SOLUTION: Find three consecutive positive odd intergers such that the square of the smallest exceeds twice the largest by 7 | ONLY AN ALGEBRAIC SOLUTION WILL BE ACCEPTED?|
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Question 734765: Find three consecutive positive odd intergers such that the square of the smallest exceeds twice the largest by 7 | ONLY AN ALGEBRAIC SOLUTION WILL BE ACCEPTED?| Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Find three consecutive positive odd intergers such that the square of the smallest exceeds twice the largest by 7
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1st: 2x-3
2nd: 2x-1
3rd: 2x+1
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Equation:
(2x-3)^2 = 2(2x+1)+7
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4x^2 -12x + 9 = 4x+9
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4x^2 -16x = 0
Factor:
4x(x-4) = 0
x = 0 or x = 4
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If x = 0::
1st: -3
2nd: -1
3rd: 1
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If x = 4
1st: 5
2nd: 7
3rd: 9
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Since you want positive odd only the x=4 set is acceptable.
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Cheers,
Stan H.
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