Solve the following inequality on the set of real numbers:

$$
\sqrt{x+6}>x-6 .
$$
Let's plot the functions

$$
f(x)=\sqrt{x+6} \quad(\text { Domain: } x \geq-6) \quad \text { and } \quad g(x)=x-6 \quad(\text { Domain: } x \in \mathbf{R})
$$

in the same coordinate system.

![](https://cdn.mathpix.com/cropped/2024_05_02_f8b7dd0c4231739d08aag-1.jpg?height=623&width=634&top_left_y=304&top_left_x=735)

The elements of the solution set of the inequality can only be chosen from real numbers not less than -6. $f(-6)=0, g(-6)=-12$, so the inequality holds at $x=-6$.

Both functions are strictly increasing, but except for an initial small segment, $f(x)$ grows more slowly than $g(x)$. Let's calculate where $g(x)$ "catches up" with $f(x)$. The

$$
\sqrt{x+6}=x-6 \quad(x \geq-6)
$$

equation is traditionally solved by squaring both sides, which leads to a consequential equation. The

$$
x+6=(x-6)^{2}
$$

equation will contain the root(s) of the original equation, but false roots may also appear. (Squaring is not an equivalent transformation.) The resulting

$$
x^{2}-13 x+30=0
$$

quadratic equation has roots: $x_{1}=10, x_{2}=3$. The latter is a false root. The original equation is satisfied by $x_{1}=10$, but not by $x_{2}=3$.

Based on all this, the solution to the inequality is:

$$
-6 \leq x<10
$$

Hegedús Dalma (Siófok, Perczel Mór Gymn., I. o.t.)

Note. Many fell into the trap because they did not check the roots, so they did not realize that 3 is a false root. Perhaps the false root is more easily revealed if the $\sqrt{x+6}=x-6$ equation is written in the form

$$
(x+6)-\sqrt{x-6}-12=0
$$

This is a quadratic equation in $\sqrt{x+6}$. According to the quadratic formula,

$$
\sqrt{x+6}=\frac{1 \pm \sqrt{1+48}}{2}=\left\{\begin{array}{c}
4 \\
-3
\end{array}\right.
$$

Since $\sqrt{x+6} \geq 0$, only $\sqrt{x+6}=4$, i.e., $x=10$ is valid.