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

$$\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$$
 
Put x-1 = t^2 
so the equation becomes 
$$\sqrt{t^2+4-4t} + \sqrt{t^2+9-6t} =1$$
complete the squares . 
EDIT: 
My answer was incorrect. Thanks to egreg and Samatix. 
