243. Find the real roots of the equation

$$
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
$$

(all square roots are considered positive),
243. We have:

$$
\begin{aligned}
x+3-4 \sqrt{x-1}= & x-1-4 \sqrt{x-1}+4= \\
& =(\sqrt{x-1})^{2}-4 \sqrt{x-1}+4=(\sqrt{x-1}-2)^{2}
\end{aligned}
$$

and similarly

$$
x+8-6 \sqrt{x-1}=x-1-6 \sqrt{x-1}+9=(\sqrt{x-1}-3)^{2}
$$

Thus, our equation can be written as

$$
\sqrt{(\sqrt{x-1}-2)^{2}}+\sqrt{(\sqrt{x-1}-3)^{2}}=1
$$

or, since we consider all roots to be positive,

$$
|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1
$$

where $|y|$ denotes the absolute value of the number $y$.

Now let's consider several cases.

$1^{\circ}$. If $\sqrt{x-1}-2 \geqslant 0, \sqrt{x-1}-3 \geqslant 0$, i.e., if $\sqrt{x-1} \geqslant 3, \quad x-1 \geqslant 9, \quad x \geqslant 10, \quad$ then. $\quad|\sqrt{x-1}-2|=$ $=\sqrt{x-1}-2, \quad|\sqrt{x-1}-3|=\sqrt{x-1}-3 \quad$ and the equation
takes the form

$$
\sqrt{x-1}-2+\sqrt{x-1}-3=1
$$

from which

$$
2 \sqrt{x-1}=6, \quad x-1=9, \quad x=10
$$

$2^{\circ}$. If $\sqrt{x-1}-2 \geqslant 0, \quad \sqrt{x-1}-3 \leqslant 0, \quad$ i.e., if $\sqrt{x-1} \geqslant 2, \quad x \geqslant 5, \quad$ but $\quad \sqrt{x-1} \leqslant 3, \quad x \leqslant 10, \quad$ then $|\sqrt{x-1}-2|=\sqrt{x-1}-2, \quad|\sqrt{x-1}-3|=-\sqrt{x-1}+3$ and the equation becomes an identity

$$
\sqrt{x-1}-2-\sqrt{x-1}+3=1
$$

Thus, our equation is satisfied by all values of $x$ between $x=5$ and $x=10$.

$3^{\circ}$. If $\sqrt{x-1}-2 \leqslant 0, \sqrt{x-1}-3 \leqslant 0$, i.e., if $\sqrt{x-1} \leqslant 2, x \leqslant 5$, then $|\sqrt{x-1}-2|=-\sqrt{x-1}+2,|\sqrt{x-1}-3|=$ $=-\sqrt{x-1}+3$ and the equation takes the form

$$
-\sqrt{x-1}+2-\sqrt{x-1}+3=1
$$

from which

$$
2 \sqrt{x-1}=4, x-1=4, \quad x=5
$$

$4^{\circ}$. The case $\sqrt{x-1}-2 \leqslant 0, \sqrt{x-1}-3 \geqslant 0$ is obviously impossible.

Thus, the solution to this equation is all values of $x$ between $x=5$ and $x=10$, i.e., $5 \leqslant x \leqslant 10$.