For the inequality $x^2 - 3 > 0$, we have

\begin{align} x^2 - 3 & = (x+\sqrt 3)(x- \sqrt 3) > 0 \end{align}


\begin{align} x > -\sqrt 3 \end{align}


\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 $.



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


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