Question

The length of a rectangle is 5 yd less than twice the width, and the area of the rectangle is 52 yd^2. Find the dimensions of the rectangle.
Please help ASAP!

Answer 1

Let the width be w.
The length is 5 yd less than twice the width, so the length is 2w - 5.
The area is LW, and the area is 52 yd^2

LW = A

(2w - 5)(w) = 52

2w^2 - 5w = 52

2w^2 - 5w - 52 = 0

2w^2 - 13w + 8w - 52 = 0

w(2w - 13) + 4(2w - 13) = 0

(2w - 13)(w + 4) = 0

2w - 13 = 0   or   w + 4 = 0

2w = 13   or   w = -4

w = 6.5   or   w = -4

Since a width cannot be a negative number, discard the solution w = -4.

The width is 6.5 yd.
The length is 2w - 5 = 2(6.5) - 5 = 13 - 8 = 8
The length is 8 yd.

The length is 8 yd, and the width is 6.5 yd.