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Find the three consecutive positive integers such that the sum of the square of the first and the product of the other two is 46.

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5, 6, 7

2, 3, 4

1, 2, 3

4, 5, 6

answer is B.

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Detailed Solution

It is given that the sum of the square of the first and the product of the other two consecutive positive integer is 46.
Let’s assume that

m

m + 1

m + 2

are the three consecutive positive integers.
According to the given condition,

⟹ m^{2} + (m + 1) (m + 2) = 46

⟹ m^{2} + m (m + 2) + 1 (m + 2) = 46

⟹ m^{2} + m^{2} + 2 m + m + 2 = 46

⟹ {2 m}^{2} + 3 m + 2 - 46 = 0

⟹ {2 m}^{2} + 3 m - 44 = 0

⟹ {2 m}^{2} + 11 m - 8 m - 44 = 0

⟹ {2 m}^{2} - 8 m + 11 m - 44 = 0

⟹ 2 m (m - 4) + 11 (m - 4) = 0

⟹ (2 m + 11) (m - 4) = 0

⟹ 2 m + 11 = 0 and m - 4 = 0

⟹ m = - \frac{11}{2} and m = 4

We will choose the value

m = 4

because they are positive integers.
So,

⟹ m = 4

⟹ m + 1 = 4 + 1 = 5

⟹ m + 2 = 4 + 2 = 6

∴

The integers are 4, 5, and 6 respectively.
Hence, option (2) is correct.

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