Solve 2x^2>x

$$2 { x }^{ 2 } > x$$

Solve for x

$$x\in \left(-\infty,0\right)\cup \left(\frac{1}{2},\infty\right)$$

Solution Steps

Subtract x from both sides.

$$2x^{2}-x>0$$

Factor out x.

$$x\left(2x-1\right)>0$$

For the product to be positive, x and x-\frac{1}{2} have to be both negative or both positive. Consider the case when x and x-\frac{1}{2} are both negative.

$$x<0 x-\frac{1}{2}<0$$

The solution satisfying both inequalities is x<0.

$$x<0$$

Consider the case when x and x-\frac{1}{2} are both positive.

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

The solution satisfying both inequalities is x>\frac{1}{2}.

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

The final solution is the union of the obtained solutions.

$$x<0\text{; }x>\frac{1}{2}$$