SOLUTION: Please help with this problem. I have been doing fine with consecutive integer problems until the "squares" part was added. Q: The sum of the squares of two consecutive odd int

Question 48853: Please help with this problem. I have been doing fine with consecutive integer problems until the "squares" part was added.
Q: The sum of the squares of two consecutive odd integers decreased by the product of the integers is the same as 67.
I have been setting it up like this, but think I may be starting wrong since I don't seem to get answers that make sense
x^2+(x+2)^2 - x(x+2) = 67
Found 2 solutions by stanbon, checkley71:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Q: The sum of the squares of two consecutive odd integers decreased by the product of the integers is the same as 67.
Comment:
Even numbers must be symbolized as multiples of two, e.g. 2x, 2(x+1), etc.
Odd numbers are always 1 more than some even number. They must be symbolized
as 2x+1 or 2x+3 etc.
--------------------
So for your problem
Two consecutive odd integers are
1st: 2x+1
2nd: 2x+3
EQUATION:
(2x+1)^2+(2x+3)^2-(2x+1)(2x+3)=67
Solve this equation for x.
Then find 2x+1 and 2x+3
Cheers,
Stan H.
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
your formula is correct x~2+(x+2)~2-x(x+2)=67 or x~2+x~2+4x+4-x~2-2x=67 or
x~2-2x-63=0 or (x-7)(x-9)=0 or x=7 & x=9
proof 7~2+9~2-7*9=67 or 49+81-63=67 or 130-63=67 or 67=67
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