SOLUTION: the dimensions of a rectangle are such that its length is 5 inches more than its width. If teh length were doubled and the width were decreases by 2 inches the area would be increa

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Question 397926: the dimensions of a rectangle are such that its length is 5 inches more than its width. If teh length were doubled and the width were decreases by 2 inches the area would be increases by 52 inches squared what are the length and width of teh rectangel
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
the dimensions of a rectangle are such that its length is 5 inches more than its width. If the length were doubled and the width were decreased by 2 inches the area would be increased by 52 inches squared. What are the length and width of the rectangel?
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Original dimensions:
Let the width be "x".
Then the length is "x+5"
Area is x(x+5) = x^2+5x.
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New dimensions:
width = x+2
length = 2(x+5)
Area is 2(x+5)(x+2) = 2(x^2+7x+10) = 2x^2+14x+20
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Equation:
New Area - Old Area = 52 sq. in.
2x^2+14x+20 -(x^2+5x) = 52
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x^2+9x+20 = 52
x^2+8x-32 = 0
Use Quadratic formula to get a positive solution:
x = 2.93 inches (width)
x+5 = 7.93 inches (length)
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Cheers,
Stan H.