RATIONALIZE THIS:

- anonymous

RATIONALIZE THIS:

- Stacey Warren - Expert brainly.com

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- schrodinger

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- anonymous

\[\frac{ \sqrt[3]{x + 24} - 3 }{ \sqrt{x + 1} - 2 }\]

- terenzreignz

I assume what you want is to get rid of the radical on the denominator?

- anonymous

so sorry for the late reply.
but yes.

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## More answers

- terenzreignz

Well, the trick I use here basically involves 'not caring about the numerator'
Put it this way: Generic fraction with a radical on the denominator...
\[\Large \frac{N}{\sqrt{r}\pm k}\]

- terenzreignz

N is the numerator (which I don't care about at the moment)
and we have \(\large \sqrt{r}\pm k\) in the denominator... and we want to remove the radical...
catch me so far?

- anonymous

yep. i tried that. actually I've reached this\[\frac{ \sqrt[3]{x+24-3} }{\sqrt{x+1}-2} \times \frac{ \sqrt[3]{x+24+3} }{\sqrt{x+1}+2} \times \frac{ \sqrt{x+1}+2 }{\sqrt{x+1}+2 }\]

- anonymous

yep. i understand you :)

- terenzreignz

Oh wait... something's wrong...

- terenzreignz

YOU paid attention to the numerator... I told you not to :P
All you do is multiply both the numerator and denominator by the *conjugate* of the denominator... (fancy word, but all it means is that you change the - to a +)

- anonymous

what? omg. sorry. we were taught to do that in school. Haha

- anonymous

its all about finding the limits. my answer was indeterminate so now, I had to like rationalize it or something

- terenzreignz

Oh... okay... then honestly, I don't know how to go about this... (something specific to your school instruction?)

- terenzreignz

Could you post the original question? :3

- radar

|dw:1378043824271:dw|

- anonymous

|dw:1378044536364:dw|

- anonymous

thankyou @radar i'll try my best to understand it. :)

- terenzreignz

|dw:1378044634136:dw|
You forgot something?

- anonymous

oh shiz. Haha. sorry. yeah. forgot that. omg. thankyou

- terenzreignz

Tricky... let's see...

- anonymous

lol. yes, it is.
my teacher said that first you have to like substitute 3 in the equation and if your answer is like indetreminate or something. you now then have to rationalize or factor it out. :)

- radar

The denominator should be free of any radicals, but that numerator don't look so good, very messy. There may be a way to simplify it further, but with a 3rd root radical and a 2nd root radical, I am at a loss, since the terms inside those radicals are different.

- radar

Oh I see this is a limit question, that was not clear in the original posting!

- terenzreignz

I imagine monsieur L'HÃ´pital got his inspiration from annoying problems such as these ^_^

- anonymous

@radar sorry for the confusion. :)

- radar

Not your fault, now that I see it, I'm still confused, the denominator approaches 0 as x approaches 3, but that is before the expression has been radicalized

- radar

|dw:1378045423529:dw| error found

- anonymous

I had the same problem too. my teacher told us to substitute x first but all ig ot was a 0/0; an indeterminate.

- asnaseer

Sine this limit is of the form 0/0, you will need to apply L'HÃ´pital's rule

- asnaseer

*Since

- anonymous

what rule is that? so sorry. we haven't learn that yet. :)

- radar

What ever 0/0 is that is the limit my guess infinity, what would that be @asnaseer ?

- terenzreignz

Tempting, no? @asnaseer ?
XD

- asnaseer

@radar - the limit is definitely not 0/0

- asnaseer

@lenarancillo - what methods have you been taught?

- terenzreignz

rationalizing, apparently.
Anyway, the limit goes to a rational number, all right...
I just tried L'HÃ´pital.

- radar

(6-6 +6-6)/(3-3) looks like 0/0 but I have not simplified the numerator.

- asnaseer

@radar - L'HÃ´pital's rule is designed for situations like this

- radar

How does the rule work?

- asnaseer

@lenarancillo - have you been taught that when you get an indeterminate form, then you can change the limit by differentiating the numerator and the denominator?

- asnaseer

@radar - see this http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule

- terenzreignz

@asnaseer I think we need to proceed assuming differentiation has not yet been taught.
I think I have already found the solution ^_^
@ienarancillo are you ready?
Brace yourself ^_^

- asnaseer

well - its difficult to teach if the asker does not respond to questions being asked.

- radar

Thanks for the link. @asnaseer

- terenzreignz

Granted. Still, @ienarancillo Are you ready?
I proceed only at your signal.

