Questions on Word Problems: Problems with consecutive odd even integers answered by real tutors!

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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: the sum of the reciprocals of two consecutive odd integers is 20/99. find the two integers.
pleasee helpp
Answer by Alan3354(1427) 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+2 = (20n^2 + 40n)/99
Multiply by 99
198n + 198 = 20n^2 + 40n
Put in quadratic form
20n^2 - 158n - 198 = 0
10n^2 - 79n - 99 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax^2+bx+c=0 (in our case 10x^2+-79x+-99 = 0) has the following solutons:

x[12] = (b+-sqrt( b^2-4ac ))/2\a

For these solutions to exist, the discriminant b^2-4ac should not be a negative number.

First, we need to compute the discriminant b^2-4ac: b^2-4ac=(-79)^2-4*10*-99=10201.

Discriminant d=10201 is greater than zero. That means that there are two solutions: x[12] = (--79+-sqrt( 10201 ))/2\a.

x[1] = (-(-79)+sqrt( 10201 ))/2\10 = 9
x[2] = (-(-79)-sqrt( 10201 ))/2\10 = -1.1

Quadratic expression 10x^2+-79x+-99 can be factored:
10x^2+-79x+-99 = (x-9)*(x--1.1)
Again, the answer is: 9, -1.1. Here's your graph:
graph( 500, 500, -10, 10, -20, 20, 10*x^2+-79*x+-99 )

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