2. Solve in the set of real numbers the equation:

$$
\sqrt{|x-1|}+\sqrt{|x-2015|}=\sqrt{2014}
$$

(Problem E:14587 from G.M. nr. 12/2013)
2. Case I: $x2014 \Rightarrow$

$\sqrt{|x-2015|}>\sqrt{2014}$, so the equation has no solution.

$2 \mathrm{p}$

Case II: $1 \leq \mathrm{x} \leq 2015 \Rightarrow \mathrm{x}-1 \geq 0 \Rightarrow|\mathrm{x}-1|=\mathrm{x}-1$

$$
\Rightarrow x-2015 \leq 0 \Rightarrow|x-2015|=-x+2015
$$

After squaring the equation becomes: $\mathrm{x}-1+2 \sqrt{|x-1| \cdot|x-2015|}-\mathrm{x}+2015=2014 \Leftrightarrow$ $|x-1| \cdot|x-2015|=0 \Rightarrow x=1$ or $x=2015$. $2 \mathrm{p}$

Case III: $\mathrm{x}>2015 \Rightarrow \mathrm{x}-1>2014 \Rightarrow|\mathrm{x}-1|>2014 \Rightarrow \sqrt{|x-1|}>\sqrt{2014}$, so the equation has no solution.

The solution to the equation is: $\mathrm{x} \in\{1,2015\}$.