SOLUTION: The difference of the squares of two consecutive negative odd integers is 40. Find the numbers. I tried to set it up like this: n=1st integer n-1=2nd integer n2-(n-1)2=40 (the

Question 415857: The difference of the squares of two consecutive negative odd integers is 40. Find the numbers.
I tried to set it up like this: n=1st integer n-1=2nd integer
n2-(n-1)2=40 (the 2's are squares, but I can't do it on my keyboard)
but then I get stuck.
Found 3 solutions by Alan3354, scott8148, htmentor:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
The difference of the squares of two consecutive negative odd integers is 40. Find the numbers.
I tried to set it up like this: n=1st integer n-1=2nd integer
n2-(n-1)2=40 (the 2's are squares, but I can't do it on my keyboard)
but then I get stuck.
---------------------
The 2nd number should be n-2, since both numbers are odd.

4n - 4 = 40
4n = 44
n = 11
--> 11 & 9
or -11 & -9
------------
PS Use n^2 for squared, the shifted 6.

Answer by scott8148(6628)   (Show Source): You can put this solution on YOUR website!
both integers are ODD , so they are 2 numbers apart
also, the more negative number will have the larger square

FOILing ___ (n^2 - 4n + 4) - n^2 = 40

-4n = 36 ___ n = -9

n - 2 = -11
Answer by htmentor(1343)   (Show Source): You can put this solution on YOUR website!
We have two consecutive ODD integers, n and n+2.
Difference of the squares:

Solving for n gives ->
This gives n = -11. Therefore the other integer, n+2, is -9.
Check:
(-11)^2 - (-9)^2 = 40
121 - 81 = 40

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