Example 2. Solve the equation:
a) $\sqrt{1+3 x}=1-x$
b) $\sqrt{1+3 x}=x-1$
Solution. a) Squaring both sides of the equation, we get $1+3 x=1-2 x+x^{2}$, i.e., the equation

$$
x^{2}-5 x=0
$$

which is a consequence of the original equation. We find the roots of this equation: $x_{1}=0$ and $x_{2}=5$.

Let's check if these roots satisfy the original equation.

Let $x=0$; then the original equation turns into a true numerical equality. Therefore, $x_{1}=0$ is a root of the original equation.

Let $x=5$; then in the left part of the original equation we have $\sqrt{1+3 \cdot 5}=4$, and in its right part $1-5=-4$. Since $4 \neq-4$, $x_{2}=5$ is not a root of the original equation.

Thus, $x_{1}=0$ is the only root of the original equation.

b) Squaring both sides of the equation, we get

$$
1+3 x=x^{2}-2 x+1
$$

i.e., the equation

$$
x^{2}-5 x=0
$$

which is a consequence of the original equation.

We find the roots of this equation: $x_{1}=0$ and $x_{2}=5$. By substituting each of these roots into the original equation, we establish that the root $x_{1}=0$ is extraneous, while the root $x_{2}=5$ satisfies it.

Thus, the number $x_{2}=5$ is the only root of the original equation.

Note that a consequence of equations a) and b) is the same equation $x^{2}-5 x=0$, which has two roots: $x_{1}=0$ and $x_{2}=5$. The root $x_{1}=0$ is a root of equation a), but extraneous for equation b); the root $x_{2}=5$ is a root of equation b), but is extraneous for equation a).