Two different solutions for the irrational inequality $(x-1) \sqrt{x^2-x-2} \geq 0$. Upon solving this problem with the concept that I have understood about solving inequalities until now, I got the following range for $x$ :-
$(x-1) \sqrt{x^2-x-2} \geq 0$
$\Rightarrow(x-1) \sqrt{(x-2)(x+1)} \geq 0$
$\Rightarrow(x-1)^2 (x-2)(x+1)\geq 0 \text{ ; upon squaring both sides }$
$\Rightarrow x\in(-\infty,-1]\cup[2,\infty)$
This is the solution range that I got for the given inequality and I have confirmed this solution by checking this in Wolfram Alpha. But this solution only comes up if I give the modified problem statement as $(x-1)^2 (x-2)(x+1)\geq 0$ i.e. after squaring both sides, on the website as you can see here.
And I thought that this should be the correct answer but when I give the original problem statement into the website, it gives the result as $x\in[2,\infty)$ as you can see here.
Now I am confused as to which one is the correct solution for this problem. Why there are two different solutions to the same problem? How a simple act of squaring both sides is changing the solutions completely? Please help me on this !!!
Thanks in advance !!!

If you're looking for a full solution: $(x-1)\sqrt{x^2-x-2}\ge 0$ automatically means that $x-1\ge 0$ (since a square root must be non-negative) or $x^2-x-2=0$ (the inequality is also valid when the square root is zero). In this case we get $x\ge 1$, or $x=-1$. However we also need $x^2-x-2\ge 0$ such that we are not taking the square root of a negative number. In this case we have $$x^2-x-2\ge 0\longrightarrow (x-2)(x+1)\ge 0\longrightarrow x\le -1\text{ or }x\ge 2$$
Combining these constraints with $x\ge 1$ or $x=-1$, we get $x=-1\text{ or }x\in [2, \infty)$.
