Task B-1.2. (20 points)

Solve the equation in the set of real numbers

$$
|x+3|+2 \sqrt{x^{2}+2 x+1}=7
$$
Solution. First, note that $x^{2}+2 x+1=(x+1)^{2}$, so $\sqrt{x^{2}+2 x+1}=|x+1|$. The equation becomes $|x+3|+2|x+1|=7$.

We need to consider three cases:

$1^{\circ} x \in\langle-\infty,-3\rangle ; \quad 2^{\circ} x \in[-3,-1\rangle ; \quad 3^{\circ} x \in[-1, \infty\rangle$

(2 points)

$1^{\circ} x \in\langle-\infty,-3\rangle$

$$
\begin{aligned}
-x-3-2 x-2 & =7 \\
x & =-4 \in\langle-\infty,-3\rangle
\end{aligned}
$$

(5 points)

$2^{\circ} x \in[-3,-1\rangle$

$$
\begin{aligned}
x+3-2 x-2 & =7 \\
x & =-6 \notin[-3,-1\rangle
\end{aligned}
$$

(5 points)

$3^{\circ} x \in[-1, \infty\rangle$

$$
\begin{aligned}
x+3+2 x+2 & =7 \\
x & =\frac{2}{3} \in[-1, \infty\rangle
\end{aligned}
$$

(5 points)

The equation has two solutions: $x=-4, x=\frac{2}{3}$.

Note. If a student writes $\sqrt{\left|x^{2}+2 x+1\right|}=x+1$, they will get the third case, and if they find the correct $x$, they receive 5 points.