Why is $\sqrt{-x}*\sqrt{-x}=-x?$

Q1 – Why is $\sqrt{-x}*\sqrt{-x}=-x?$

Q2 – I was thinking it would be:
$\sqrt{-x}*\sqrt{-x}=\sqrt{-x*-x}=\sqrt{x^2}$ but apparently not (why not?)

Q3 – What are the formal algebra rules to use? Can I calculate this without using i such as in: $\sqrt{-x}*\sqrt{-x}=i\sqrt{x}*i\sqrt{x}=-\sqrt{x^2}=-x$.

Solutions Collecting From Web of "Why is $\sqrt{-x}*\sqrt{-x}=-x?$"

By definition, a square root of $u$ is something that, when squared, gives $u$. Hence, if $\sqrt{-x}$ exists, then $\sqrt{-x}\cdot\sqrt{-x}=-x$ by definition.

Now, the principal square root of a nonnegative real number $u$ is the nonnegative real number whose square is $u$. We denote this by $\sqrt u.$ What this means is that, for $\sqrt{-x}$ to be defined, we need $-x$ to be a nonnegative real number, which means that $x$ is nonpositive real. Now, if $x=0,$ this is no problem, and you can say that $$\sqrt{-x}\cdot\sqrt{-x}=x,\tag{$\star$}$$ since $-0=0.$ If $x$ is positive, then the left hand side of $(\star)$ isn’t even defined, so $(\star)$ is false. If $x$ is negative, then the right hand side of $\star$ is a negative number, while the left hand side is the square of a positive number, so is positive, and so $(\star)$ is again false.

However, we can conclude that, if $\sqrt{-x}$ is defined (that is, if $x$ is nonpositive real), then $$\sqrt{-x}\cdot\sqrt{-x}=\sqrt{x^2}.$$ How can this be? Well, remember that a principal square root has to be nonnegative real, so for any real $u,$ we have in general that $$\sqrt{u^2}=\lvert u\rvert.$$ In particular, then, since $x$ is nonpositive real, then $$\sqrt{-x}\cdot\sqrt{-x}=\sqrt{x^2}=\lvert x\rvert=-x.$$

$\sqrt{-x}*\sqrt{-x}=(\sqrt{-x})^2=-x$ (the square root and square cancels out)

The square root function is not uniquely defined, but is rather multivalued. Therefore by declaring that $\sqrt{1}=1$ you are making a branch cut and eliminating the -1. By changing the choice of branch you can obtain the correct answer and use the property that @StephenMontgomery-Smith mentioned (which is not valid for the entire domain of complex numbers but can be restricted to a certain domain).

The square root function is defined for $\left\{ x \in \mathbb{R}: x \geq 0\right\}$. When you have $-x$ under the square root, the implication is that $x$ is negative.
The square root function doesn’t work the same way for complex numbers as it does for reals.

Here is a “proof” that $1=-1$:

$$1=\sqrt{1}=\sqrt{1^2}=\sqrt{(-1)^2}=\sqrt{(-1)(-1)}=\underline{\sqrt{-1}\sqrt{-1}}=i^2=-1$$

The problem is in the underlined step. The rule of $\sqrt{xy}=\sqrt{x}\sqrt{y}$ only works when $x,y>0$.