Scenario I: For x = 0 ,
It is evident that x = 0 satisfies the equation.
Scenario II: For x > 0,
considering : | x | (6x² + 1) = 5x²
⇒ x(6x² + 1) = 5x²
⇒ 6x² + 1 - 5x = 0.
Solving the quadratic equation, we find x = \(\frac{1}{2},\frac{1}{3}\) both of which are valid solutions.
Scenario III: For x < 0
considering : | x | (6x² + 1) = 5x²
⇒ -x(6x² + 1) = 5x²
⇒ 6x² + 5x + 1 = 0
Solving the quadratic equation, we obtain x = \(-\frac{1}{2},-\frac{1}{3}\), both of which are valid solutions.
Hence, there are a total of 5 solutions.