Question

The number of real values of $x$ satisfying the equation $\frac{|x-5||x-2|}{|x+9|}=4$ is

Answer 1

$\frac{|x-5||x-2|}{|x+9|}=4\Rightarrow \frac{|(x-5)(x-2)|}{|x+9|}=4\because |a||b|=|ab|\Rightarrow \left|\frac{(x-5)(x-2)}{x+9}\right|=4(\because \frac{|a|}{|b|}=\left|\frac{a}{b}\right|)\Rightarrow \frac{(x-5)(x-2)}{x+9}=\pm 4$

Case 1:
$\Rightarrow \frac{(x-5)(x-2)}{x+9}=4$
$\Rightarrow (x-5)(x-2)=4(x+9)$
$\Rightarrow x^2-7x+10-4x-36=0$
$\Rightarrow x^2-11x-26=0$
$\Rightarrow x^2-13x+2x-26=0$
​​​​​​​$\Rightarrow (x-13)(x+2)=0$
​​​​​​​​​​​​​​$\Rightarrow x=13\text{or}x=-2$

Case 2:
$\Rightarrow \frac{(x-5)(x-2)}{x+9}=-4$
$\Rightarrow (x-5)(x-2)=-4(x+9)$
$\Rightarrow x^2-7x+10+4x+36=0$
​​​​​​​$\Rightarrow x^2-3x+46=0D=9-4\times 46<0$
So, no possible values.

Hence, the required number of values is $2$.