SOLUTION: The length of a rectangular floor is 3 meters less than three times its width. If a diagonal of the rectangle is 13 meters, find the length and width of the floor.

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Question 628789: The length of a rectangular floor is 3 meters less than three times its width. If a diagonal of the rectangle is 13 meters, find the length and width of the floor.
Answer by graphmatics(170) About Me  (Show Source):
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Let w be the width of the rectangle. Then the length of the rectangle is given by 3w-3. For a right triangle with diagonal c and sides a and b c*c=a*a+b*b. So for this rectangle with a diagonal of 13 meters. 13*13=w*w+(3w-3)*(3w-3) 169=w*w+9*w*w-18w+9 10*w*w-18*w-160=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 10x%5E2%2B-18x%2B-160+=+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=%28-18%29%5E2-4%2A10%2A-160=6724.

Discriminant d=6724 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--18%2B-sqrt%28+6724+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-18%29%2Bsqrt%28+6724+%29%29%2F2%5C10+=+5
x%5B2%5D+=+%28-%28-18%29-sqrt%28+6724+%29%29%2F2%5C10+=+-3.2

Quadratic expression 10x%5E2%2B-18x%2B-160 can be factored:
10x%5E2%2B-18x%2B-160+=+10%28x-5%29%2A%28x--3.2%29
Again, the answer is: 5, -3.2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+10%2Ax%5E2%2B-18%2Ax%2B-160+%29

So if the width is 5 the length is 12.