SOLUTION: The width of a rectangle is 1 less than twice its length. If the area of the rectangle is 33 cm^2, what is the length of the diagonal?

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Question 211150: The width of a rectangle is 1 less than twice its length. If the area of the rectangle is 33 cm^2, what is the length of the diagonal?
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
L=w+1
wL=A
w(w+1)=33
w^2+w-33=0
w=5.266... Quadratic formula
L=6.266...
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a^2+b^2=c^2
(5.266)^2+(6.266)^2=67
sqrt(67)cm is the length of the diagonal.
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Ed
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B-33+=+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=%281%29%5E2-4%2A1%2A-33=133.

Discriminant d=133 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-1%2B-sqrt%28+133+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%281%29%2Bsqrt%28+133+%29%29%2F2%5C1+=+5.2662812973354
x%5B2%5D+=+%28-%281%29-sqrt%28+133+%29%29%2F2%5C1+=+-6.2662812973354

Quadratic expression 1x%5E2%2B1x%2B-33 can be factored:
1x%5E2%2B1x%2B-33+=+1%28x-5.2662812973354%29%2A%28x--6.2662812973354%29
Again, the answer is: 5.2662812973354, -6.2662812973354. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B-33+%29