## Task B-2.3.

Solve the equation

$$
\sqrt{x^{2}+2 x+1}+\sqrt{x^{2}-2 x+1}=2024
$$
## Solution.

Notice that the expressions under both square roots are squares of binomials, so the given equation can be written as

$$
\begin{gathered}
\sqrt{(x+1)^{2}}+\sqrt{(x-1)^{2}}=2024, \text { or } \\
|x+1|+|x-1|=2024
\end{gathered}
$$

We will solve this equation by intervals $x \leqslant-1, -1 < x \leqslant 1, x > 1$. For $x \leqslant-1$, we solve the equation $-x-1-x+1=2024$. Its solution is $x=-1012$ and it satisfies the condition $x \leqslant-1$, so it is a solution to the original equation.

For $-1 < x \leqslant 1$, the equation $x+1-x+1=2024$ simplifies to $2=2024$, which has no solution.

For $x > 1$, we solve the equation $x+1+x-1=2024$. Its solution is $x=1012$ and it satisfies the condition $x>1$, so it is a solution to the original equation.

Therefore, the solutions to the original equation are -1012 and 1012.