2. Zadatak

Najprije uočimo $x^2+2x+1=(x+1)^2$, pa je $\sqrt{x^2+2x+1}=|x+1|$. Jednadžba postaje $|x+3|+2|x+1|=7$.

Treba promatrati tri slučaja:

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

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

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

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

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

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

$$\begin{aligned}x+3+2x+2&=7\\x&=\frac23\in[-1,\infty\rangle\end{aligned}$$

Jednadžba ima dva rješenja: $x=-4$, $x=\dfrac23$.