3A. In the set of real numbers, solve the inequality

$$
\sqrt{x^{2}-x-12}>7 x
$$
Solution. First, we need to determine the domain in which the square root is defined, i.e., the set where $x^{2}-x-12 \geq 0$. This is equivalent to $(x-4)(x+3) \geq 0$, from which $x \in(-\infty,-3] \cup[4,+\infty)$.

We will consider two cases. For $x < 0$, the inequality $x < \sqrt{x^{2}-x-12}$ holds for every $x$ for which the root is defined, i.e., on the domain. Therefore, every $x \in(-\infty,-3]$ is a solution to the original inequality. For $x \geq 0$, we can square the inequality, yielding $x^{2}-x-12 > 49x^{2} \Leftrightarrow 48x^{2}+x+12 < 0$. However, the equivalent inequality has no real solutions. Finally, the solutions to the original inequality are all $x \in(-\infty,-3]$.