Square root easy problem I'm studying the representation on the Cartesian Plane of function that use the square root. I've noticed that the equation $y=\sqrt x$ is different than $y^2=x$ when represented in the plane because the square root of a quantity is always positive. So I could generalize writing
If $\sqrt x = p(x)$ then $x = {\left| \ p(x)\ \right |}^2$  for every $x \in \mathbb R^+$
Is this correct?
 A: This is true. But it looks more complicated that it needs to be, because $|p(x)|^2$ is always the same as simply $p(x)^2$ (when $p(x)$ is a real number). So you can just say
$$ \text{If }\sqrt x=p(x)\text{ then }x = p(x)^2 $$
However, what is not true is the other direction:
$$ \tag{Not true in general!}\text{If }x=p(x)^2\text{ then }\sqrt x = p(x) $$
because the left-hand side may be true when $p(x)$ is negative. So there we must say
$$ \text{If }x=p(x)^2\text{ then }\sqrt x = |p(x)| $$
A: If $\sqrt x=y$, then you can say $x=y^2$. You don't need to set the absolute value here, since you just squared both sides.
What you do need to be careful is the other way around. That is, if $x=y^2$, then you cannot say that $\sqrt{x} = y$. You can only say $\sqrt{x} = |y|$.
A: There is nothing wrong with what you have written. But I would like to bring to your notice that
$$
|p(x)|^{2} = p(x)^{2}
$$
since your function $p(x)$ and $x$ are real.
Also I would like to remove your confusion. The difference between $y = \sqrt{x}$ and $y^{2} = x$ is that the second equation has 2 possible values of $y$ i.e. $\sqrt{x}$ and $-\sqrt{x}$, where as the first equation has only one.
So now I think you got the idea.
If you are given that 
$$
y^{2} = x
$$
it would imply that
$$
|y| = \sqrt{x}.
$$
A: Almost yes, but a negative number would create an imaginary result. So, we could say true for all non-negative integers or [0, infinity)
