#### 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.

#### Solution

Let three consecutive integer be x, (x + 1)* *and (x + 2)

Then according to question

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

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

2x^{2} + 3x + 2 - 46 = 0

2x^{2} + 3x - 44 = 0

2x^{2} - 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.

Is there an error in this question or solution?

Solution 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. Concept: Solutions of Quadratic Equations by Factorization.