SOLUTION: find two consecutive intgers such that the square of the sum of the two integers is 11 more than the smaller integer.

Question 183587This question is from textbook
: find two consecutive intgers such that the square of the sum of the two integers is 11 more than the smaller integer. This question is from textbook

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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.
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