Rationalizing the Denominator

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September 10th 2008, 01:26 PM
xJenniferx

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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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September 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

If you can't see the image then click this link.
ImageShack - Hosting :: 91076157en5.png

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

I think you forgot $\sqrt{xy}$ in the numerator and you can't go further...
September 10th 2008, 01:42 PM
xJenniferx

so i stop at 4 / xy and forget the (2)(2)/xy?
September 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 $\sqrt{xy}$ in the numerator, though.

Have you talked about absolute values yet in class? If so, think of how $\sqrt{x^2}$ isn't exactly the same as $x$ . If you haven't gotten that far yet in your class, never mind; you did it just fine.
September 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 $\frac{4 \sqrt{xy}}{xy}$ , not [tex]\frac{4}{xy}[/Math]
September 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
September 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 $\sqrt{xy}$ at the very beginning, you already "supposed" xy>0.
But it's always better to write this down (Wink)
September 10th 2008, 01:57 PM
Nox

EDIT: Moo's right. Haha. Neither $x$ nor $y$ could be negative (technically, both could be negative... but they'd cancel out in the last step), so absolute value signs are redundant.
September 10th 2008, 01:58 PM
Moo

Quote:

Originally Posted by Nox

It's tricky. Since you multiplied $\sqrt{xy}$ by $\sqrt{xy}$ , you'd have $\sqrt{x^2y^2}$ , which comes out to | $x$ |*| $y$ |. 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|.
September 10th 2008, 02:03 PM
Nox

I know. :P I just didn't think of it before. See the EDIT.