How do you solve #x^2-4x<-3# using a sign chart?

1 Answer
Dec 13, 2016

Answer:

The answer is #x in ] 1,3 [#

Explanation:

Rewrite the equation as, #x^2-4x+3<0#

Let #f(x)=x^2-4x+3=(x-1)(x-3)#

The domain of #f(x)# is #D_f(x)=RR#

Now, we can do our sign chart

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

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

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

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

So,

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