<s>[INST] What is the number of real solutions of the following? $ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $ What is the number of real solutions of the following? 
$$ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $$
My solution:
$$ \sqrt{x + 3 - 4\sqrt{x-1}} + \sqrt{x + 8 - 6\sqrt{x-1}} = 1 $$
$$ \implies \sqrt{(\sqrt{x-1}-2)^2} + \sqrt{(\sqrt{x-1}-3)^2} = 1 $$
$$ \implies (\sqrt{x-1}-2) + (\sqrt{x-1}-3) = 1 $$
$$ \implies \sqrt{x-1} = 3$$
So, $ x = 10$ is the only solution.
But the answer key (and Wolfram alpha too) says there are infinite number of solutions to this equation. Where I am going wrong?
 [/INST] It happens, for the choices of the arguments of the radicals in this problem, that we can make this a bit less of a headache to think about by noting that setting $ \ u = \sqrt{x-1} \ \Rightarrow \ u^2 = x - 1 \ $  reduces the original equation to 
$$ \sqrt{(u^2 + 4) - 4u} \ + \ \sqrt{(u^2 + 9) - 6u} \ = \ 1 \ \Rightarrow  \ | u - 2 | \ + \ | u - 3 | = 1 \ . $$
[To this point, this is similar to Adriano's argument.]
Since these terms must be positive or zero, we can choose, say, $ \ | u - 2 | = a \ $   and $ \  | u - 3 | = 1 - a \ , $  with $ \ 0 \le a \le 1 \ . $  Two of the possible equations, $ \ u - 2 = a \ $ and $ \ 3 - u = 1 - a \ $  produce $ \ u = 2 + a \ $ , while using $ \ 2 - u \ = a \ \Rightarrow \ u = 2 - a \ $ or $ \ u - 3 \ \Rightarrow \ 1 - a \ \Rightarrow \ u = 4 - a \ $ are not mutually consistent results.  
So we have the single interval,
$$ 0 \ \le \ a \ \le \ 1 \ \Rightarrow \ 2 \ \le \ u = 2 + a \ \le \ 3 \ . $$
(Graphing $ \ | u - 2 | \ + \ | u - 3 | = 1 \ $ confirms this.)  From this, we have 
$$4 \ \le \ u^2 = x - 1 \ \le \ 9 \ \Rightarrow \  5 \ \le \ x  \ \le \ 10 \ . $$
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