SOLUTION: The length of a rectangle is 5 cm more than 2 times its width. If the area of the rectangle is 75 cm2, find the dimensions of the rectangle to the nearest thousandth.

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Question 74859: The length of a rectangle is 5 cm more than 2 times its width. If the area of the rectangle is 75 cm2, find the dimensions of the rectangle to the nearest thousandth.
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
Let the width = x cm.
Two times width = 2x cm.
5cm more than this value = (2x + 5) cm
So, the length = (2x + 5) cm.

Hence area = Length X Width = %282x+%2B+5%29x+=+2x%5E2+%2B+5x cm%5E2.

But, according to the problem this area is 75 cm%5E2.

So, 2x%5E2+%2B+5x+=+75
2x%5E2+%2B+5x+-+75+=+0
2x%5E2+%2B+15x+-+10x+-+75+=+0
x%282x+%2B+15%29+-+5%282x+%2B+15%29+=+0
%28x-5%29%282x%2B15%29=0

So either (x-5) = 0 or (2x+15) = 0
i.e. either x = 5 or x = -7.5

But width of a rectangle (x) cannot be negative.
So negative answer is discarded.
Hence x = 5.

So the width = 5 cm and length = 2x5 + 5 = 15 cm.