In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions? In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions?
I tried shifting the second term to the rhs and squaring.Even after that i'm left with square roots.No idea how to proceed.Help!

\begin{align}
&\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}} = 1\\ \implies &\sqrt{(x-1)-4\sqrt{x-1} + 4}+\sqrt{(x-1)-6\sqrt{x-1}+9}=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\tag{1}
\end{align}
This calls for casework:

1. $\quad\sqrt{x-1}\geq 3$
$(1)\implies \sqrt{x-1}-2 + \sqrt{x-1}-3 = 1\implies \sqrt{x-1} = 3\implies x=10$

2. $\quad 2\leq\sqrt{x-1} < 3$
$(1)\implies \sqrt{x-1}-2 - \sqrt{x-1}+3 = 1\implies 1 = 1$ and thus all $x$ such that $2\leq\sqrt{x-1} < 3$, i.e. $x\in [5,10\rangle$

3. $\sqrt{x-1} < 2$
$(1)\implies -\sqrt{x-1}+2 - \sqrt{x-1}+3 = 1\implies \sqrt{x-1} = 2 \implies x = 5$, but this $x$ doesn't satisfy our condition 3 (although it does satisfy condition 2, and is already included as a solution)

Taking union, we conclude that any $x\in[5,10]$ is a solution to the equation.
