How do you solve #x^2-2x-8<0# using a sign chart?

1 Answer
Dec 24, 2016

Answer:

The answer is #x in ] -2,4 [#

Explanation:

We factorise the expression

#x^2-2x-8=(x+2)(x-4)#

Let, #f(x)=(x+2)(x-4)#

Now we can make the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaa)##4##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+2##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

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

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

Therefore,

#f(x)<0#, when #x in ] -2,4 [#