SOLUTION: The length of a rectangle is 4 inches longer than twice its width. The area of the rectangle is 96 square inches. What is the length and the width of the rectangle?

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Question 607218: The length of a rectangle is 4 inches longer than twice its width. The area of the rectangle is 96 square inches. What is the length and the width of the rectangle?
Answer by asuar010(338) About Me  (Show Source):
You can put this solution on YOUR website!
l=2w+4 and the area is 96 so if we set up te equation as l*w=96 and we substitute we obtain 2w^2+4w-96=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aw%5E2%2Bbw%2Bc=0 (in our case 2w%5E2%2B4w%2B-96+=+0) has the following solutons:

w%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%2A2%2A-96=784.

Discriminant d=784 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+784+%29%29%2F2%5Ca.

w%5B1%5D+=+%28-%284%29%2Bsqrt%28+784+%29%29%2F2%5C2+=+6
w%5B2%5D+=+%28-%284%29-sqrt%28+784+%29%29%2F2%5C2+=+-8

Quadratic expression 2w%5E2%2B4w%2B-96 can be factored:
2w%5E2%2B4w%2B-96+=+2%28w-6%29%2A%28w--8%29
Again, the answer is: 6, -8. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B4%2Ax%2B-96+%29


and we take the positive answer as the solution for w so l=2(6)+4=12+4=16