# Rationalizing the Denominator

• Sep 10th 2008, 01:26 PM
xJenniferx
Rationalizing the Denominator
This is a problem that i'm not sure if i finished it or does it still got more steps? thanks

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• Sep 10th 2008, 01:39 PM
Moo
Hello,
Quote:

Originally Posted by xJenniferx
This is a problem that i'm not sure if i finished it or does it still got more steps? thanks

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$\displaystyle \frac{4}{\sqrt{xy}}\cdot\frac{\sqrt{xy}}{\sqrt{xy} }=\frac{4}{xy}$ ??

I think you forgot $\displaystyle \sqrt{xy}$ in the numerator and you can't go further...
• Sep 10th 2008, 01:42 PM
xJenniferx
so i stop at 4 / xy and forget the (2)(2)/xy?
• Sep 10th 2008, 01:44 PM
Nox
I don't think you needed to reduce 4 to "(2)(2)," but that's right for the most part. You did forget the $\displaystyle \sqrt{xy}$ in the numerator, though.

Have you talked about absolute values yet in class? If so, think of how $\displaystyle \sqrt{x^2}$ isn't exactly the same as $\displaystyle x$. If you haven't gotten that far yet in your class, never mind; you did it just fine.
• Sep 10th 2008, 01:44 PM
Moo
Quote:

Originally Posted by xJenniferx
so i stop at 4 / xy and forget the (2)(2)/xy?

Yes

But it's $\displaystyle \frac{4 \sqrt{xy}}{xy}$, not [tex]\frac{4}{xy}[/Math]
• Sep 10th 2008, 01:46 PM
xJenniferx
if we did do absolute value what do i need to change? can't find where i will have to do with x^2
• Sep 10th 2008, 01:54 PM
Moo
Quote:

Originally Posted by xJenniferx
if we did do absolute value what do i need to change? can't find where i will have to do with x^2

By writing $\displaystyle \sqrt{xy}$ at the very beginning, you already "supposed" xy>0.
But it's always better to write this down (Wink)
• Sep 10th 2008, 01:57 PM
Nox
EDIT: Moo's right. Haha. Neither $\displaystyle x$ nor $\displaystyle y$ could be negative (technically, both could be negative... but they'd cancel out in the last step), so absolute value signs are redundant.
• Sep 10th 2008, 01:58 PM
Moo
Quote:

Originally Posted by Nox
It's tricky. Since you multiplied $\displaystyle \sqrt{xy}$ by $\displaystyle \sqrt{xy}$, you'd have $\displaystyle \sqrt{x^2y^2}$, which comes out to |$\displaystyle x$|*|$\displaystyle y$|. My guess is, if you're under 10th or 11th grade or so and/or you haven't gotten to Algebra II, you probably haven't discussed this, and you really needn't worry about it.

no.
It is enough that xy is positive. You don't need to take the absolute values of x and y (though it is not incorrect). I mean it is not useful to write |x| |y|.
• Sep 10th 2008, 02:03 PM
Nox
I know. :P I just didn't think of it before. See the EDIT.