SOLUTION: A rectangle has the dimensions 1080/x^2-5x+6 by x-2/x. If the area is 20 units^2, find the value(s) for x. I got 1060/x(x-3) but I don't think that is right. Help would be great

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Question 610525: A rectangle has the dimensions 1080/x^2-5x+6 by x-2/x. If the area is 20 units^2, find the value(s) for x.
I got 1060/x(x-3) but I don't think that is right. Help would be greatly appreciated!
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangle has the dimensions 1080/x^2-5x+6 by x-2/x. If the area is 20 units^2, find the value(s) for x.
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[1080/x^2-5x+6] * [x-2/x] = 20
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[1080/(x=3)(x-2)] * [x-2/x] = 20
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[1080/(x-3)] * [1/x] = 20
Multiply thru by x(x-3) to get:
1080 = 20x(x-3)
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20x^2-60x-1080 = 0
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x^2 - 3x - 54 = 0
(x-9)(x+6) = 0
x = 9 or x = -6
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Cheers,
Stan H.