<s>[INST] 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?
 [/INST] 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
$$\sqrt{\frac x{y^2}}=\frac{\sqrt{x}}{\sqrt{y^2}}=\frac{\sqrt{x}}{-y}.$$
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