Solve for $x$ : $\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$? I want to solve the following equation for $x$ :
$$\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$$
My approach:  
Let the given eq.:
$$\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9 \tag {i}$$
On rearranging, we get:
$$\sqrt{x-6} \,  + \, \sqrt{x+6} = 9 \, - \, \sqrt{x-1} $$
On Squaring both sides, we get:
$$(x-6) \,  + \, (x+6) + 2 \,\,. \sqrt{(x^2-36)} = 81 + (x-1)\, - 18.\, \sqrt{x-1}$$
$$\implies 2x + 2 \,\,. \sqrt{(x^2-36)}= 80 + x - 18.\, \sqrt{x-1}$$
$$\implies x + 2 \,\,. \sqrt{(x^2-36)}= 80  - 18.\, \sqrt{x-1} \tag{ii}$$
Again we are getting equation in radical form.
But, in Wolfram app, I am getting its answer as $x=10$, see it in WolframAlpha. 
So, how to solve this equation? Please help...

Hint:
Clearly,the  LHS is increasing function of $x$
so,we cannot have multiple roots
For real solution, $x\ge6$
Also, $3\sqrt{x+6}>9>3\sqrt{x-6}$
$\implies x+6>9>x-6\iff3< x<15\implies6\le x<15$
