Rationalizing the denominator of $\frac{\sqrt {2}}{\sqrt {x-3}}$

Question

ok, so im reviewing for a math test and the following question is from the practice final exam.

Rationalize the denominator in the example: $$\frac{\sqrt {2}}{\sqrt {x-3}}$$

After multiplying both the numeration and denominator by the conjugate of the denominator, I got $$\frac{\sqrt {2x+6}}{x-3}$$

But, in the answer key the answer is $$\frac{\sqrt {2x-6}}{x-3}$$

The problem looks quite simple, but I'm not sure what is the answer.

Answer 1

$$\frac{\sqrt{2}}{\sqrt{x-3}}=\frac{\sqrt{2}\cdot\sqrt{x-3}}{\sqrt{x-3}\cdot\sqrt{x-3}}=\frac{\sqrt{2\cdot(x-3)}}{\left(\sqrt{x-3}\right)^2}=\frac{\sqrt{2x-6}}{x-3}$$

Answer 2

You multiply by the conjugate of the denominator when the denominator has two different square root radicals or a square root radical and a rational term. Compare the following:

$\color{maroon}{\dfrac{\sqrt2}{\sqrt{x-3}}\not=\dfrac{\sqrt{2(x+3)}}{x-3}}$

$\color{blue}{\dfrac{\sqrt2}{\sqrt x-\sqrt3}=\dfrac{\sqrt{2}(\sqrt x+\sqrt3)}{x-3}}$