What is the domain of $({\sqrt x})^2$?

Question

I have the following question with me:

Find the middle point of solution of the inequality.$$x^2+2(\sqrt x)^2-3\le0$$

I went through the following process: $$x^2+2x-3\le0$$ $$(x+3)(x-1)\le0$$ $$x\in[-3,1]$$ But $$\sqrt x\ge0\implies x\ge0$$ So, $$x\in[0,1]$$ After I went through the question again, I pointed out that the negative numbers also satisfy the inequality. I asked my teacher but he said that square of a number can't give you a negative number if we are working in $\Bbb R$. I understood that this is possible when we work in Complex numbers but the problem is that Wolfram too is showing the domain of x to be $[-3,1]$ and the graph of this inequality also confirms it. When I checked for the domain of $(\sqrt x)^2$ in Wolfram Alpha then it showed $x\ge0$. So, I have been confused as to what should I consider the solution and what not to if we are working in Real Numbers' Domain?

Answer 1

Negative numbers do not satisfy the inequality for the reason that your teacher said. So any "solution" for $x < 0$ is not valid, and the domain of $(\sqrt{x})^2$ is hence $x \geq 0$. The solution of the inequality is hence [0, 1]. This is assuming we are working in the real space.

and the graph of this inequality also confirms it

Desmos does not support that statement: