Question 1060305: The sum of the squares of two consecutive positive odd integers is 103 more than the product of these two integers.Find the integers Found 3 solutions by Alan3354, rothauserc, stanbon:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! The sum of the squares of two consecutive positive odd integers is 103 more than the product of these two integers. Find the integers
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n^2 + (n+2)^2 = n*(n+2) + 103
Solve for n
You can put this solution on YOUR website! let x and x+2 be the two odd integers
:
x^2 + (x+2)^2 = x * (x+2) + 103
:
x^2 +x^2+4x+4 = x^2 +2x +103:
:
x^2 +2x -99 = 0
:
(x+11) * (x-9) = 0
:
x = -11 or x = 9
:
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There are two solutions
:
9 and 11
:
-11 and -9
:
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:
You can put this solution on YOUR website! The sum of the squares of two consecutive positive odd integers is 103 more than the product of these two integers.Find the integers
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1st odd:: 2x-1
2nd odd:: 2x++1
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Equation:
(2x-1)^2 + (2x+1)^2 = (2x-1)(2x+1)+103
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4x^2 - 4x +1 + 4x^2 + 4x + 1 = 4x^2-1+103
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8x^2 +2 = 4x^2+102
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4x^2 = 100
x^2 = 25
x = 5 or x = -5
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1st solution:: 2x-1 = 2*5-1 = 9 and 2x+1 = 11
2nd solution:: 2x-1 = 2*-5-1 = -11 and 2x+1 = 2(-5)+1 = -9
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Cheers,
Stan H.
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