case I : x=0. Clearly, x=0 satisfy the equation. case II : x>0 |x|(6x2+1)=5x2 ⇒x(6x2+1)=5x2 ⇒6x2+1−5x=0 On solving the quadratic equation, we get x=‌
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(both valid) Case III: x<0 |x|(6x2+1)=5x2 ⇒‌‌−x(6x2+1)=5x2 ⇒‌‌6x2+5x+1=0 On solving the quadratic equation, we get x=‌