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?

It's not so bad; after you substitute $y=\sqrt{x-2}$ you will have
$$\sqrt{y^2 - 8y + 16 } + \sqrt{y^2 - 10y+ 25 } = 3$$
and you should be able to take it from here without any squaring of both sides.
