SOLUTION: The width of a rectangle is 2 less than twice its length. If the area of the rectangle is 106cm^2 , what is the length of the diagonal?

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Question 51223: The width of a rectangle is 2 less than twice its length. If the area of the rectangle is 106cm^2 , what is the length of the diagonal?
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
length = l
width = 2l - 2
l(2l - 2) = 106
2l^2 - 2l - 106 = 0
l^2 - l - 53 = 0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation al%5E2%2Bbl%2Bc=0 (in our case 1l%5E2%2B-1l%2B53+=+0) has the following solutons:

l%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%2A1%2A53=-211.

The discriminant -211 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -211 is + or - sqrt%28+211%29+=+14.5258390463339.

The solution is

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B53+%29

No rectangle exists.