Q:

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

A:

Straight, you can get the following equation : 
$$
\sqrt{(2-\sqrt{x-1})^2} + \sqrt{(3-\sqrt{x-1})^2} =1
$$
which leads to the following equation :
$$
|2-\sqrt{x-1}| + |3-\sqrt{x-1}| =1
$$
Then you will have three cases to discuss :

case : $\sqrt{x-1} \leq2$  (equivalent to $x\leq5$) :

$\sqrt{x-1} = 2$ then $x = 5$

case : $\sqrt{x-1} >2$ and $\sqrt{x-1} <3$  (equivalent to $5<x<10$) :

The equation below can be written : 
$$
\sqrt{x-1}-2 + 3-\sqrt{x-1} =1
$$
equivalent to :
$
1=1
$
The solutions belongs to $]5,10[$

case : $\sqrt{x-1} \geq3$  (equivalent to $x\geq10$) :

$\sqrt{x-1} = 3$ then $x = 10$
The solutions belongs to $[5,10]$

