Find $x$ given $\sqrt{x+14-8\sqrt{x-2}}$ + $\sqrt{x+23-10\sqrt{x-2}} = 3$ If $\sqrt{x+14-8\sqrt{x-2}}$  + $\sqrt{x+23-10\sqrt{x-2}} = 3$, then what is the value of $x$?
Is there an easy way to solve such equation, instead of squaring on both sides and replacing $\sqrt{x-2}$ with a different variable?

Condition: $x\geq 2$.
You have $$LHS = \sqrt{(\sqrt{x-2}-4)^2} + \sqrt{(\sqrt{x-2}-5)^2}.$$
So, if $\sqrt{x-2} < 4$, or $x<18$, then $$LHS = 4-\sqrt{x-2} + 5-\sqrt{x-2} = 9-2\sqrt{x-2} = 3 = RHS.$$
If $4 \leq \sqrt{x-2} < 5$, or $18\leq x < 27$, then $$LHS = \sqrt{x-2} -4 + 5-\sqrt{x-2} = 1 \neq 3 = RHS.$$
If $\sqrt{x-2} \geq 5$, or $x \geq 27$, then $$LHS = \sqrt{x-2}  - 4+ \sqrt{x-2} -5= 2\sqrt{x-2} - 9 = 3 = RHS.$$
