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

**Question:**

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

**Solution:**

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

According to the given condition,

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

$\Rightarrow x^{2}+x^{2}+3 x+2=46$

$\Rightarrow 2 x^{2}+3 x-44=0$

$\Rightarrow 2 x^{2}+11 x-8 x-44=0$

$\Rightarrow x(2 x+11)-4(2 x+11)=0$

$\Rightarrow(2 x+11)(x-4)=0$

$\Rightarrow 2 x+11=0$ or $x-4=0$

$\Rightarrow x=-\frac{11}{2}$ or $x=4$

∴ *x* = 4 (*x* is a positive integer)

When *x* = 4,*x* + 1 = 4 + 1 = 5*x* + 2 = 4 + 2 = 6

Hence, the required integers are 4, 5 and 6.