How do you solve #x^2>4# using a sign chart?

1 Answer
Dec 16, 2016

Answer:

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

Explanation:

Let #f(x)=x^2-4#

Let's rewrite the equation as

#x^2-4=(x+2)(x-2)>0#

Now we can do the sign chart

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

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