<s>[INST] 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?
 [/INST] $$\sqrt{x+14-8\sqrt{x-2}}  +\sqrt{x+23-10\sqrt{x-2}} = 3$$
$$\sqrt{(\sqrt{x-2}-4})^2+\sqrt{(\sqrt{x-2}-5)^2}=3$$
Then 
$$|\sqrt{x-2}-4|+|\sqrt{x-2}-5|=3$$
Let $\sqrt{x-2}\ge5 (x\ge27)$. Then 
$$\sqrt{x-2}-4+\sqrt{x-2}-5=3$$
$$\sqrt{x-2}=6$$
Let $\sqrt{x-2}\le4 (2\le x\le18)$. Then 
$$4-\sqrt{x-2}-\sqrt{x-2}+5=3$$
$$\sqrt{x-2}=3$$
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