SOLUTION: if the length of one side of a square is doubled and the length of an adjacent side is decreased by 3, the area of the resulting rectangle exceeds the area of the original square b

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Question 570236: if the length of one side of a square is doubled and the length of an adjacent side is decreased by 3, the area of the resulting rectangle exceeds the area of the original square by 16. The length of a side of the original square is?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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if the length of one side of a square is doubled and the length of an adjacent side is decreased by 3,
the area of the resulting rectangle exceeds the area of the original square by 16.
The length of a side of the original square is?
:
Let s = length of the side of the original square
then
s^2 = original area
:
and the new rectangle dimensions will be
2s by (s-3)
:
New area - old area = 16
2s(s-3) - s^2 = 16
2s^2 - 6s - s^2 = 16
2s^2 - s^2 - 6s - 16 = 0
s^2 - 6s - 16 = 0
Factors to
(s-8)(s+2) = 0
positive solution
s = 8 units, the length of the original square
:
:
Check this by finding the areas
16(8-3) = 80
8*8 = 64
-------------
differ; 16