SOLUTION: How do you find two consecutive odd integers whose product is 36?

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Question 879728: How do you find two consecutive odd integers whose product is 36?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
n (n+2)= 36
n^2 + 2n - 36 = 0 Not going to work..solution does not yield an Integer.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B2x%2B-36+=+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=%282%29%5E2-4%2A1%2A-36=148.

Discriminant d=148 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+148+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+148+%29%29%2F2%5C1+=+5.08276253029822
x%5B2%5D+=+%28-%282%29-sqrt%28+148+%29%29%2F2%5C1+=+-7.08276253029822

Quadratic expression 1x%5E2%2B2x%2B-36 can be factored:
1x%5E2%2B2x%2B-36+=+1%28x-5.08276253029822%29%2A%28x--7.08276253029822%29
Again, the answer is: 5.08276253029822, -7.08276253029822. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B2%2Ax%2B-36+%29