Here's the problem: The length of a rectangle is four meters less than twice the width. If the area of the rectangle of the square is 96 meters, what is the width?
2007-12-16 06:17:35 · 4 answers · asked by KaylaKK 2 in Science & Mathematics ➔ Mathematics
Area = A
Width = W
Length = L
2w-4 =L
A = L * W
96 = (2W-4)
2W^2 - 4W - 96 = 0
W^2 - 2W - 48 = 0
(W-8)(W+6) = 0
W = 8
2007-12-16 06:28:59 · answer #1 · answered by Apathy 2 · 0⤊ 0⤋
Let w = the width
2w-4 = length
Area = length x width
96 = w(2w-4)
2w^2 - 4w - 96 = 0
w^2 - 2w - 48 = 0
(w-8)(w+6) = 0
w = 8 or w = -6
width is 8 meters (can't be a negative six!!)
that's it! :)
2007-12-16 14:23:44 · answer #2 · answered by Marley K 7 · 0⤊ 0⤋
Let w = width
Then 2w-4 = length
Area = l*w = w(2w-4) = 2w^2-4w
So 2w^2-4w = 96
w^2-2w-48 = 0
(w-8)(w+6) = 0
w = 8 = width in meters
2007-12-16 14:25:34 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
L=2W-4
A=96m
A=L*W
A= (2W-4)*W
96=2W^2-4W
2W^2-4W-96=0
W^2-2W-48=0
I will allow you to take it from there! I am not going to do the whole thing for you. I hope you can use algebra to figure the rest out!
2007-12-16 14:26:35 · answer #4 · answered by Andy H 3 · 0⤊ 0⤋