SOLUTION: The area of a rectangular conference table is 51 square feet. If its length is 4 feet longer than its width, find the dimensions of the table. Round each dimension to the nearest t

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Question 450354: The area of a rectangular conference table is 51 square feet. If its length is 4 feet longer than its width, find the dimensions of the table. Round each dimension to the nearest tenth of a foot, if necessary.
Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
x(x+4)=51
x2+4x-51=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B4x%2B-51+=+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=%284%29%5E2-4%2A1%2A-51=220.

Discriminant d=220 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+220+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%284%29%2Bsqrt%28+220+%29%29%2F2%5C1+=+5.41619848709566
x%5B2%5D+=+%28-%284%29-sqrt%28+220+%29%29%2F2%5C1+=+-9.41619848709566

Quadratic expression 1x%5E2%2B4x%2B-51 can be factored:
1x%5E2%2B4x%2B-51+=+1%28x-5.41619848709566%29%2A%28x--9.41619848709566%29
Again, the answer is: 5.41619848709566, -9.41619848709566. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B4%2Ax%2B-51+%29

Throwing out the negative answer, we get the dimensions of the table to be
5.42ft. by 9.42ft..