SOLUTION: A rectangle is 4 feet longer than it is wide, and its area is 20 square feet. Find its dimension to the nearest tenth of a foot.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A rectangle is 4 feet longer than it is wide, and its area is 20 square feet. Find its dimension to the nearest tenth of a foot.       Log On


   

Question 876711: A rectangle is 4 feet longer than it is wide, and its area is 20 square feet. Find its dimension to the nearest tenth of a foot.
Answer by LinnW(1048) About Me  (Show Source):
You can put this solution on YOUR website!
set W = width
set L = W + 4
area = L*W
20 = (W + 4)(W)
20 = W^2 + 4W
add -20 to each side
0 = W^2 + 4W -20
Using quadratic equation we get
W = -2%281+%2B+sqrt%286%29%29 or 2%28+sqrt%286%29+-+1+%29
Since we need a positive length, the answer is 2%28+sqrt%286%29+-+1+%29
= 2.90 ft
Let's see if this reasonable.
W = 2.90 , L = 2.90 + 4 = 6.90
2.90 * 6.90 = 20.01 , so we are good