Given that, $ \left | x \right | (6x^{2}+1) = 5x^{2} $
We know that, $\left | x \right | = \left\{\begin{matrix} x\; ;& x \geq 0 \\ -x \; ;& x<0 \end{matrix}\right.$
$\textbf{Case 1:} \; x \geq 0 $
Now, $x (6x^{2} + 1) = 5x^{2} $
$ \Rightarrow 6x^{3} + x = 5x^{2} $
$ \Rightarrow 6x^{3} – 5x^{2} + x = 0 $
$ \Rightarrow x( 6x^{2} -5x +1) = 0$
$ \Rightarrow x ( 6x^{2} -3x – 2x +1) = 0 $
$ \Rightarrow x [3x (2x-1) -1(2x-1)] = 0 $
$ \Rightarrow x [ (3x-1)(2x-1)] = 0 $
$ \Rightarrow x=0 , (3x-1)=0 , (2x-1)=0 $
$ \boxed{x=0,x=\frac{1}{3},x=\frac{1}{2}} $
$\therefore$ The number of solutions $=3$
$\textbf{Case 2:}\; x < 0$
Now, $ -x(6x^{2}+1) = 5x^{2} $
$ \Rightarrow – 6x^{3} – x – 5x^{2} = 0 $
$\Rightarrow 6x^{3} + 5x^{2} + x = 0 $
$ \Rightarrow x (6x^{2} + 5x + 1) = 0 $
$\Rightarrow x (6x^{2} + 3x + 2x + 1) = 0 $
$ \Rightarrow x [3x(2x+1) +1(2x+1)] = 0$
$ \Rightarrow x [(3x+1)(2x+1)] = 0 $
$\Rightarrow x=0,(3x+1)=0,(2x+1)=0 $
$ \boxed {x=0,x=\frac{-1}{3},x=\frac{-1}{2}} $
$\therefore$ The number of solutions $=3$
So, the total number of solutions $ = 2 + 3 = 5.$
Correct Answer $:5$