SOLUTION: If the length of a rectangle is 3 feet longer than the width and the diagonal is 15 feet, then what are the length and width?

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Question 349686: If the length of a rectangle is 3 feet longer than the width and
the diagonal is 15 feet, then what are the length and width?
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Let the width be w and the length be l
w%5E2+%2B+l%5E2=+15} pythagorian theorem
w+=+l%2B3
w%5E2+%2B+%28w%2B3%29%5E2+=+15
w%5E2+%2B+w%5E2+%2B+6w+%2B+9+=+15
2w%5E2+%2B+6w+-+6+=+0
w%5E2+%2B+3w+-+3+=+0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B3x%2B-3+=+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-3=21.

Discriminant d=21 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+21+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%283%29%2Bsqrt%28+21+%29%29%2F2%5C1+=+0.79128784747792
x%5B2%5D+=+%28-%283%29-sqrt%28+21+%29%29%2F2%5C1+=+-3.79128784747792

Quadratic expression 1x%5E2%2B3x%2B-3 can be factored:
1x%5E2%2B3x%2B-3+=+1%28x-0.79128784747792%29%2A%28x--3.79128784747792%29
Again, the answer is: 0.79128784747792, -3.79128784747792. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B3%2Ax%2B-3+%29

The width cannot be nagative, so it must be 0.79
Which makes the length 3.79
Check your answer. Does
%283.79%29%5E2+%2B+%280.79%29%5E2+=+15?