# How do you solve z^2-16<0 using a sign chart?

May 18, 2017

Solution: $- 4 < z < 4$ In interval notation: $\left(- 4 , 4\right)$

#### Explanation:

${z}^{2} - 16 < 0 \mathmr{and} \left(z + 4\right) \left(z - 4\right) < 0$. Critical points are $z = - 4 , z = 4$

When  z < -4 ;  sign of $\left(z + 4\right) \left(z - 4\right) = \left(-\right) \cdot \left(-\right) = \left(+\right) i . e > 0$

When  -4 < z < 4 ;  sign of $\left(z + 4\right) \left(z - 4\right) = \left(+\right) \cdot \left(-\right) = \left(-\right) i . e < 0$

When  z > 4 ;  sign of $\left(z + 4\right) \left(z - 4\right) = \left(+\right) \cdot \left(+\right) = \left(+\right) i . e > 0$

Solution: $- 4 < z < 4$
In interval notation: $\left(- 4 , 4\right)$ [Ans]