Finding Domain with sqrt on bottom of Fraction

[airforceone]

How do I do this?

Suppose I have: 1/(sqrt(x^2 - 4))

I know that there should not be a negative square root and a zero in the bottom of a fraction, so I tried this:

x^2 - 4 >= 0
x^2 >= 4
x >= 2 , -2
D: {x: x >= 2, -2}

But then I got confused that the bottom of a fraction can't be zero also, so I tried this:

x^2 - 4 <> 0
x^2 <> 4
x <> 2, -2
D: {x: x <> 2, -2}

The correct answer is:

D: {x: x > 2 or x < 2}

How did they do it algebraically? I already know how it's right.

 

[pka]

Suppose I have: 1/(sqrt(x^2 - 4))
The correct answer is:
D: {x: x > 2 or x < 2}
How did they do it algebraically? I already know how it's right.

Correct because division by zero is not allowed.

\(\displaystyle \begin{array}{l}
x^2 - 4 > 0 \\
(x - 2)(x + 2) > 0 \\
x < - 2\quad \vee \quad x > 2 \\
\left( { - \infty , - 2} \right) \cup \left( {2,\infty } \right) \\
\end{array}\)