# 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?

You have $$3$$ intervals to consider: $$(-\infty,-\sqrt3), (-\sqrt3,\sqrt3)$$ and $$(\sqrt3,\infty)$$. The sign of the product $$(x-\sqrt3)(x+\sqrt3)$$ is constant on each interval. (That's by continuity, since the function $$f(x)=x^2-3$$ must pass through zero to change sign.)

Use a test point in each interval: \begin{align} (-\infty,-\sqrt3): f(x)&\gt0\\ (-\sqrt3,\sqrt3): f(x)&\lt0\\ (\sqrt3,\infty): f(x)&\gt0\end{align}

So actually you get $$x\lt -\sqrt3 \color{blue}{\text{ or }} x\gt\sqrt3$$.

HINT

Remember $$ab > 0$$ if $$a,b > 0$$ or $$a,b < 0$$

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: $$(-\infty, -\sqrt{3}) \cup (\sqrt{3}, \infty)$$.