<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] 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.
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