The rectangle has length that is 5 feet less than twice its width and it has an area of 85.9 square feet. What are the dimensions to the nearest tenth of a foot?
Let W = width of rectangle
Then L = 2W - 5 (length that is 5 feet less than twice its width)
Area is A = WL
85.9 = WK = W(2W - 5)
85.9 = 2w^2 - 5W
2W^2 - 5W - 85.9 = 0
Using thhe quadratic formula,
W = [5 +/- sqrt(5^2 - 4*2*(-85.9))]/(2*2)
W =[5 +/- sqrt(25 + 687.2)]/4
W = [5 +/- sqrt(712.2)]/4
W = 1.25 +/- 13.34
W = 1.25 + 6.67 (only positve solutions allowed)
Width = 7.9 feet