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Question 485024: The sum of the reciprocals of two even integer is 9/40. What are the integers?
Found 2 solutions by chessace, MathTherapy:
Answer by chessace(471)   (Show Source):
You can put this solution on YOUR website! This is a fun problem!
1/a + 1/b =
b/ab + a/ab =
(a+b)/ab = 9/40
So there is a constant c, not necessarily 1 such that
a+b = 9c and ab = 40c
b=9c-a
a(9c-a)= 40c
9ac - a^2 - 40c = 0
a^2 - 9ca + 40c = 0
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At this point a very interesting thing happened.
Because I was in a rush and have been out of HS for a LONG time, I wrote
"This is no good if c>0"
Later I went back and used the Quadratic Formula to find:
c=2 a=8 b=10
I then recalled enjoying this same problem in HS.
It's possible that at that time the problem was stated "... 2 consecutive even intergers..." but let's assume not.
I'm now a much better mathematian than I was in HS.
In particular, I don't think it would have occured to me to look for a negative c after success with c=2.
This is somewhat of a warning in that if you turn this into a HS Math teacher, it may raise 2 red flags that it was not done by you.
Back to the original solution.
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Let d = -c
a^2 + 9da - 40d = 0
To factor this, we need 2 numbers whose difference is 9d and product is 40d.
There might be a direct way to do this, but good old trial and error (d = 1 2 3 4)
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Again, had I had time to find the more direct way, I would have skipped all odd d.
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which finds d=4, numbers 4 and 40: product 160, difference 36.
a^2 + 36a - 160 = 0
(a + 40)(a - 4) = 0
a = -40, 4.
b = -36 - a = -36 - (-40 or 4) = 4 or -40; of course the same set (excellent double check).
The 2 numbers are 4 and -40
Triple check: 1/4 - 1/40 = 10/40 - 1/40 = 9/40
Answer by MathTherapy(10552)  (Show Source):
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