Which expression is equivalent to $\frac{\sqrt{3+x}}{\sqrt{3-x}}$ for all $x$ such that $-3 < x < 3$

Question

Which expression is equivalent to $$\frac{\sqrt{3+x}}{\sqrt{3-x}}$$ for all $x$ such that $-3 < x < 3$?

This is a question in the ACT practice book. The answer is $$\frac{\sqrt{9-x^2}}{3-x}$$

But why? Why not just $$\frac{3+x}{3-x}$$?

Answer 1

$$\frac{\sqrt{3+x}}{\sqrt{3-x}} = \frac{\sqrt{3+x}\sqrt{3-x}}{\sqrt{3-x}\sqrt{3-x}}$$

$$= \frac{\sqrt{(3+x)(3-x)}}{\left(\sqrt{3-x}\right)^2}$$

$$= \frac{\sqrt{9-x^2}}{3-x}$$

Your expression can't be correct. Take for example $x = 1$

In general:

$$\frac{\sqrt{a}}{\sqrt{b}} =\sqrt\frac{a}{b} \neq \frac{a}{b}$$