Question

The area of a rectangle is 28 ft2 , and the length of the rectangle is 1 ft more than twice the width. find the dimensions of the rectangle.

Answer 1

W=x
L=2x+1
Area=(2x+1)x=28
2x²+x=28
2x²+x-28=0

(2x-7)(x+4)=0
x=3.5, -4. (3.5 makes more sense)
2x+1=8
area = 28 ft²
width = 3.5 feet | length = 8 feet.

Answer 2

The length is 8feet and the width of the rectangle is 3.5feet

Let the width of the rectangle be represented by w

Let the length of the rectangle be represented by = 2w + 1

Therefore, based on the information given, the equation to solve the question will be:

w(2w + 1) = 28

2w² + w = 28

2w² + w - 28 = 0

2w² + 8w - 7w - 28 = 0

2w(w + 4) - 7(w + 4) = 0

2w - 7 = 0

2w = 0 + 7

2w = 7

w = 7/2

w = 3.5

The width is 3.5 feet

The length will be:

= 2w + 1

= 2(3.5) + 1

= 7 + 1

= 8

The length is 8feet and the width is 3.5feet

Read related link on:

https://brainly.com/question/4193637