Question

A two-digit number is such that the product of the digits is $12$. When $36$ is added to the number the digits interchange their places. Formulate the quadratic equation whose root(s) is (are) digit(s) of the number.

Answer 1

Let the ten's digit of the number be $x$

It is given that the product of digits is $12$

Then unit's digit$=\frac{12}{x}$
Given number can be written as $=10x+\frac{12}{x}$
If $36$ is added to the number the digits interchange their places

$\therefore 10x+\frac{12}{x}+36=10\times \frac{12}{x}+x$

$\Rightarrow 10x-x-\frac{120-12}{x}+36=0$
$\Rightarrow 9x-\frac{108}{x}+36=0$

$\Rightarrow 9x^2+36x-108=0$

$\Rightarrow x^2+4x-12=0$ (divided throughout by $9$)

hence, required quadratic equation is $x^2+4x-12=0$