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...

I'm not entirely sure if this approach will find all possible $x$ in any such question, nevertheless here it is giving the correct answer and is relatively short.
Since the RHS $=9$ is an integer, all three terms in the RHS under the radicals must be perfect squares. $$\therefore, x-6=p^2$$
$$x-1=q^2$$ $$x+6=r^2$$
where $p,q,r$ are integers. $$\Rightarrow x=p^2+6=q^2+1=r^2-6$$
Here we need value of only one of $p,q,r$ to find $x$
$\Rightarrow (p+q)(p-q)=-5$
$\Rightarrow (p+q),(p-q)=(-1)(5), (1,-5), (5,-1), (-5,1)$
$\Rightarrow p=±2$
Since $x=p^2+6$,
$$\therefore \fbox{x=10}$$
