How do you solve #x^2+6x-16>0# using a sign chart?

1 Answer
Dec 24, 2016

Answer:

The answer is #x in ] -oo,-8 [ uu ]2, +oo[ #

Explanation:

Let's factorise the expression

#x^2+6x-16=(x+8)(x-2)#

and let #f(x)=(x+8)(x-2)#

Now we can do the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-8##color(white)(aaaa)##2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+8##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-2##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 ] -oo,-8 [ uu ]2, +oo[ #