Solve 16a^2-4a(2a+1)>0

$$16 a ^ { 2 } - 4 a ( 2 a + 1 ) > 0$$

Solve for a

$$a\in \left(-\infty,0\right)\cup \left(\frac{1}{2},\infty\right)$$

Solution Steps

Multiply -1 and 4 to get -4.

$$16a^{2}-4a\left(2a+1\right)>0$$

Use the distributive property to multiply -4a by 2a+1.

$$16a^{2}-8a^{2}-4a>0$$

Combine 16a^{2} and -8a^{2} to get 8a^{2}.

$$8a^{2}-4a>0$$

Factor out a.

$$4a\left(2a-1\right)>0$$

For the product to be positive, a and a-\frac{1}{2} have to be both negative or both positive. Consider the case when a and a-\frac{1}{2} are both negative.

$$a<0 a-\frac{1}{2}<0$$

The solution satisfying both inequalities is a<0.

$$a<0$$

Consider the case when a and a-\frac{1}{2} are both positive.

$$a-\frac{1}{2}>0 a>0$$

The solution satisfying both inequalities is a>\frac{1}{2}.

$$a>\frac{1}{2}$$

The final solution is the union of the obtained solutions.

$$a<0\text{; }a>\frac{1}{2}$$