100k views
5 votes
The length of a rectangular room is one less than twice the width. The area of the room is 28 square feet. Find the dimensions of the room.

User DickieBoy
by
8.5k points

2 Answers

3 votes
Equation: x(2x-1)=28

Create a quadratic equation: 2x^2 -x=28
2x^2 -x-28=0

Find roots: plug into quadratic formula, x=4 or x=-7/2

Since x cannot be negative, x=4

Plug into original equation to find the width, width=7

Check: Plug into original equation or just multiply the two using the formula for area
7*4=28
28=28

Length is 4 ft, width is 7 feet
User Afiya
by
8.0k points
2 votes

The first step to solve this problem is to represent variables for the width and the length:

Let w = width of the rectangle

2w – 1 = length of the rectangle

The formula to compute for the area of the rectangle is:

A = LW

Substituting the values and variables to the formula:

28 = w (2w – 1)

2w^2 – w = 28

2w^2 – w – 28 = 0

Solve the quadratic equation:

(2w + 7)(w – 4) = 0

w = -7/2 or w = 4

You cannot use the -7/2 because there is no negative measurement.

W = 4 feet

L = 2(4) – 1 = 7 feet

Therefore the dimension of the rectangle is 4 feet by 7 feet.

User Scott Wisniewski
by
7.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories