Question 183587
find two consecutive intgers such that the square of the sum of the two integers is 11 more than the smaller integer.
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1st: x
2nd: x+1
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Equation:
(x + x+1)^2 = x + 11

(2x+1)^2 = x + 11

4x^2 + 4x + 1 = x + 11
4x^2 + 3x - 10 = 0
4x^2 + 8x - 5x - 10 = 0
4x(x + 2) - 5(x + 2) = 0

(x+2)(4x-5) = 0
x = -2 or x = 5/4
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Cheers,
Stan H.