<s>[INST] Solve the inequality $x^2 - 3 > 0$ For the inequality $x^2 - 3 > 0$, we have 
\begin{align} x^2 - 3 & = (x+\sqrt 3)(x- \sqrt 3) > 0 \end{align}
Therefore,
\begin{align} x > -\sqrt 3 \end{align}
and
\begin{align} x > \sqrt 3 \end{align}
But, this is clearly wrong as we should get $x < -\sqrt 3$ and $x > \sqrt 3$ as the two intervals. What have I done wrong?
 [/INST] First treat the inequality as if it were to be an equation: $ x^{2} - 3 = 0 $. Thus, $$ x^{2} = 3 $$ and solving for $x$, $$x =\pm \sqrt{3}$$
This makes $x = -\sqrt{3}$ and $x = \sqrt{3}$ the "roots" of that equation.
Now to solve the $inequality$ $ x^{2} - 3 > 0 $, keep in mind the aforementioned "roots" into the following cases:
Case #1: When $x < -\sqrt{3}$, the inequality holds true since $x^{2} - 3$ is positive.
Case #2: $-\sqrt{3} < x < \sqrt{3}$ would make the inequality false because $x^{2} - 3$ is negative.
Case #3: When $x > \sqrt{3}$, the inequality holds true since $x^{2} - 3$ is also positive.
Thus the solutions to the inequality are $x < -\sqrt{3} \quad \textbf{or} \quad x > \sqrt{3}$.
The solutions expressed in interval notation: $(-\infty, -\sqrt{3}) \cup (\sqrt{3}, \infty)$.
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