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.

Answer 1

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.