Maybe, vectors will help

$\color{#20A900}{LHS:}C=\sqrt{x-2}+\sqrt{4-x},2\le x\le 4$ $\implies C^2=(\not x-2)+(4-\not x)+2\color{#D61F06}{\sqrt{(x-2)(4-x)}}$ We know, by $GM\le AM~~(\because2\le x\le 4)$ , $\color{#D61F06}{\sqrt{(x-2)(4-x)}}\le \dfrac{(x-2)+(4-x)}2=1$

$\therefore C^2\le2+2(\color{#D61F06}{1}\color{#333333})=4$

Hence $0\le C\le 2 \text{ or}\color{#3D99F6}{LHS\le 2}$ $\color{#20A900}{RHS:}(x-3)^2+2\color{#3D99F6}\ge 2$ Therefore $LHS=RHS =2$ This occurs when $\boxed{x=3}$ .

√(x-2) + √(4-x) = (x – 3)^2 + 2. On the LHS 2≤x≤4, On the RHS 2≤y≤∞

Let see on the LHS, y=√(x-2) + √(4-x), dy/dx = ( √(4-x)-√(x-2) )/(2√(x-2) . √(4-x)), the stationery point x=3,y=2. And it is also the solution of the equation.