- anonymous

@asnaseer so sorry for the late reply. but yeah, it think i have learned that

- terenzreignz

Oh.. well that simplifies things all right XD

- anonymous

ok guys. haha. im ready. :)

- asnaseer

ok - if you have then use that method - it is known as l'Hopitals rule

- asnaseer

but lets see the alternative method being proposed by @terenzreignz - I am intrigued. :)

- terenzreignz

Well, you better brace yourselves :/
Okay, let's have a look at this:
\[\large \frac{\sqrt[3]{x+24}-3}{\sqrt{x+1}-2}\]

- anonymous

ok guys. im holding on

- terenzreignz

Sorry, too big, scaling down..
\[ \frac{\sqrt[3]{x+24}-3}{\sqrt{x+1}-2}\cdot \frac{\sqrt[3]{(x+24)^2}+3\sqrt[3]{x+24}+9}{\sqrt{x+1}+2}\cdot \frac{\sqrt{x+1}+2}{\sqrt[3]{(x+24)^2}+3\sqrt[3]{x+24}+9}\]

- terenzreignz

Basically conjugates... things that remove denominators...

- terenzreignz

err... remove radicals... not denominators.

- anonymous

uhm, i'm kinda lost basically after you wrote the given. how did the numerator end that way?

- terenzreignz

Don't panic, use these two facts:
\[\Large (p-q)(p^2+pq+q^2)=p^3-q^3\]and\[\Large (p-q)(p+q)=p^2-q^2\]

- terenzreignz

For the numerator, I simply took p = \(\large \sqrt[3]{x+24}\) and q to be 3

- terenzreignz

Anyway, believe it or not (using those two facts I posted), THIS bit:
\[\color{red}{\frac{\sqrt[3]{x+24}-3}{\sqrt{x+1}-2}\cdot \frac{\sqrt[3]{(x+24)^2}+3\sqrt[3]{x+24}+9}{\sqrt{x+1}+2}}\cdot \frac{\sqrt{x+1}+2}{\sqrt[3]{(x+24)^2}+3\sqrt[3]{x+24}+9}\]

- terenzreignz

Is just equal to
\[\Large \frac{x-3}{x-3}=1\]

- terenzreignz

And cancels out.

- asnaseer

@lenarancillo - @terenzreignz basically did this:\[\frac{\sqrt[3]{x+24}-3}{\sqrt{x+1}-2}\cdot\frac{a}{b}\cdot\frac{b}{a}\]

- asnaseer

so the a's and b's cancel out

- asnaseer

@terenzreignz - you Sir are a genius! :)

- terenzreignz

Basically multiplying by 1, but in a creative manner so as to make life easier... :D
\[\cancel{\frac{\sqrt[3]{x+24}-3}{\sqrt{x+1}-2}\cdot \frac{\sqrt[3]{(x+24)^2}+3\sqrt[3]{x+24}+9}{\sqrt{x+1}+2}}^1\cdot \frac{\sqrt{x+1}+2}{\sqrt[3]{(x+24)^2}+3\sqrt[3]{x+24}+9}\]

- terenzreignz

And now, what's left, it's far easier to evaluate the limit of THIS, as x goes to 3, right? :P
\[\Large \frac{\sqrt{x+1}+2}{\sqrt[3]{(x+24)^2}+3\sqrt[3]{x+24}+9}\]

- terenzreignz

And @asnaseer no... I'm a student... I'm somehow still used to being *forbidden* from using L'Hopital XD

- asnaseer

lol!

- terenzreignz

Although, L'Hopital DOES save you from the agony of even writing all that down.
Are you certain you are allowed to use it @ienarancillo ? ^_^

- anonymous

aww, I think its making sense to me. lol

- terenzreignz

Oh it is?
I LOVE making sense!
<3

- radar

If you are getting 4/27 for an answer, I used the link and got the same.

- terenzreignz

Yes!
Awwww yeaaa

- asnaseer

@lenarancillo - you now have two methods that you can use to solve this - take your pick! :)

- terenzreignz

And now... I'm off to sleep. I have an exam in about 11 hours.
Catch you guys later, maybe :D
--------------------------------
Terence out

- anonymous

omg. thankyou so much for helping me

- anonymous

so 4/27 is the answer? lol. im trying to figure out the eqaution you used guys. :)

- asnaseer

yes :)

- radar

Have you learned how to take derivatives, the L'Hopital method is viaable, or if not terenzreignz gave a step by step method you can follow.

- radar

@ienarancillo Can you do derivatives?

- radar

You only need to take the derivative of the numerator of the numerator
and then take the derivative of the denominator treating each as a separate function.

- radar

|dw:1378047642481:dw|

- anonymous

uhm. honestly, idk a thing about the derivatives. or maybe im not familiar with the word but am familiar with how to do it?

- radar

If you haven't done them yet, I guess it is better to wait for further training. Use the method that terenzreignz used. Unless you are curious about the derivative, I will show you.

- anonymous

aww, okay. Haha.

- radar

Just in case you're interested:|dw:1378048263552:dw|

- anonymous

ohhh

- radar

|dw:1378048434813:dw|

- radar

Note the \[27^{2/3}=\sqrt[3]{27^{2}}\]

- radar

Or 9

- radar

I think it was simpler this way, and the thread that @ansaseer posted is a good reference.

- radar

Good luck with your studies, I learned something this morning too.

- anonymous

@radar thankyiu so much for ur help. really means a lot

- anonymous

thanks so much. i can finally understand it. :)

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