Let's modify the inequality by subtracting #4# from each side. Then
#x^2 – 4 > 0#
We start by finding the critical numbers.
Set #f(x) = x^2 – 4 = 0# and solve for #x#.
#(x+2)(x - 2) = 0#
#x+2 = 0# or #x-2 = 0#
#x = -2# or #x = +2#
The critical numbers are #-2# and #+2#.
Now we check for positive and negative regions.
We have three regions to consider: (a) #x < -2#; (b) #-2< x <2#; and (c) #x >2#.
Case (a): Let #x = -3#.
Then #f(-3) = (-3)^2 - 4 = 9-4 = 5#
#f(x) > 0# when #x < -2#.
Case (b): Let #x = 0#.
Then #f(0) = 0^2 -4 = 0-4 = -4#
#f(x) < 0# when #-2 < x < 2#
Case (c): Let #x = 3#.
Then #f(3) = 3^2 - 4 = 9-4 = 5#
#f(x) > 0# when #x > 2#.
If #x^2 -4 > 0# when #x < -2# and when #x > 2#,
#x^2 >4# when #x < -2# and when #x > 2#.
