Solve for $x$ : $\sqrt{x-6} \, + \, \sqrt{x-1} \, + \, \sqrt{x+6} = 9$

Question

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…

Answer 1

Hint:

Clearly,the LHS is increasing function of $x$

so,we cannot have multiple roots

For real solution, $x\ge6$

Also, $3\sqrt{x+6}>9>3\sqrt{x-6}$

$\implies x+6>9>x-6\iff3< x<15\implies6\le x<15$