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Chapter 1: Problem 145

The length of a rectangular sign is 3 feet longer than the width. If the sign's area is 54 square feet, find its length and width.

Short Answer

Expert verified
The dimensions of the rectangle are width = 6 feet and length = 9 feet.

Step by step solution

01

Write down the equations

From the problem, we know that the area of a rectangle is given by the product of its length and width. So, we can write the following equation: \( width \times length = 54 \). In addition, we are told that the length is 3 feet longer than the width. This gives us a second equation: \( length = width + 3 \). Now we have a system of two equations.
02

Substitute the second equation into the first one

Substitute the second equation into the first equation, we have \( width \times (width + 3) = 54 \), which simplifies to \( width^2 + 3width - 54 = 0 \).
03

Solve the quadratic equation

This is a quadratic equation of the form \( ax^2 + bx + c = 0 \). It can be factored and solved. The factors are \( (width -6)(width + 9) = 0 \). Solving, we find the possible values for the width: width = 6 or width = -9. However, width cannot be negative so we discard -9. Thus, width = 6.
04

Substitute width value into the second equation to find length

Substituting width = 6 into the equation \( length = width + 3 \) we get \( length = 6 + 3 = 9 \).

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