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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.

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    $\begingroup$ Is it $\frac{\sqrt{2}}{\sqrt{x}-3}$ or $\frac{\sqrt{2}}{\sqrt{x-3}}$? If it is the second, then the answer key is right. $\endgroup$ Commented May 16, 2016 at 14:12
  • $\begingroup$ Thanks for replying, yes is the second, but where did the -6 came from? $\endgroup$
    – Gavriel
    Commented May 16, 2016 at 14:18
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    $\begingroup$ Multiplying the $\sqrt{2}$ on top by $\sqrt{x-3}$, we get $\sqrt{2x-6}$. $\endgroup$ Commented May 16, 2016 at 14:20
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    $\begingroup$ No, multiplying by its conjugate is (I think) only for denominators such as $\sqrt{x}-3$. Otherwise, if it's under a radical, we can just square it to rationalize the denominator $\endgroup$
    – Frank
    Commented May 16, 2016 at 14:21
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    $\begingroup$ We multiply by $\sqrt{x}+3$ if the denominator is $\sqrt{x}-3$, but it isn't. $\endgroup$ Commented May 16, 2016 at 14:22

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

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The answer key is right (it's you): You must have multiplied numerator by $\sqrt{x-3}$, not $\sqrt{x+3}$.

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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:

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

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

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