How do you solve #3x^2+16x<-5#?

1 Answer
Nov 9, 2016

Answer:

The answer is #x in] -5,-1/3 [#

Explanation:

Let's rewrite the inequality #f(x)=3x^2+16x+5=(3x+1)(x+5)<0#
Let's do a sign chart
#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaa)##-1/3##color(white)(aaaa)##+oo#
#color(white)(aaaa)##x+5##color(white)(aaaaa)##-##color(white)(aaaaa)##+##color(white)(aaaaa)##+#
#color(white)(aaaa)##3x+1##color(white)(aaaa)##-##color(white)(aaaaa)##-##color(white)(aaaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaaaa)##-##color(white)(aaaaa)##+#
So the answer is #x in] -5,(-1/3) [#
graph{3x^2+16x+5 [-22.8, 22.83, -11.4, 11.4]}