SOLUTION: The length of a rectangle is 3 centimeters less than its width. What are the dimensions of the rectangle if its area is 134 .

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Question 1029035: The length of a rectangle is 3 centimeters less than its width. What are the dimensions of the rectangle if its area is 134 .
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 3 centimeters less than its width. What are the dimensions of the rectangle if its area is 134 .
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L = W + 3
L*W = 134
w*(w+3) = 134
w^2 + 3w - 134 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B3x%2B-134+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%283%29%5E2-4%2A1%2A-134=545.

Discriminant d=545 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+545+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%283%29%2Bsqrt%28+545+%29%29%2F2%5C1+=+10.1726175299288
x%5B2%5D+=+%28-%283%29-sqrt%28+545+%29%29%2F2%5C1+=+-13.1726175299288

Quadratic expression 1x%5E2%2B3x%2B-134 can be factored:
1x%5E2%2B3x%2B-134+=+%28x-10.1726175299288%29%2A%28x--13.1726175299288%29
Again, the answer is: 10.1726175299288, -13.1726175299288. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B3%2Ax%2B-134+%29

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W =~ 10.1726
L =~ 13.1726