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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.
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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) = 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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