Question

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

• A. 12, 13, 14
• B. 11, 12, 13
• C. 5, 6, 7
• D. 10, 11, 12

Answer 1

The correct option is D 10, 11, 12

Let the three consecutive positive integers be $x$, $x+1$ and $x+2$.
Square of first integer = $x^2$
Product of other two integers = $(x+1)(x+2)$

Given that,
Sum = 232
$\Rightarrow x^2+(x+1)(x+2)=232$

We know that
$(x+a)(x+b)=x^2+(a+b)x+ab$

$\Rightarrow x^2+(x^2+3x+2)=232$

$\Rightarrow 2x^2+3x-230=0$

Using quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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

$\Rightarrow x=\frac{-3\pm \sqrt{3^2-4\times 2\times -230}}{4}$
$=\frac{-3\pm \sqrt{1849}}{4}$
$=\frac{-3\pm 43}{4}$
$\Rightarrow x=10,-11.5$

Since '$x$' is a positive integer, the negative value is discarded.
$\therefore$ x = 10

Hence the numbers are 10, 11 and 12.