SOLUTION: if the length and width of 4 x 2 inc rectangle are each increased by the same amount. The area of the new rectangle will be twice the old. What are the dimensions to two decimal pa

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Question 782694: if the length and width of 4 x 2 inc rectangle are each increased by the same amount. The area of the new rectangle will be twice the old. What are the dimensions to two decimal palaces of the new rectangle?
Answer by stanbon(75887) About Me  (Show Source):
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if the length and width of 4 x 2 inc rectangle are each increased by the same amount. The area of the new rectangle will be twice the old. What are the dimensions to two decimal places of the new rectangle?
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New dimensions:
length = 4+x
width = 2+x
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New Area:
(4+x)(2+x) = x^2+6x+8
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Equation::
new area = 2*old area
x^2+6x+8 = 2(4*2)
x^2 + 6x -8 =0
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x = [-6 +-sqrt(36-4*-8)]/2
x = [-6 +- sqrt(68)]/2
Positive solution:
x = -3 + (1/2)sqrt(68) = 5.25
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New Dimensions:
length: 4 + x = 9.25
width:: 2 + x = 7.25
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Cheers,
Stan H.
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