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

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Answer 1 · 2017-05-18T12:28:35

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

$z^{2} - 16 < 0 or (z + 4) (z - 4) < 0$ $z^2 -16 < 0 or (z+4) (z-4) < 0$ . Critical points are $z = - 4, z = 4$ $z=-4 , z= 4$

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

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

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

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

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