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

Related questions
See all questions in Polynomial Inequalities
Impact of this question
2128 views around the world
You can reuse this answer
Creative Commons License