SOLUTION: Find the dimensions of a rectangle if the perimeter is 14in and the area is 12cm square

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Question 894754: Find the dimensions of a rectangle if the perimeter is 14in and the area is 12cm square
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find the dimensions of a rectangle if the perimeter is 14in and the area is 12cm square
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P = 14 in = 14*2.54 cm
P = 35.56 cm
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P = 2W + 2L = 34.56
L + W = 17.28 --> L = 17.28 - W
L*W = 12
W*(17.28 - W) = 12
W^2 - 17.28W + 12 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-17.28x%2B12+=+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=%28-17.28%29%5E2-4%2A1%2A12=250.5984.

Discriminant d=250.5984 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--17.28%2B-sqrt%28+250.5984+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-17.28%29%2Bsqrt%28+250.5984+%29%29%2F2%5C1+=+16.5551500301637
x%5B2%5D+=+%28-%28-17.28%29-sqrt%28+250.5984+%29%29%2F2%5C1+=+0.724849969836327

Quadratic expression 1x%5E2%2B-17.28x%2B12 can be factored:
1x%5E2%2B-17.28x%2B12+=+%28x-16.5551500301637%29%2A%28x-0.724849969836327%29
Again, the answer is: 16.5551500301637, 0.724849969836327. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-17.28%2Ax%2B12+%29

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W = 1 solution
L = the other