When is $\sqrt{x/y^2}$ equal to $\sqrt{x}/y$? The solution to the quadratics is given by
$r = -\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$, which is shortened to
$r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$, but I'm wondering how if this is justified, given that $4a^2$ can be negative if $a \in \mathbb{C}$, and $\sqrt{\dfrac{x}{y}} \neq \dfrac{\sqrt{x}}{\sqrt{y}}$ if $x$ and $y$ are negative, but given that we have $a^2$, is this justified?
Is $\sqrt{x/y^2} = \sqrt{x}/y$? Is $r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$ always true?
 A: When $y> 0$, you have
$$\sqrt{\frac x{y^2}}=\frac{\sqrt{x}}{\sqrt{y^2}}=\frac{\sqrt{x}}{y}.$$
When $y<0$, you have
$$\sqrt{\frac x{y^2}}=\frac{\sqrt{x}}{\sqrt{y^2}}=\frac{\sqrt{x}}{-y}.$$
A: $$\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$$
Is "the solution" to $ax^2 + bx + c$, provided that $a,b,c$ are real numbers and $a \ne 0$.
$4a^2 \ge 0$ for all real numbers $a$. Since we are excluding $a = 0$, then $4a^2 > 0$. We can then say 
$$\sqrt{4a^2} = 
\begin{cases}
    2a,  & \text{if $a \ge 0$} \\
    -2a, & \text{if $a < 0$}
\end{cases}$$
The real problem has to do with $b^2 - 4ac$ since it can be positive, negative, or $0$. But, in all three cases, we will have
$$\sqrt{\dfrac{b^2-4ac}{4a^2}}= \pm \dfrac{\sqrt{b^2-4ac}}{2a}$$
so we end up with
\begin{align}
    \dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}
    &= \dfrac{b}{2a}\pm \left( \pm \dfrac{\sqrt{b^2-4ac}}{2a} \right) \\
    &= \dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a} \\
\end{align}
By the way, for any real number $x$, $\sqrt{x^2} = |x|$.
A: Since, in general, for all $\;z\in\Bbb R\;,\;\;\sqrt{z^2}=|z|\;$, we have that
$$\sqrt{y^2}=y\iff y\ge 0$$
and thus
$$\sqrt\frac x{y^2}=\frac{\sqrt x}y\iff y>0\;\;\text{(because clearly it can't be in this case}\;\;y=0)$$
A: The two possible values of $\sqrt{4a^2}$ for complex $a$ differ only by a sign (one is $2a$, the other is $-2a$), so the set of solutions given by the formula
$$r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$$
is the same as that given by the formula
$$r = -\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}.$$
(The "plus" in one formula may or may not correspond to "plus" in the other one, but this doesn't matter.)
A: That's a good question. 
But for the sake of the quadratic equation we don't need $\sqrt{x/y^2} = \sqrt{x}/y$.  We just need $\sqrt{x/y^2} = \pm \sqrt{x}/y$. 
If $y^2 = c \ne 0$ then $y = \{x_1,x_2\}$ where $x_1 = -x_2$.  This is true for all $c$, real or complex.  Arbitrarily declare one of them, say $x_1$, to be the $\sqrt{c}$ then we can always have $\sqrt{y^2} = \pm y$.  Always. 
(If $y^2 = 0 \implies y = 0$ and $0 = \pm 0$.)
If $y^2 = c = r*e^{i\theta} \in \mathbb C$ then $y = \{\sqrt{r}e^{i(\theta/2},\sqrt{r}e^{i\theta/2 + \pi}\}$ = $\pm \sqrt{r} e^{i\theta/2}$
So $\sqrt{x/y^2} = \pm \sqrt{x}/y$.  So equation is good even for complex numbers.
