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?
 A: Separate the square roots and square both sides.  If you isolate the term $\sqrt{x+3-4\sqrt{x-1}}$, after simplifying you will obtain:
$$\sqrt{x+8-6\sqrt{x-1}}=3-\sqrt{x-1}$$
Squaring both sides again gives:
$$x+8-6\sqrt{x-1}=9-6\sqrt{x-1}+x-1$$
Which holds for all valid $x$.  It follows that the original equation is true for any $x$ in the domain of the left hand side.  This is $5\le x\le 10$.
A: To see where the infinite solutions come from, first note that:
$$
\sqrt{(\sqrt{x-1}-2)^2} + \sqrt{(\sqrt{x-1}-3)^2} = 1 \iff 1=|\sqrt{x-1}-2| + |\sqrt{x-1}-3|
$$
Now consider the case where $5\le x\le10$. This implies that:
$$
\sqrt{x-1}-2\ge\sqrt{5-1}-2=0 \qquad \text{and} \qquad\sqrt{x-1}-3\le\sqrt{10-1}-3=0
$$
Thus we obtain:
$$
1=|\sqrt{x-1}-2| + |\sqrt{x-1}-3| = (\sqrt{x-1}-2) - (\sqrt{x-1}-3) = 1
$$
Since we obtained a tautology ($1=1$), it follows that any $x\in[5,10]$ satisfies the equation.
A: 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 
$$4 \ \le \ u^2 = x - 1 \ \le \ 9 \ \Rightarrow \  5 \ \le \ x  \ \le \ 10 \ . $$
