<s>[INST] How to solve problems involving roots. $\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$ How to solve problems involving roots. If we square them they may go to fourth degree.There must be some technique to solve this.
$$\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$$
 [/INST] Put $y=\sqrt{x-1}$, getting $x=y^2+1$; thus the equation becomes
$$
\sqrt{y^2-4y+4}+\sqrt{y^2-6y+9}=1
$$
which should ring a bell.

 It becomes $|y-2|+|y-3|=1$ that can be treated without resorting to squaring; divide it into cases:
 If $y<2$, the equation becomes $2-y+3-y=1$, or $2y=4$, that means $y=2$, absurd.
 If $2\le y\le 3$, the equation becomes $y-2+3-y=1$, an identity.
 If $y>3$, the equation becomes $y-2+y-3=1$, or $2y=6$ and $y=3$, absurd. 
 Therefore the solutions are all the numbers $x$ such that $2\le\sqrt{x-1}\le3$, that is $4\le x-1\le 9$ or $5\le x\le 10$.

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