SOLUTION: The length of a rectangle is 8ft greater than it's width. If the width were doubled, the area of the rectangle would be 84 sq. ft. greater than it is now. Find the dimensions of th

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Question 220: The length of a rectangle is 8ft greater than it's width. If the width were doubled, the area of the rectangle would be 84 sq. ft. greater than it is now. Find the dimensions of the deck.
Found 2 solutions by KMST, MathTherapy:
Answer by KMST(5484) About Me  (Show Source):
You can put this solution on YOUR website!
I suspect that the shape of the deck you only mention in the last sentence is the rectangle mentioned in the first sentence.
w = width of the deck, in feet
w%2B8 = length of the deck, in feet
If the width were 2w , with the same length, the area in square feet, would be

EDIT: Oops!
Everything from here down is wrong.
Thanks to MathTherapy for posting the correct answer to the problem.
I misread the problem as
"If the width were doubled, the area of the rectangle would be 84 sq. ft",
instead of "... would be 84 sq. ft. greater than it is now."
2w%2A%28w%2B8%29=84-->w%2A%28w%2B8%29=42-->w%5E2%2B8w=42-->w%5E2%2B8W%2B16=42%2B16-->%28w%2B4%29%5E2=58-->w%2B4=%22+%22+%2B-+sqrt%2858%29-->w=-4+%2B-+sqrt%2858%29
The negative result cannot be a width,
so highlight%28w=sqrt%2858%29-4%29=3.616%28rounded%29%29 is the width of the deck
The length of the deck is
w%2B8=sqrt%2858%29-4=highlight%28sqrt%2858%29%2B4%29=11.616%28rounded%29 .
The area of the deck, in square feet, is
%28sqrt%2858%29-4%29%28sqrt%2858%29-4%29=58-16=highlight%2842%29 ,
which could have been calculated as half of what the area would be with double the width:
84sqft%2F2=42sqft

Answer by MathTherapy(10898) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 8ft greater than it's its width. If the width were doubled, the area of the rectangle would be 84 sq. ft.
greater than it is now. Find the dimensions of the deck.
******************
Let actual width be W
Then actual length = W + 8
Actual area: W(W + 8) = W%5E2+%2B+8W

Hypothetical width (doubled): 2W
Hypothetical area: 2W(W + 8) = 2W%5E2+%2B+16W
Increase in actual area to arrive at hypothetical area: 84 sq ft
We then get: 2W%5E2+%2B+16W+=+W%5E2+%2B+8W+%2B+84
 2W%5E2+%2B+16W+-+W%5E2+-+8W+-+84+=+0
                    W%5E2+%2B+8W+-+84+=+0
            (W - 6)(W + 14) = 0
   W, or Actual width = 6 ft     OR     W = - 14 (IGNORE)
Actual length: W + 8 = 6 + 8 = 14 ft