# 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 !!!

• you cannot simply square both sides. $(x-1)^2\ge 0$ does not imply that $x-1\ge 0$ Jul 16, 2021 at 16:46
• 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? Jul 16, 2021 at 16:49
• Ok...Can someone please put down the way to solve this problem then with pen and paper? That will be really helpful. Jul 16, 2021 at 16:53
• $x\ge0$ 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
– user876009
Jul 16, 2021 at 16:54
• You should see wolframalpha.com/input/… Jul 16, 2021 at 17:06

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

• This is helpful. I should be careful in squaring the inequalities from the next time. Jul 16, 2021 at 17:03
• You missed the solution $x=-1$, sqrt can be zero, not just positive. Jul 16, 2021 at 17:08
• @Macavity right. Thanks for pointing that out Jul 16, 2021 at 17:10

$$x\ge0$$ 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.

• 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. Jul 16, 2021 at 16:51
• @Ganit it gave you that solution only after you put in your invalid "modified" inequality. Jul 16, 2021 at 16:52
• 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. Jul 16, 2021 at 16:55
• "$x\ge0$ is obviously not the same as $x^2\ge0$" Can you give some example to prove your this statement
– user876009
Jul 16, 2021 at 16:58
• The argument in the sqrt can be zero, hence $x=-1$ works too. Jul 16, 2021 at 17:08

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!