3. Zadatak

Uvedimo supstituciju $t=x-1$. Tada jednadžba prelazi u

$$(t+1)^2+\left(\frac{t+1}{t}\right)^2=8,$$ $$t^2+2t+1+1+\frac2t+\frac1{t^2}=8,$$ $$t^2+2+\frac1{t^2}+2\cdot\left(t+\frac1t\right)-8=0,$$ $$\left(t+\frac1t\right)^2+2\cdot\left(t+\frac1t\right)-8=0.$$

Rješavanjem ove jednadžbe dobivamo

$$t+\frac1t=2\qquad\text{ili}\qquad t+\frac1t=-4.$$

Slijedi

$$t^2-2t+1=0\qquad\text{ili}\qquad t^2+4t+1=0,$$

odnosno

$$t_1=1\quad\Longrightarrow\quad x_1=2,$$ $$t_2=-2-\sqrt3\quad\Longrightarrow\quad x_2=-1-\sqrt3,$$ $$t_3=-2+\sqrt3\quad\Longrightarrow\quad x_3=-1+\sqrt3.$$

Rješenja su svi $x\in\{2,-1-\sqrt3,-1+\sqrt3\}$.