Solve x^2>4x

$${ x }^{ 2 } > 4x$$

Solve for x

$$x\in \left(-\infty,0\right)\cup \left(4,\infty\right)$$

Solution Steps

Subtract 4x from both sides.

$$x^{2}-4x>0$$

Factor out x.

$$x\left(x-4\right)>0$$

For the product to be positive, x and x-4 have to be both negative or both positive. Consider the case when x and x-4 are both negative.

$$x<0 x-4<0$$

The solution satisfying both inequalities is x<0.

$$x<0$$

Consider the case when x and x-4 are both positive.

$$x-4>0 x>0$$

The solution satisfying both inequalities is x>4.

$$x>4$$

The final solution is the union of the obtained solutions.

$$x<0\text{; }x>4$$