SOLUTION: The area of a rectangle is 63 ft squared. the length is 2 ft longer than the width. what are the dimentions of the rectangle.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: The area of a rectangle is 63 ft squared. the length is 2 ft longer than the width. what are the dimentions of the rectangle.      Log On


   



Question 254716: The area of a rectangle is 63 ft squared. the length is 2 ft longer than the width. what are the dimentions of the rectangle.
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
Let W = width and L = length.
the length is 2 ft longer than the width translates into
L = W + 2
A = L x W
A = (W+2)*W = 63
We can see by inspection that W = 7 and W+2 = 9 = L, but I will go through the process any way.
w^2 + 2W = 63
W^2 + 2W - 63 = 0
(W+9)(W-7) = 0
W = -9 which is not possible
W = 7.
If W = 7, then L = 9.