SOLUTION: Given three consecutive odd numbers such that the square of the second number is 104 less than the square of the third number. Find those numbers.
Question 619378: Given three consecutive odd numbers such that the square of the second number is 104 less than the square of the third number. Find those numbers. Found 2 solutions by stanbon, bosco20106:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Given three consecutive odd numbers such that the square of the second number is 104 less than the square of the third number. Find those numbers.
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1st: 2x-3
2nd: 2x-1
3rd: 2x+1
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Equation:
(2x-1)^2 = (2x+1)^2 -104
4x^2-4x+1 = 4x^2+4x+1 - 104
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8x = 104
x = 13
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1st: 2x-3 = 23
2nd: 25
3rd: 27
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Cheers,
Stan H.
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You can put this solution on YOUR website! Let the odd numbers be n, n+2 and n+4.
=> (n+2)² +104 = (n+4)²
Expanding the brackets;
n²+4n+4+104 = n²+8n+16
Re-arranging the terms;
n²-n²+4n-8n=16-4-104
-4n=-92
n=23
Therefore, the odd numbers are: 23, (23+2), (23+4)
i.e. 23, 25, 27
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