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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?

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2 Answers 2

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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.

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    $\begingroup$ 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. $\endgroup$
    – Arthur
    Aug 25, 2022 at 5:36
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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.

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