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The question just asks me to simplify this expression:

$$\frac{3 - \sqrt{x}}{x - 9}.$$

I'm stuck. I tried rationalizing, but I'm not sure if that's the correct method. Any help is appreciated. Thanks!

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Can you please clarify what is within the square root and what is the denominator? –  Dre Sep 15 '13 at 1:57
    
\frac{3 - \sqrt{x}}{x - 9} –  jayson Sep 15 '13 at 2:05
    
Its just (3-√x)/(x-9) thanks –  jayson Sep 15 '13 at 2:07
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up vote 1 down vote accepted

Is this what you mean by simplifying that expression?

$\frac{3-\sqrt{x}}{x-9}=\frac{3-\sqrt{x}}{x-9}\frac{3+\sqrt{x}}{3+\sqrt{x}}=\frac{9-x}{x-9}\frac{1}{3+\sqrt{x}}=\frac{-1}{3+\sqrt{x}}$

Usually we apply this process in reverse to rationalize.

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Oh yeah, I messed up my signs, thank you very much –  jayson Sep 15 '13 at 2:19
    
You're very welcome. –  user71352 Sep 15 '13 at 2:21
    
So if we have to let g(x) be the result from the simplification. Is domain of f equal to domain of g? –  jayson Sep 15 '13 at 2:37
    
The domains will not quite be the same. $f(x)=\frac{3-\sqrt{x}}{x-9}$ is technically defined (by this formula) on the positive reals excluding $9$ since there is a singularity there. $g(x)=\frac{-1}{3+\sqrt{x}}$ is defined on the positive reals. The above simplification shows we can enlarge the domain of $f$ to the positive reals by defining $f(9)=\frac{-1}{6}$. –  user71352 Sep 15 '13 at 2:45
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