Find all real numbers x, y, z, u and v in $\sqrt{x}+\sqrt{y}+2\sqrt{z-2}+\sqrt{u}+\sqrt{v}=x+y+z+u+v$ And thanks in advance for your answers. Sorry if the text is badly formatted, I'm new here. Anyway, here is the question:
Find all real numbers x, y, z, u and v in $\sqrt{x}+\sqrt{y}+2\sqrt{z-2}+\sqrt{u}+\sqrt{v}=x+y+z+u+v$
Could I use the method of completing the square (I'm familiar with that) or is there some better way?
 A: HINT.-You have always $2\sqrt{z-2}-z\le -1$ and $x-\sqrt x$ takes its minumun at $x=\frac 14$ this minimun being equal to $-\frac 14$. Since 
$$2\sqrt{z-2}-z=(x-\sqrt x)+(y-\sqrt y)+(u-\sqrt u)+(v-\sqrt v)$$ it follows that the only solution is $$(x,y,z,u,v)=(\frac14,\frac14,3,\frac14,\frac14)$$
A: Yes, you can solve your equation by completing the squares.
$$\sqrt{x}+\sqrt{y}+2\sqrt{z-2}+\sqrt{u}+\sqrt{v}=x+y+z+u+v\tag{*1}$$
For each variable, subtract its appearance in LHS from corresponding term  in RHS, you have:
$$
\begin{array}{rl}
x - \sqrt{x} &= (\sqrt{x}-\frac12)^2 - \frac14\\
y - \sqrt{y} &= (\sqrt{y}-\frac12)^2 - \frac14\\
z - 2\sqrt{z-2} &= (\sqrt{z-2}-1)^2 + 1\\
u - \sqrt{u} &= (\sqrt{u}-\frac12)^2 - \frac14\\
v - \sqrt{v} &= (\sqrt{v}-\frac12)^2 - \frac14\\
\end{array}
$$
Now sum over both sides and compare result with $(*1)$, you find:
$$\verb/RHS/(*1) - \verb/LHS/(*1) = 
(\sqrt{x}-\frac12)^2 +
(\sqrt{y}-\frac12)^2 +
(\sqrt{z-2}-1)^2 +
(\sqrt{u}-\frac12)^2 +
(\sqrt{v}-\frac12)^2$$
This leads to
$$\begin{cases}
\sqrt{x} = \sqrt{y} = \sqrt{u} = \sqrt{v} = \frac12,\\
\sqrt{z-2} = 1
\end{cases}
\quad\implies\quad (x,y,z,u,v) = \left(\frac14,\frac14,3,\frac14,\frac14\right)$$
