SOLUTION: The length of a rectangle sign is 4 feet shorter than twice the width. The area of the rectangle is 48 squared feet. Could you help me find the dimensions of the sign? I tried s

Algebra ->  Rectangles -> SOLUTION: The length of a rectangle sign is 4 feet shorter than twice the width. The area of the rectangle is 48 squared feet. Could you help me find the dimensions of the sign? I tried s      Log On


   

Question 578999: The length of a rectangle sign is 4 feet shorter than twice the width. The area of the rectangle is 48 squared feet. Could you help me find the dimensions of the sign?
I tried setting it up as
l(w)= 48sq.ft
2w-4=l
But it's not quite working.
Help please and thanks :)
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
your equations are okay

substituting ___ (2w - 4)w = 48

dividing by 2 ___ (w - 2)w = 24 ___ w^2 - 2w - 24 = 0

factoring ___ (w - 6)(w + 4) = 0

w + 4 = 0 ___ w = -4 ___ negative value not realistic

w - 6 = 0 ___ w = 6

l = 2(6) - 4 ___ l = 8