Solve a^2-16a<0

$$a ^ { 2 } - 16 a < 0$$

Solve for a

$$a\in \left(0,16\right)$$

Solution Steps

Factor out a.

$$a\left(a-16\right)<0$$

For the product to be negative, a and a-16 have to be of the opposite signs. Consider the case when a is positive and a-16 is negative.

$$a>0 a-16<0$$

The solution satisfying both inequalities is a\in \left(0,16\right).

$$a\in \left(0,16\right)$$

Consider the case when a-16 is positive and a is negative.

$$a-16>0 a<0$$

This is false for any a.

$$a\in \emptyset$$

The final solution is the union of the obtained solutions.

$$a\in \left(0,16\right)$$