Solve 1/2(x(4x-8))<x^2

$$\frac{ 1 }{ 2 } (x(4x-8)) < { x }^{ 2 }$$

Solve for x

$$x\in \left(0,4\right)$$

Solution Steps

Use the distributive property to multiply \frac{1}{2}x by 4x-8.

$$2x^{2}-4x<x^{2}$$

Subtract x^{2} from both sides.

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

Combine 2x^{2} and -x^{2} to get x^{2}.

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

Factor out x.

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

For the product to be negative, x and x-4 have to be of the opposite signs. Consider the case when x is positive and x-4 is negative.

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

The solution satisfying both inequalities is x\in \left(0,4\right).

$$x\in \left(0,4\right)$$

Consider the case when x-4 is positive and x is negative.

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

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(0,4\right)$$