How do I use a sign chart to solve #x^2<4#?

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Answer 1 · 2015-06-30T17:34:24

#x^2 < 4# when #-2 < x < 2#.

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#.

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If #x^2 -4 < 0# when #-2 < x < 2#, then

#x^2 < 4# when #-2 < x < 2#.