SOLUTION: Here is the question: If the length of a rectangular field is 8 feet more than double its width. Find the dimensions of the filed if its area is 540 square feet? I know y

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Question 10210: Here is the question:

If the length of a rectangular field is 8 feet more than double its width. Find the dimensions of the filed if its area is 540 square feet?

I know you have to use A = l * w, but how do you start. I started like this:
540 = (8 + x) ( x^2)
would it then be : 540 = 8x + x^2?
Then do you have to subtract 540 from the left side to bring it to the right?

8x + x^2 - 540 = 0

I'm lost. Please help. I appreciate all your help. Thank you.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Well, you've got the right idea!
A = L*W but L = 2W+8 Because the length (L) is 8 ft more (+8) than twice the width (2W).
%282W%2B8%29%28W%29+=+540
2W%5E2+%2B+8W+=+540 Subtract 540 from both sides.
2W%5E2+%2B+8W+-+540+=+0 Solve this quadratic for W. Factor a 2 to simplify it a bit.
w%5E2+%2B+4W+-+270+=+0 Since this doesn't factor you can solve it by using the quadratic formula: W=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a
Here, a = 1, b = 4, and c = -270.
W+=+%28-4%2B-sqrt%284%5E2-4%2A1%2A%28-270%29%29%29%2F%282%2A1%29
W+=+%28-4%2B-sqrt%2816%2B1080%29%29%2F2
W+=+%28-4%2B-sqrt%281096%29%29%2F2
W+=+-2%2B%281%2F2%29%2833.106%29 or W+=+-2-%281%2F2%29%2833.106%29
W+=+-2%2B16.55 or W+=+-2-16.55
W+=+14.55 or W+=+-18.55 Discard the 2nd solution as width can only be positive. Note: sqrt%281096%29+=+33.1058907145 approx, so I rounded it to 33.106 Finally:
The length = 2W+8 = 2(14.55)+8 = 29.1+8 = 37.1 ft.
The width = 14.55 ft.
Check:
A = L*W = (37.1)(14.55) = 539.805 sq.ft. which is pretty darn close to 540 sq.ft.