How do you solve #x^2+4x+4>=9# using a sign chart?

1 Answer
Narad T.
Oct 23, 2016

the answer is #x in(-oo,-5)uu(1,+oo)#

Explanation:

#x^2+4x+4>=9#
so #x^2+4x-5>=0#
Factorising #(x-1)(x+5)>=0#
The values we are looking at are #x=1# and #x=-5#

the sign chart is the following

#x##color(white)(aaaaaaaaaaa)##-oo##color(white)(aaaaaa)##-5##color(white)(aaaaaa)##1##color(white)(aaaaaa)##+oo#
#x-1##color(white)(aaaaaaaaaaaaa)##-##color(white)(aaaaaa)##-##color(white)(aaaaa)##+#
#x+5##color(white)(aaaaaaaaaaaaa)##-##color(white)(aaaaaa)##+##color(white)(aaaaa)##+#
#(x-1)(x+5)##color(white)(aaaaaa)##+##color(white)(aaaaaa)##-##color(white)(aaaaa)##+#

From the sign chart, we see that # x in (-oo,-5)# for the function to be positive
From the sign chart, we see that # x in (1,+oo)# for the function to be positive

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