23. Find the sum of all the integral values of $x$ that satisfy
$$
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
$$
23. Answer: 45
The equation can be rewritten as $\sqrt{(\sqrt{x-1}-2)^{2}}+\sqrt{(\sqrt{x-1}-3)^{2}}=1$. If $\sqrt{x-1} \geq 3$, it reduces to $\sqrt{x-1}-2+\sqrt{x-1}-3=1$ i.e. $\sqrt{x-1}=3$ giving $x=10$.
If $\sqrt{x-1} \leq 2$, it reduces to $2-\sqrt{x-1}+3-\sqrt{x-1}=1$ i.e. $\sqrt{x-1}=2$ giving $x=5$.
If $2<\sqrt{x-1}<3$, i.e. $5<x<10$, it reduces to $\sqrt{x-1}-2+3-\sqrt{x-1}=1$ which is true for all values of $x$ between 5 and 10 .
Hence the sum of all integral solutions is $5+6+7+8+9+10=45$.