<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] 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 $5\le x\le 10$.
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