# Can we write square roots in a fraction separately? eg, $\sqrt{\frac{9-x^2}{x-2}}$ vs $\frac{\sqrt{9-x^2}}{\sqrt{x-2}}$

I was doing some questions related to functions when I came across this question where they gave us two functions as $$\sqrt{\frac{9-x^2}{x-2}}\quad\text{and}\quad\frac{\sqrt{9-x^2}}{\sqrt{x-2}}$$ I can see that those two are a little different, but I have learned that we write $$\sqrt\frac{a}{b}=\frac{\sqrt a}{\sqrt b}$$ I am a little confused: is there a specific condition in which we can apply this?

Hint: $$\sqrt{\frac{-3}{-2}}$$ is defined, while $$\frac{\sqrt{-3}}{\sqrt{-2}}$$ is not (in case we work only with real numbers).
In your case, for $$\sqrt{\frac{9-x^2}{x-2}}$$ to be defined, you need $$\frac{9-x^2}{x-2} \geq 0 \Leftrightarrow x \in (-\infty, -3] \cup [2, 3]$$ and for $$\frac{\sqrt{9-x^2}}{\sqrt{x-2}}$$ you need both $$9 - x^2 \geq 0$$ and $$x-2\geq 0$$, which means $$x \in [2, 3]$$. So, for $$x = -4$$ you have $$\sqrt{\frac{9-x^2}{x-2}} = \sqrt{\frac{-7}{-6}} = \sqrt{\frac{7}{6}}$$ and $$\frac{\sqrt{9-x^2}}{\sqrt{x-2}} = \frac{\sqrt{-7}}{\sqrt{-6}}$$ which is undefined.
• I would argue even when working with complex numbers it doesn't make sense. Yes, negative numbers have square roots, but $\sqrt{-3}$ is still very dubious. Aug 25, 2022 at 5:36
For positive real numbers we have that $$\sqrt{xy}= \sqrt{x}\sqrt{y}$$. Now if I set $$y = 1/z$$ this becomes $$\sqrt{x/z} = \sqrt{x}/\sqrt{z}$$. The conditions here are that they be real numbers and everything is well-define, ie no square roots of negative numbers and no division by zero. Complex numbers can complicate this as well since the square root is multivalued there.