Example 19. Solve the equation

$$
\sqrt{x+11}=x-1
$$
Solution.

The first method. The domain of definition (DOD) of equation (7) is given by the condition $x \geqslant-11$. Considering that

$$
f(x)=g(x) \Rightarrow f^{2}(x)=g^{2}(x)
$$

we have

$$
\text { (7) } \begin{aligned}
\Rightarrow x+11=x^{2}-2 x+1 \Leftrightarrow x^{2}-3 x-10=0 \Leftrightarrow \\
\Leftrightarrow(x-5)(x+2)=0 \Leftrightarrow\left[\begin{array}{l}
x=5 \\
x=-2
\end{array}\right.
\end{aligned}
$$

## 14 CH. 1. EQUIVALENT EQUATIONS AND INEQUALITIES

Thus, the solutions of the original equation are among the numbers $x_{1}=5$ and $x_{2}=-2$.

Before making the check, let us draw the reader's attention to a common mistake. Transitioning from the given equation to its consequence, roots are found. Then, it is checked whether the found roots belong to the DOD of the original equation. The roots that do not belong to the DOD are discarded, and the remaining ones (belonging to the DOD of the original equation) are listed in the answer. This is where the mistake lies. It is not enough to check whether the found roots belong to the DOD of the equation. It is necessary to check whether the roots of the consequence, which belong to the DOD of the original equation, satisfy the original equation itself. This is confirmed by the given example.

Indeed, both roots of the system satisfy the DOD, but the number 5 satisfies equation (7), while the number -2 does not satisfy it. Therefore, equation (7) has a single root $x=5$.

The second method. The equation $\sqrt{\alpha(x)}=\beta(x)$ is equivalent to the system

$$
\left\{\begin{array}{l}
\beta(x) \geqslant 0 \\
\alpha(x)=\beta^{2}(x)
\end{array}\right.
$$

since $\beta^{2}(x) \geqslant 0$, and the equality $\alpha(x)=\beta^{2}(x)$ imposes a non-negativity condition on $\alpha(x)$.

When solving equations of type (7) by equivalent transitions, it is not necessary to find the DOD of this equation, but it is mandatory to impose the non-negativity condition on the function $\beta(x)$.

Considering this, for the given example, we have

(7) $\Leftrightarrow\left\{\begin{array}{l}x-1 \geqslant 0, \\ x+11=x^{2}-2 x+1\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geqslant 1, \\ x^{2}-3 x-10=0\end{array} \Leftrightarrow\right.\right.$

$$
\Leftrightarrow\left\{\begin{array}{l}
x \geqslant 1 \\
{\left[\begin{array}{l}
x=-2, \\
x=5
\end{array}\right.}
\end{array} \Leftrightarrow x=5\right.
$$

Thus, $x=5$ is the only solution to equation (7).