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

Let the three consecutive positive integers be x, x + 1, x + 2

According to the given condition,

x^{2} + (x + 1)(x + 2) = 46

x^{2} + x^{2} + 3x + 2 = 46

2x^{2} + 3x – 44 = 0

Using the splitting middle term – the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 2 b = 3 c = – 44

= 2. – 44 = – 88

And either of their sum or difference = b

= 3

Thus the two terms are 11 and – 8

Sum = 11 – 8 = 3

Product = 11. – 8 = – 88

2x^{2} + 3x – 44 = 0

2x^{2} + 11x – 8x – 44 = 0

x(2x + 11) – 4(2x + 11) = 0

(2x + 11)(x – 4) = 0

x = 4 or – 11/2

x = 4 (x is a positive integers)

When x = 4

x + 1 = 4 + 1 = 5

x + 2 = 4 + 2 = 6

Hence the required integers are 4, 5, 6

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