29. Give three real roots of $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$.
(ans. Any three numbers in $[5,10]$.
$$
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
$$
can be written
$$
\sqrt{(x-1)-4 \sqrt{x-1}+4}+\sqrt{(x-1)-6 \sqrt{x-1}+9}=\sqrt{(\sqrt{x-1}-2)^{2}}+
$$
$\sqrt{(\sqrt{x-1}-3)^{2}}=1$ hence $|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$. There are four cases. (I) $\sqrt{x-1}-2 \geq 0$ and $\sqrt{x-1}-3 \geq 0$ which implies $x \geq 10 \Rightarrow x=10$. (II) $\sqrt{x-1}-2 \leq 0$ and $\sqrt{x-1}-3 \leq 0$ which implies $x \leq 5 \Rightarrow x=5$. (III) $\sqrt{x-1}-2 \geq 0$ and $\sqrt{x-1}-3 \leq 0$ implies $5 \leq x \leq 10$ and (IV) $\sqrt{x-1}-2 \leq 0$ and $\sqrt{x-1}-3 \geq 0$, which is impossible. Thus, any $x \in[5,10]$ solves the equation.)