# How do solve x^2>4 and write the answer as a inequality and interval notation?

Jan 6, 2018

$x < - 2$ or $x > 2$
Interval notation:
$\left(- \infty , - 2\right) \cup \left(2 , \infty\right)$

#### Explanation:

To solve, all we need to do is take the square root of both sides to get $x$ by itself:
$\sqrt{{x}^{2}} > \sqrt{4}$

$| x | > 2$

Which is true either when $x > 2$ or when $x < - 2$

To express the answer in interval notation,we think about which interval satisfies this inequality. It's simply all numbers less than $- 2$, and all numbers greater than $2$. This is the interval from $- \infty$ to $- 2$ and from $2$ to $\infty$. Note that the interval does not include $2$ or $- 2$ themselves.

We can write the interval like so:
$\left(- \infty , - 2\right) \cup \left(2 , \infty\right)$

The $\cup$ ($U$ for union) symbol means to combine the two intervals.

Jan 9, 2018

$x < - 2 \vee x > 2$

$x \in \left(- \infty , - 2\right) \cup \left(2 , \infty\right)$

#### Explanation:

Given:

${x}^{2} > 4$

Subtract $4$ from both sides to get:

${x}^{2} - 4 > 0$

Factor the left hand side to find:

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

The left hand side is positive if both of the factors are positive or both of the factors are negative.

Hence:

$x < - 2 \text{ }$ or $\text{ } x > 2$

In interval notation:

$x \in \left(- \infty , - 2\right) \cup \left(2 , \infty\right)$
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