Solve 1/3x^2>x

$$\frac { 1 } { 3 } x ^ { 2 } > x$$

Solve for x

$$x\in \left(-\infty,0\right)\cup \left(3,\infty\right)$$

Solution Steps

Subtract x from both sides.

$$\frac{1}{3}x^{2}-x>0$$

Factor out x.

$$\frac{1}{3}x\left(x-3\right)>0$$

For the product to be positive, x and x-3 have to be both negative or both positive. Consider the case when x and x-3 are both negative.

$$x<0 x-3<0$$

The solution satisfying both inequalities is x<0.

$$x<0$$

Consider the case when x and x-3 are both positive.

$$x-3>0 x>0$$

The solution satisfying both inequalities is x>3.

$$x>3$$

The final solution is the union of the obtained solutions.

$$x<0\text{; }x>3$$