SOLUTION: The length of a rectangle is one less than twice the rectangle's width. The rectangle's area equals 703 in^2. Find the rectangle's width (in inches)

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Question 993402: The length of a rectangle is one less than twice the rectangle's width. The rectangle's area equals 703 in^2. Find the rectangle's width (in inches)
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
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W=width; L=length=2W-1
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A+LW
703in^2=(2W-1)(W)
703=2W^2-W
0=2W^2-W-703
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aW%5E2%2BbW%2Bc=0 (in our case 2W%5E2%2B-1W%2B-703+=+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=%28-1%29%5E2-4%2A2%2A-703=5625.

Discriminant d=5625 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+5625+%29%29%2F2%5Ca.

W%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+5625+%29%29%2F2%5C2+=+19
W%5B2%5D+=+%28-%28-1%29-sqrt%28+5625+%29%29%2F2%5C2+=+-18.5

Quadratic expression 2W%5E2%2B-1W%2B-703 can be factored:
2W%5E2%2B-1W%2B-703+=+2%28W-19%29%2A%28W--18.5%29
Again, the answer is: 19, -18.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-1%2Ax%2B-703+%29

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ANSWER: The width is 19 inches.