Question

a parking lot had a rectangular are of 40,000 yd. the length is 200 yd more than twice the width. what are the dimension of the lot?

Answer 1

First I would draw a rectangle.
The length is 200 yards more than twice the length
So set length to 2x+200 and the width is x
The area is 40,000 so: (2x+200)(x)=40000 Since L(w)=Area
(2x+200)(x)=40000
2x^2+200x=40000
Subtract 40,000 from both sides and set to zero
2x^2+200x-40,000=0
Factor (may be easier to factor out a 2 first.
2(x^2+100x-200)=0
2(x-100)(x+200)=0