Question

Three consecutive positive integers are taken such that sum of square of the first number and the product of other two numbers is 232. Find the numbers?

Answer 1

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

Given that, sum of square of the first number and the product of other two numbers is 232
$\Rightarrow x^2+(x+1)(x+2)=232$
$\Rightarrow x^2+(x^2+3x+2)=232$
$\Rightarrow 2x^2+3x-230=0$
Using quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here $a=2,b=3$ and $c=-230$

$\Rightarrow x=\frac{-3\pm \sqrt{3^2-4\times 2\times -230}}{4}$
$=\frac{-3\pm \sqrt{1849}}{4}$
$=\frac{-3\pm 43}{4}$
$\Rightarrow x=10\text{or}-11.5$
Ignoring negative value, since $x$ is a positive integer.
$\therefore$ x = 10.
Hence, the numbers are 10, 11 and 12.