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Question
There are three consecutive integers such that the square of the first increased by the product of the first increased by the product of the others the two gives 154. What are the integers?
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Solution
Let three consecutive integer be x , (x + 1) and (x + 2)
Then according to question
x2 + (x + 1)(x + 2) = 154
x2 + x2 + 3x + 2 = 154
2x2 - 3x + 2 - 154 = 0
2x2 + 3x - 152 = 0
2x2 - 16x - 19x - 152 = 0
2x(x - 8) + 19(x - 8) = 0
(x - 8)(2x + 19) = 0
(x - 8) = 0
x = 8
Or
2x + 19 = 0
2x = -19
x = -19/2
Since, x being a positive number, so x cannot be negative.
Therefore,
Whenx = 8then other positive integer
x + 1 = 8 + 1 = 9
And
x + 2 = 8 + 2 = 10
Thus, three consecutive positive integer be 8, 9, 10.
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