SOLUTION: how to find two consecutive odd integers such that the square of the greater decreased by the lesser is 212

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Question 174042: how to find two consecutive odd integers such that the square of the greater decreased by the lesser is 212
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
lets call our two consecutive odd integers a and a+2
:
%28a%2B2%29%5E2-a=212
:
a%5E2%2B4a%2B4-a=212
:
a%5E2%2B3a-208=0
:
%28a-13%29%28a%2B16%29=0
a= 13,-16....since -16 is not odd we throw it out
:
integerssystem%2813%2C15%29
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B3x%2B-208+=+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=%283%29%5E2-4%2A1%2A-208=841.

Discriminant d=841 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+841+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%283%29%2Bsqrt%28+841+%29%29%2F2%5C1+=+13
x%5B2%5D+=+%28-%283%29-sqrt%28+841+%29%29%2F2%5C1+=+-16

Quadratic expression 1x%5E2%2B3x%2B-208 can be factored:
1x%5E2%2B3x%2B-208+=+%28x-13%29%2A%28x--16%29
Again, the answer is: 13, -16. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B3%2Ax%2B-208+%29