SOLUTION: the sum of the reciprocals of two consecutive odd integers is 20/99. find the two integers. pleasee helpp

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Question 146812: the sum of the reciprocals of two consecutive odd integers is 20/99. find the two integers.
pleasee helpp
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
1/n + 1/(n+2) = 20/99
Multiply by n(n+2)
n+2 + n = 20n(n+2)/99
2n%2B2+=+%2820n%5E2+%2B+40n%29%2F99
Multiply by 99
198n+%2B+198+=+20n%5E2+%2B+40n
Put in quadratic form
20n%5E2+-+158n+-+198+=+0
10n%5E2+-+79n+-+99+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 10x%5E2%2B-79x%2B-99+=+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=%28-79%29%5E2-4%2A10%2A-99=10201.

Discriminant d=10201 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--79%2B-sqrt%28+10201+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-79%29%2Bsqrt%28+10201+%29%29%2F2%5C10+=+9
x%5B2%5D+=+%28-%28-79%29-sqrt%28+10201+%29%29%2F2%5C10+=+-1.1

Quadratic expression 10x%5E2%2B-79x%2B-99 can be factored:
10x%5E2%2B-79x%2B-99+=+%28x-9%29%2A%28x--1.1%29
Again, the answer is: 9, -1.1. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+10%2Ax%5E2%2B-79%2Ax%2B-99+%29

-1.1 is not a positive integer, so discard it.
n = 9
n+2 = 11
It does work, 1/9 + 1/11 = 20/99