How do you solve the inequality #x^2-3x-18>0#?

1 Answer
Narad T.
Nov 5, 2016

The answer is #( -oo < x<-3)##uu##(6< x<+oo )#

Explanation:

We factorise the expression #x^2-3x-18=(x+3)(x-6)#
then we do a sign chart
#color(white)(aaaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-3##color(white)(aaaa)##6##color(white)(aaa)##+oo#
#color(white)(aaaaa)##x+3##color(white)(aaaaa)##-##color(white)(aa)##0##color(white)(aa)##+##color(white)(aa)##+#
#color(white)(aaaaa)##x-6##color(white)(aaaaa)##-##color(white)(aaaaa)##-##color(white)(aa)##+#
#color(white)(a)##(x-6((x+3)##color(white)(aa)##+##color(white)(aaaaa)##-##color(white)(aa)##+#

So #x^2-3x-18>0# when

# -oo < x<-3 # and # 6< x<+oo #

Related questions
See all questions in Polynomial Inequalities
Impact of this question
6851 views around the world
You can reuse this answer
Creative Commons License