## Problem 3

Solve the equation in the set $\mathbb{R}$: $\sqrt{x-2}+\sqrt[3]{4-x}=2$.
## Problem 3

CE: $x \geq 2 \Rightarrow x \in[2, \infty)$, let $\sqrt{x-2}=u \geq 0$ and $\sqrt[3]{4-x}=v \Rightarrow u+v=2 \Rightarrow u=2-v$ (1).

$\left\{\begin{array}{l}u^{2}=x-2 \\ v^{3}=4-x\end{array} \Rightarrow u^{2}+v^{3}=2\right.$ $\qquad$
From (1) and (2) $\Rightarrow v^{3}+v^{2}-4 v+2=0 \Leftrightarrow(v-1)\left(v^{2}+2 v-2\right)=0 \Rightarrow v_{1}=1, v_{2}=-1+\sqrt{3}, v_{3}=-1-\sqrt{3}$

$u_{1}=1 \Rightarrow x_{1}=3 \geq 2 ; u_{2}=3-\sqrt{3} \Rightarrow x_{2}=14-6 \sqrt{3} \geq 2, u_{3}=3+\sqrt{3} \Rightarrow x_{3}=14+6 \sqrt{3} \geq 2$. . $.2 p$