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

1 Answer
May 18, 2017

Answer:

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

Explanation:

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

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

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

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

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