Question

The length of a rectangle is 5 yd less than twice the width, and the area of the rectangle is 33 yd?. Find the dimensions of the rectangle.

Answer 1

Explanation:

L=2W-5

A=LW, using L from above gives you

A=(2W-5)W

A=2W^2-5W

2W^2-5W-A=0, since A=33yd^2

2W^2-5W-33=0

2W^2+6W-11W-33=0

2W(W+3)-11(W+3)=0

(2W-11)(W+3)=0, since W>0

W=11/2

W=5.5yd, since L=2W-5

L=6yd

So the dimensions are W by L, 5.5yd by 6yd