How do you solve #z^2>=4(z+3)# using a sign chart?

1 Answer
Jul 30, 2016

Answer:

#z<=-2# or #z>=6#

Explanation:

#z^2>=4(z+3)# can be simplified to

#z^2>=4z+12# or

#z^2-4z-12>=0# or

#z^2-6z+2z-12>=0# or

#z(z-6)+2(z-6)>=0# or

#(z+2)(z-6)>=0#

AS this gives critical values of #z# as #-2# and #6#,

the sign chart of it is

#color(white)(xxxxxxxxxxx)# #-2# #color(white)(xxxxx# #6#

#z+2##color(white)(xxxxxx)##-##color(white)(xxxxxx)##+##color(white)(xxxxxx)##+#

#z-6##color(white)(xxxxxx)##-##color(white)(xxxxxx)##-##color(white)(xxxxxx)##+#
#z^2-4z-12##color(white)(x)##+##color(white)(xxxxxx)##-##color(white)(xxxxxx)##+#

Hence solution is #z<=-2# or #z>=6#

graph{x^2-4x-12 [-10, 10, -15, 15]}