SOLUTION: a school is building a rectangle soccer field that has an area of 6000 square yards. the soccer field must be 40 yards longer than its width. determine algebraically the dimensions

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Question 1016172: a school is building a rectangle soccer field that has an area of 6000 square yards. the soccer field must be 40 yards longer than its width. determine algebraically the dimensions of the soccer field.
Found 2 solutions by LinnW, macston:
Answer by LinnW(1048) About Me  (Show Source):
You can put this solution on YOUR website!
Length = width + 40
Set w = width
Area = length * width
6000 = (w + 40)(w)
6000 = w^2 + 40w
add -6000 to each side
0 = w^2 + 40w -6000
0 = (w-60)(w+100)
w = 60 or w = -100
Since we need a positive width, w = 60 and length = 100

Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
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W=width; L=length=W+40yds
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LW=6000yd^2
(W+40yd)(W)=6000yd^2
W^2+40W=6000
W^2+40W-6000=0
(W-60)(W+100)=0
W-60=0 -OR- W+100=0
W=60 -OR- W=-100
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ANSWER 1: The width is 60 yards.
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L=W+40yds-60yds+40yds=100yds
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ANSWER 2: The length is 100 yards.
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