Solve 4lambda^2-4lambda*8>0

$$4 \lambda ^ { 2 } - 4 \lambda \cdot 8 > 0$$

Solve for λ

$$\lambda \in \left(-\infty,0\right)\cup \left(8,\infty\right)$$

Solution Steps

Multiply 4 and 8 to get 32.

$$4\lambda ^{2}-32\lambda >0$$

Factor out \lambda .

$$4\lambda \left(\lambda -8\right)>0$$

For the product to be positive, \lambda and \lambda -8 have to be both negative or both positive. Consider the case when \lambda and \lambda -8 are both negative.

$$\lambda <0 \lambda -8<0$$

The solution satisfying both inequalities is \lambda <0.

$$\lambda <0$$

Consider the case when \lambda and \lambda -8 are both positive.

$$\lambda -8>0 \lambda >0$$

The solution satisfying both inequalities is \lambda >8.

$$\lambda >8$$

The final solution is the union of the obtained solutions.

$$\lambda <0\text{; }\lambda >8$$