How do solve #x^2+x<12# and write the answer as a inequality and interval notation?

1 Answer
Jan 16, 2017

Answer:

The answer is #x in ] -4,3 [#
#-4 < x < 3#

Explanation:

The inequation is

#x^2+x-12<0#

Let's factorise the expression

#x^2+x-12=(x-3)(x+4)#

Let #f(x)=x^2+x-12#

Now, we can do the sign chart

#color(white)(aaaa)##x##color(white)(aaaaa)##-oo##color(white)(aaaa)##-4##color(white)(aaaa)##3##color(white)(aaaa)##-oo#

#color(white)(aaaa)##x+4##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-3##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 ] -4,3 [#

or #-4 < x <3#