SOLUTION: The width of a rectangle is 2 units less than the length. If the area is 48 square units, then find the dimensions of the rectangle.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: The width of a rectangle is 2 units less than the length. If the area is 48 square units, then find the dimensions of the rectangle.      Log On


   



Question 1089290: The width of a rectangle is 2 units less than the length. If the area is 48 square units, then find the dimensions of the rectangle.
Found 2 solutions by addingup, MathTherapy:
Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
L x W = Area
L x (L-2) = 48
L^2-2L = 48
The coefficient of L is -2. Divide it by 2 and you get -1. Square -1 and add to both sides to make the perfect square:
L^2-2L+(-1)^2 = 48+(-1)^2 = L^2-2L+1 = 49
(L-1)^2 = 49
sqrt(L-1)^2 = sqrt49
L-1 = 7 or L-1 = -7
L = 8 or L = -6
We are not looking for a negative, so let's try the 8:
L x (L-2) = 48
8 x (8-2) = 48 Correct.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

The width of a rectangle is 2 units less than the length. If the area is 48 square units, then find the dimensions of the rectangle.
You just need to find TWO (2) factors of 48 with a DIFFERENCE of 2. That's all!!