How do you solve #x^2+4x+3<=0#?

1 Answer
Feb 28, 2017

Answer:

The solution is #x in [-3,-1]#

Explanation:

Let's factorise the inequality

#x^2+4x+3<=0#

#(x+1)(x+3)<=0#

Let, #f(x)=(x+1)(x+3)#

Now, we can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-3##color(white)(aaaa)##-1##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+3##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x+1##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)<=0# when #x in [-3,-1]#