Question 382431: Find three consecutive positive odd integers such that the product of the first and third is equal to 1 less than twice the second. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Find three consecutive positive odd integers such that the product of the first and third is equal to 1 less than twice the second.
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1st: 2x-1
2nd: 2x+1
3rd: 2x+3
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Equation:
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4x^2+4x-3 = 4x+1
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4x^2-4 = 0
4(x+1)(x-1) = 0
x = -1 or x = 1
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If x = 1:
1st: 2x-1 = 1
2nd: 3
3rd: 5
Checking:
(2x-1)(2x+3) = 2(2x+1)-1
1*5 = 2(3)-1
5=5
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OR
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If x = -1:
1st:2x-1 = -3
2nd: 2x+1 = -1
3rd: 2x+3 = 1
Checking:
(2x-1)(2x+3) = 2(2x+1)-1
(-3)(1) = 2(-1)-1
-3 = -3
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Cheers,
Stan H.
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