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Question
Three consecutive positive integers are such that the sum of the square of the first and the product of other two is 46, find the integers.
Solution
Let three consecutive integer be x, (x + 1) and (x + 2)
Then according to question
x2 + (x + 1)(x + 2) = 46
x2 + x2 + 3x + 2 = 46
2x2 + 3x + 2 - 46 = 0
2x2 + 3x - 44 = 0
2x2 - 8x + 11x - 44 = 0
2x(x - 4) + 11(x - 4) = 0
(x - 4)(2x + 11) = 0
x - 4 = 0
x = 4
Or
2x + 11 = 0
2x = -11
x = -11/2
Since, x being a positive number, so x cannot be negative.
Therefore,
When x = 4 then other positive integer
x + 1 = 4 + 1 = 5
And
x + 2 = 4 + 2 = 6
Thus, three consecutive positive integer be 4, 5, 6.
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