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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 $:5$
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