SOLUTION: Determine three consecutive even negative integers, if the square of the largest integer is 48 less than the sum of the squares of the two smaller integers.

Question 696219: Determine three consecutive even negative integers, if the square of the largest integer is 48 less than the sum of the squares of the two smaller integers.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Determine three consecutive even negative integers, if the square of the largest integer is 48 less than the sum of the squares of the two smaller integers.
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1st: 2x-2
2nd: 2x
3rd: 2x+2
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Equation:
(2x-2)^2 = (2x)^2 + (2x+2)^2 - 48
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4x^2 -8x + 4 = 4x^2 + 4x^2 +8x + 4 -48
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-8x = 8x^2 + 8x - 48
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8x^2 + 16x - 48 = 0
x^2 + 2x -3 = 0
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Factor:
(x+3)(x-1) = 0
x = -3 or x = 1
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Negative Solutions:
1st: 2x-2 = -8
2nd: -6
3rd: -4
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Cheers,
Stan H.
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