Is there a difference between $\pm\sqrt{x}$ and $\sqrt{x}$ When square rooting something, do we have to put the $\pm$ sign as soon as the $\sqrt{}$ sign appears? For example, consider the following equation:-
$$x+1=5$$
Process 1:
$$\implies \sqrt{x+1}=\pm\sqrt{5}$$
Process 2:
$$\implies \sqrt{x+1}=\sqrt{5}$$
Which of the following is correct?
 A: Only the second one is correct.
If $a=b$, then $\sqrt{a}=\sqrt{b}$. Note that $x\to\sqrt{x}$ is a function from $[0,+\infty)$ to $[0,+\infty)$: you can only take the square root of a nonnegative number, and the result is always a nonnegative number.
Your process 1 is referring to something else. You write $a=\pm b$ as a shortcut for $a=b$ or $a=-b$, and both are supposed to be valid [1]. And it doesn't really appear because of the square root.
For example, say, $x^2=4$. Then, taking square root,
$$\sqrt{x^2}=2$$
$$|x|=2$$
And only then,
$x=2$ or $x=-2$, that is $x=\pm2$.
You could also write:
$$x^2-4=(x-2)(x+2)=0$$
Hence $x=2$ or $x=-2$, that is $x=\pm2$.
You have to remember that $\sqrt{x^2}$ is not $x$, but $|x|$.
[1] One might want to say that because "$2=2$ or $2=-2$" is true, one could write $2=\pm2$, but I don't think it's ever used that way.
A: Short answer.
Use the definition of the square root:
$$\sqrt{x^2}=|x|≥0$$
and
$$x^2=y^2\iff |x|=|y|.$$
Note that, $x+1=5\implies x+1≥0$.
Then you have
$$\begin{align}&x+1=5,\thinspace x+1≥0\\
\iff &\left(\sqrt{x+1}\right)^2=\left(\sqrt 5\right)^2\\
\iff &|\sqrt{x+1}|=|\sqrt 5|\\
\iff &\sqrt{x+1}=\sqrt 5.\end{align}$$
A: The $\pm 3$ symbol, for example, literally means $+3$ OR $-3$.
It only works when we work with something like  $x^2 = 5$ then taking square root of both sides we got $\pm x = \sqrt 5$
Note that originally the $\pm$ sign is on the left, not on the right, since $$\sqrt{x^2} = \pm x.$$
Then the "hidden step" is multiplying by negative and get $x = \pm \sqrt 5$. Note also that here we deal with two equalities instead of one. It just looks like only one.
But $\sqrt{4} = 2$ not $\pm 2$ since $\sqrt{x}$ mean "the non-negative square roots of $x$"
