SOLUTION: The area of a rectangle is 52 and the length is 5 less than twice the width. What are the dimensions of the rectangle?

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Question 887133: The area of a rectangle is 52 and the length is 5 less than twice the width. What are the dimensions of the rectangle?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Step 1, try the factors of 52 in pairs.
----
1*52 NG
2*26 NG
etc
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If that doesn't work:
L*W = 52
L = 2W - 5
Sub for L in the 1st eqn
(2W - 5)*W = 52
2W%5E2+-+5W+-+52+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-5x%2B-52+=+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=%28-5%29%5E2-4%2A2%2A-52=441.

Discriminant d=441 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+441+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+441+%29%29%2F2%5C2+=+6.5
x%5B2%5D+=+%28-%28-5%29-sqrt%28+441+%29%29%2F2%5C2+=+-4

Quadratic expression 2x%5E2%2B-5x%2B-52 can be factored:
2x%5E2%2B-5x%2B-52+=+%28x-6.5%29%2A%28x--4%29
Again, the answer is: 6.5, -4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-5%2Ax%2B-52+%29

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W = 6.5
L = 8