# SOLUTION: The width of a rectangle is 4 cm less than its length. If the area of the rectangle is 87cm^2 what are the dimensions of the rectangle? Round to the nearest thousandth. I have

Algebra ->  -> SOLUTION: The width of a rectangle is 4 cm less than its length. If the area of the rectangle is 87cm^2 what are the dimensions of the rectangle? Round to the nearest thousandth. I have       Log On

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Question 338556: The width of a rectangle is 4 cm less than its length. If the area of the rectangle is 87cm^2 what are the dimensions of the rectangle? Round to the nearest thousandth.
I have figured out the answer as
Length = 11.539
Width = 7.539
Just don't know the formula or how to solve, help would be greatly appreciated thank you.

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The width of a rectangle is 4 cm less than its length. If the area of the rectangle is 87cm^2 what are the dimensions of the rectangle? Round to the nearest thousandth.
I have figured out the answer as
Length = 11.539
Width = 7.539
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Area = L*W
L = W+4
Area = (W+4)*W = 87
W^2 + 4W = 87
W^2 + 4W - 87 = 0
 Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=364 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 7.53939201416946, -11.5393920141695. Here's your graph:

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W = x (ignore the negative solution)
W = 7.53939 ...
L = 11.53939...