# Three consecutive

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

According to the given condition,

x2 + (x + 1)(x + 2) = 46

x2 + x2 + 3x + 2 = 46

2x2 + 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

2x2 + 3x – 44 = 0

2x2 + 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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