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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 !!!

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    $\begingroup$ you cannot simply square both sides. $(x-1)^2\ge 0$ does not imply that $x-1\ge 0$ $\endgroup$
    – 5201314
    Jul 16, 2021 at 16:46
  • $\begingroup$ But even when I checked that problem statement in Wolfram Alpha it is also showing the same answer as I got. Does that mean Wolfram Alpha is giving a wrong solution? $\endgroup$
    – Ganit
    Jul 16, 2021 at 16:49
  • $\begingroup$ Ok...Can someone please put down the way to solve this problem then with pen and paper? That will be really helpful. $\endgroup$
    – Ganit
    Jul 16, 2021 at 16:53
  • $\begingroup$ $x\ge0$ is not equivalent to $\sqrt{x}\ge0$ as $\sqrt{x}$ might be negative too. Also in wolframalpha you inputted your third expression and not the originial question giving you the wrong answer $\endgroup$
    – user876009
    Jul 16, 2021 at 16:54
  • $\begingroup$ You should see wolframalpha.com/input/… $\endgroup$
    – Macavity
    Jul 16, 2021 at 17:06

3 Answers 3

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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 $x=-1\text{ or }x\in [2, \infty)$.

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  • $\begingroup$ This is helpful. I should be careful in squaring the inequalities from the next time. $\endgroup$
    – Ganit
    Jul 16, 2021 at 17:03
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    $\begingroup$ You missed the solution $x=-1$, sqrt can be zero, not just positive. $\endgroup$
    – Macavity
    Jul 16, 2021 at 17:08
  • $\begingroup$ @Macavity right. Thanks for pointing that out $\endgroup$
    – 5201314
    Jul 16, 2021 at 17:10
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$x\ge0$ is obviously not the same as $x^2\ge0$ !

For this problem, as a square root is positive by definition, the inequality reduces to

$$x-1\ge0,$$ provided that the argument of the square root is non-negative.

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  • $\begingroup$ Then how come the website is also giving the same solution as I got. I thought we can trust Wolfram Alpha as it is a credible source. $\endgroup$
    – Ganit
    Jul 16, 2021 at 16:51
  • $\begingroup$ @Ganit it gave you that solution only after you put in your invalid "modified" inequality. $\endgroup$
    – 5201314
    Jul 16, 2021 at 16:52
  • $\begingroup$ Even with your reduced problem statement to $x-1 \geq 0$, we get the solution range to be $x \geq 1$ which is not the correct answer. $\endgroup$
    – Ganit
    Jul 16, 2021 at 16:55
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    $\begingroup$ "$x\ge0$ is obviously not the same as $x^2\ge0$" Can you give some example to prove your this statement $\endgroup$
    – user876009
    Jul 16, 2021 at 16:58
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    $\begingroup$ The argument in the sqrt can be zero, hence $x=-1$ works too. $\endgroup$
    – Macavity
    Jul 16, 2021 at 17:08
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In your $2$nd step to the $3$rd step, you just squared a term without realizing that the negative values are extraneous and don't actually work.

Also, when you input the inequality into Wolfram Alpha, you input the incorrect inequality; your $3$rd step; not the original equation. This is why Wolfram Alpha got it "wrong".

Hope this helps!

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