How do you solve and write the following in interval notation: #(x + 6)(x + 8) ≥ 0#?

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Answer 1 · 2017-11-24T12:59:57

The solution is #x in (-oo,-8] uu [-6, +oo)#

Let #f(x)=(x+6)(x+8)#

Build a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaaaa)##-8##color(white)(aaaaaaa)##-6##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##x+8##color(white)(aaaaa)##-##color(white)(aaaa)##0##color(white)(aaaa)##+##color(white)(aaaaa)##+#

#color(white)(aaaa)##x+6##color(white)(aaaaa)##-##color(white)(aaaa)####color(white)(aaaaa)##-##color(white)(aa)##0##color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaaa)##0##color(white)(aaaa)##-##color(white)(aa)##0##color(white)(aa)##+#

Therefore,

#f(x)>=0# when # x in (-oo,-8] uu [-6, +oo)#

graph{(x+6)(x+8) [-11.54, 0.947, -2.155, 4.09]}

Answer 2 · 2017-11-24T13:23:37

#( -oo, -8 ] uu [ -6, oo )#

enter image source here

#( x + 6 )# is ZERO for #x = ( -6 )#

#( x + 6 )# is POSITIVE for #x > ( -6 )#

#( x + 8 )# is ZERO for #x = ( -8 )#

#( x + 8 )# is POSITIVE for #x > ( -8 )#

Refer to the SIGN Table

Our required Interval Notation can be written as follows:

#( -oo, -8 ] uu [ -6, oo )#