Why is $\sqrt{x^2}$ always $x$? This question has the potential to sound extremely stupid, but I've seen (and also used) countless times the idea that $\sqrt{x^2} = x$. However $x^2 = x\cdot x = (-x)\cdot(-x)$.
I know that when taking the square root of something we take both the positive and negative root. Yet when solving an equation and we're faced with $\sqrt{x^2y}$ we make it $x\sqrt{y}$. Why didn't we consider $(-x)\sqrt{y}$? Similarly, $\sqrt{x^3}$ is often changed to $x\sqrt{x}$ and not $(-x)\sqrt{-x}$ which would still give the same result if cubed? (I do understand that the latter is imaginary, but that shouldn't stop us from using it, should it?)
 A: The symbol $\sqrt{\mathstrut\quad}$ is defined to denote the non-negative square root. In other words: $\sqrt{x}$ is the number $y$ such that $y\geq 0$ and $y^2=x$. The number $y$ is then called the radical of $x$. This is a matter of definition. We need to choose one of the two values to make $\sqrt{\mathstrut\quad}$ into a function, and we like positive numbers better.
If $y = \sqrt{x}$ then $y^2 = (-y)^2 = x$. So now, what if $y = \sqrt{x^2}$?  
Since $y \geq 0$ by definition, we have $y=x$ if $x\geq0$ or $y = -x$ if $x<0$. We write this as $y = |x|$ and say $y$ is the absolute value of $x$. 
Also note that if $x$ is positive, $\sqrt{-x}$ is not well defined even if you can still find a number $y$ such that $y^2=-x$. You have to be careful: the radical is only defined for non-negative numbers.
A: Actually, $\sqrt{x^2} = |x|$. 
If $x\geq 0$, then $\sqrt{x^2} = x = |x|$. If $x<0$, then $\sqrt{x^2} = \sqrt{(-x)^2} = -x = |x|$.
A: As mentioned in comments:
$\sqrt {x^2}=|x|$
Reason is that range of square root function is non-negative. 
Assume that using $\sqrt {x^2}=x$  you write 
$\sqrt {(-2)^2}=-2$ which is clearly wrong. 
I hope you get the point.
A: Consider $(-2)^2=4=2^2$. If one could devise a definition of the square root function such that $\sqrt{x^2}=x$ for any $x$, what would be the value of 
$$
\sqrt{4}=\sqrt{(-2)^2}=\sqrt{2^2}
$$
without getting a contradiction?
This shows that it is not possible to define a function with the desired property, so we abandon the idea and define, for $x\ge0$,

$\sqrt{x}$ is the unique nonnegative number $y$ such that $y^2=x$

In particular, $\sqrt{x^2}=|x|$, because $|x|\ge0$ by definition and $|x|^2=x^2$.

As an aside, note that $\sqrt{x^3}$ makes sense only if $x\ge0$, so in this case $\sqrt{x^3}=x\sqrt{x}$ is correct. You could also say
$$
\sqrt{x^3}=\sqrt{x^2\cdot x}=\sqrt{x^2}\cdot \sqrt{x}
=|x|\sqrt{x}
$$
which would be correct too, but $|x|=x$ as $x\ge0$.
Note, instead, that if $x<0$ and $y<0$, it would be very incorrect to write
$$
\sqrt{xy}=\sqrt{x}\sqrt{y}
$$
but you can surely say $\sqrt{xy}=\sqrt{|x|}\sqrt{|y|}$.
