SOLUTION: Problem Page The area of a rectangle is 65ft^2 and the length of the rectangle is 3ft less than twice the width. Find the dimensions of the rectangle.

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Question 1112053: Problem Page The area of a rectangle is 65ft^2 and the length of the rectangle is 3ft less than twice the width. Find the dimensions of the rectangle.
Found 3 solutions by josgarithmetic, ikleyn, stanbon:
Answer by josgarithmetic(39631) About Me  (Show Source):
You can put this solution on YOUR website!
x%282x-3%29=65 for dimensions x and 2x-3

Answer by ikleyn(52928) About Me  (Show Source):
You can put this solution on YOUR website!
.
W*L = 65


W*(2W-3) = 65


2W^ - 3W - 65 = 0


W%5B1%2C2%5D = %283+%2B-+%28sqrt%289+-+4%2A2%2A%28-65%29%29%29%29%2F%282%2A2%29 = %283+%2B-+23%29%2F4.


Answer.  The width is  W = %283+%2B+23%29%2F4 = 6.5ft.


         The length is  L = 2W - 3 = 10 ft.


Check.  10*6.5  = 65.   ! Correct !

Solved.




Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Problem Page The area of a rectangle is 65ft^2 and the length of the rectangle is 3ft less than twice the width. Find the dimensions of the rectangle.
Width = W
Length = 2W-3
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Equation:
W(2W-3) = 65
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2W^2 - 3W - 65 = 0
W = [3+-sqrt(9-4*2*-65)]/4
W = [3+-sqrt(529)]/4
positive solution:
W = 6.5 ft (width)
2W-3 = 10 ft (length)
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Cheers,
Stan H.
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