of factors that can give us an area of 54 square feet:

54 2 x 27

54 6 x 9

54 18 x 3

54 54 x 1

The difference between the two factors must be 3:

The pair (6, 9) has a difference of 3.

So, the width and length are:

Width 6 feet

Length 9 feet

Conclusion

In both methods, we have arrived at the same solution: the width of the rectangle is 6 feet, and the length is 9 feet. This demonstrates the robustness of the quadratic equation approach and the simplicity of factorization in solving such problems.