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

Jun 30, 2015

${x}^{2} > 4$ when $x < - 2$ and when $x > 2$.

#### Explanation:

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 $x$.

$\left(x + 2\right) \left(x - 2\right) = 0$

$x + 2 = 0$ 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 $f \left(- 3\right) = {\left(- 3\right)}^{2} - 4 = 9 - 4 = 5$

$f \left(x\right) > 0$ when $x < - 2$.

Case (b): Let $x = 0$.

Then $f \left(0\right) = {0}^{2} - 4 = 0 - 4 = - 4$

$f \left(x\right) < 0$ when $- 2 < x < 2$

Case (c): Let $x = 3$.

Then $f \left(3\right) = {3}^{2} - 4 = 9 - 4 = 5$

$f \left(x\right) > 0$ when $x > 2$.

If ${x}^{2} - 4 > 0$ when $x < - 2$ and when $x > 2$,

${x}^{2} > 4$ when $x < - 2$ and when $x > 2$.