Solve (x-8)(x-3)<0

$$( x - 8 ) ( x - 3 ) < 0$$

Solve for x

$$x\in \left(3,8\right)$$

Solution Steps

For the product to be negative, x-3 and x-8 have to be of the opposite signs. Consider the case when x-3 is positive and x-8 is negative.

$$x-3>0 x-8<0$$

The solution satisfying both inequalities is x\in \left(3,8\right).

$$x\in \left(3,8\right)$$

Consider the case when x-8 is positive and x-3 is negative.

$$x-8>0 x-3<0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(3,8\right)$$