anonymous
  • anonymous
RATIONALIZE THIS:
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\frac{ \sqrt[3]{x + 24} - 3 }{ \sqrt{x + 1} - 2 }\]
terenzreignz
  • terenzreignz
I assume what you want is to get rid of the radical on the denominator?
anonymous
  • anonymous
so sorry for the late reply. but yes.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

terenzreignz
  • 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
  • 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
  • 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
  • anonymous
yep. i understand you :)
terenzreignz
  • terenzreignz
Oh wait... something's wrong...
terenzreignz
  • 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
  • anonymous
what? omg. sorry. we were taught to do that in school. Haha
anonymous
  • anonymous
its all about finding the limits. my answer was indeterminate so now, I had to like rationalize it or something
terenzreignz
  • terenzreignz
Oh... okay... then honestly, I don't know how to go about this... (something specific to your school instruction?)
terenzreignz
  • terenzreignz
Could you post the original question? :3
radar
  • radar
|dw:1378043824271:dw|
anonymous
  • anonymous
|dw:1378044536364:dw|
anonymous
  • anonymous
thankyou @radar i'll try my best to understand it. :)
terenzreignz
  • terenzreignz
|dw:1378044634136:dw| You forgot something?
anonymous
  • anonymous
oh shiz. Haha. sorry. yeah. forgot that. omg. thankyou
terenzreignz
  • terenzreignz
Tricky... let's see...
anonymous
  • 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
  • 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
  • radar
Oh I see this is a limit question, that was not clear in the original posting!
terenzreignz
  • terenzreignz
I imagine monsieur L'Hôpital got his inspiration from annoying problems such as these ^_^
anonymous
  • anonymous
@radar sorry for the confusion. :)
radar
  • 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
  • radar
|dw:1378045423529:dw| error found
anonymous
  • 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
  • asnaseer
Sine this limit is of the form 0/0, you will need to apply L'Hôpital's rule
asnaseer
  • asnaseer
*Since
anonymous
  • anonymous
what rule is that? so sorry. we haven't learn that yet. :)
radar
  • radar
What ever 0/0 is that is the limit my guess infinity, what would that be @asnaseer ?
terenzreignz
  • terenzreignz
Tempting, no? @asnaseer ? XD
asnaseer
  • asnaseer
@radar - the limit is definitely not 0/0
asnaseer
  • asnaseer
@lenarancillo - what methods have you been taught?
terenzreignz
  • terenzreignz
rationalizing, apparently. Anyway, the limit goes to a rational number, all right... I just tried L'Hôpital.
radar
  • radar
(6-6 +6-6)/(3-3) looks like 0/0 but I have not simplified the numerator.
asnaseer
  • asnaseer
@radar - L'Hôpital's rule is designed for situations like this
radar
  • radar
How does the rule work?
asnaseer
  • 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
  • asnaseer
@radar - see this http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule
terenzreignz
  • 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
  • asnaseer
well - its difficult to teach if the asker does not respond to questions being asked.
radar
  • radar
Thanks for the link. @asnaseer
terenzreignz
  • terenzreignz
Granted. Still, @ienarancillo Are you ready? I proceed only at your signal.
anonymous
  • anonymous
@asnaseer so sorry for the late reply. but yeah, it think i have learned that
terenzreignz
  • terenzreignz
Oh.. well that simplifies things all right XD
anonymous
  • anonymous
ok guys. haha. im ready. :)
asnaseer
  • asnaseer
ok - if you have then use that method - it is known as l'Hopitals rule
asnaseer
  • asnaseer
but lets see the alternative method being proposed by @terenzreignz - I am intrigued. :)
terenzreignz
  • 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
  • anonymous
ok guys. im holding on
terenzreignz
  • 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
  • terenzreignz
Basically conjugates... things that remove denominators...
terenzreignz
  • terenzreignz
err... remove radicals... not denominators.
anonymous
  • anonymous
uhm, i'm kinda lost basically after you wrote the given. how did the numerator end that way?
terenzreignz
  • 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
  • terenzreignz
For the numerator, I simply took p = \(\large \sqrt[3]{x+24}\) and q to be 3
terenzreignz
  • 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
  • terenzreignz
Is just equal to \[\Large \frac{x-3}{x-3}=1\]
terenzreignz
  • terenzreignz
And cancels out.
asnaseer
  • 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
  • asnaseer
so the a's and b's cancel out
asnaseer
  • asnaseer
@terenzreignz - you Sir are a genius! :)
terenzreignz
  • 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
  • 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
  • terenzreignz
And @asnaseer no... I'm a student... I'm somehow still used to being *forbidden* from using L'Hopital XD
asnaseer
  • asnaseer
lol!
terenzreignz
  • 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
  • anonymous
aww, I think its making sense to me. lol
terenzreignz
  • terenzreignz
Oh it is? I LOVE making sense! <3
radar
  • radar
If you are getting 4/27 for an answer, I used the link and got the same.
terenzreignz
  • terenzreignz
Yes! Awwww yeaaa
asnaseer
  • asnaseer
@lenarancillo - you now have two methods that you can use to solve this - take your pick! :)
terenzreignz
  • 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
  • anonymous
omg. thankyou so much for helping me
anonymous
  • anonymous
so 4/27 is the answer? lol. im trying to figure out the eqaution you used guys. :)
asnaseer
  • asnaseer
yes :)
radar
  • 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
  • radar
@ienarancillo Can you do derivatives?
radar
  • 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
  • radar
|dw:1378047642481:dw|
anonymous
  • 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
  • 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
  • anonymous
aww, okay. Haha.
radar
  • radar
Just in case you're interested:|dw:1378048263552:dw|
anonymous
  • anonymous
ohhh
radar
  • radar
|dw:1378048434813:dw|
radar
  • radar
Note the \[27^{2/3}=\sqrt[3]{27^{2}}\]
radar
  • radar
Or 9
radar
  • radar
I think it was simpler this way, and the thread that @ansaseer posted is a good reference.
radar
  • radar
Good luck with your studies, I learned something this morning too.
anonymous
  • anonymous
@radar thankyiu so much for ur help. really means a lot
anonymous
  • anonymous
thanks so much. i can finally understand it. :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.