Solve x^2>2x

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

Solve for x

$$x\in \left(-\infty,0\right)\cup \left(2,\infty\right)$$

Solution Steps

Subtract 2x from both sides.

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

Factor out x.

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

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

$$x<0 x-2<0$$

The solution satisfying both inequalities is x<0.

$$x<0$$

Consider the case when x and x-2 are both positive.

$$x-2>0 x>0$$

The solution satisfying both inequalities is x>2.

$$x>2$$

The final solution is the union of the obtained solutions.

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