Question

Three consecutive positive integer are such that the sum of square of the first and product of other two is $46$. The smallest integer is:

Answer 1

Step 1: Form an equation based on the given statement

Let us consider the three consecutive positive integers be $x,(x+1)and(x+2)$

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

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

$\Rightarrow 2x^2+3x+2–46=0$

$\Rightarrow 2x^2+3x–44=0$

Step 2: Solve the above quadratic equation

$\Rightarrow 2x^2-8x+11x–44=0$

$\Rightarrow 2x(x–4)+11(x–4)=0$

$\Rightarrow (x–4)(2x+11)=0$

$\Rightarrow x–4or(2x+11)=0$

$\Rightarrow x=4or2x=-11$

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

As $x$ being a positive number, $x$ cannot be negative.

Therefore, $x=4$

Thus the three consecutive positive integers are $4,5,6.$
Hence smallest integer is $4$