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 Posted: Wed Jan 30, 2008 9:25 pm
simplify: (8^1/6 - 9^1/4) / (3^1/2 + 2^1/2)
Not only am I the only one to get it right in my class, but apparently out of all the classes. But, it's seems sorta easy. Does anyone else??
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Posted: Wed Jan 30, 2008 11:57 pm
in theory, yes. it's fairly simple, but the numbers end up messy and, depending on the class you're in, there are a lot of people who forget how to do fraction exponents.
how's this:
[(8^(1/3))^(1/2)-(9^(1/2)^(1/2))]/[(3^(1/2)+2^(1/2))]
[2^(1/2)-3^(1/2)]/[3^(1/2)+2^(1/2)]
aka [sqrt(2)-sqrt(3)]/[sqrt(3)+sqrt(2)]
if i tried to rationalize it, i might try this, but it's been a long time, and it's late:
[sqrt(2)-sqrt(3)]/[sqrt(3)+sqrt(2)] * [sqrt(2)-sqrt(3)]/[sqrt(2)-sqrt(3)]
[[sqrt(2)-sqrt(3)]^2]/{[sqrt(2)-sqrt(3)] * [sqrt(2)+sqrt(3)]}
[[sqrt(2)-sqrt(3)]^2]/[2-3] ?
- [sqrt(2)-sqrt(3)]^2
(which, btw, is NOT "-2-3" => "1")
in other words, you're right, you probably shouldn't have been the only person that could do anything with that expression.
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Posted: Thu Jan 31, 2008 6:54 am
Doomschneider in theory, yes. it's fairly simple, but the numbers end up messy and, depending on the class you're in, there are a lot of people who forget how to do fraction exponents. how's this: [(8^(1/3))^(1/2)-(9^(1/2)^(1/2))]/[(3^(1/2)+2^(1/2))] [2^(1/2)-3^(1/2)]/[3^(1/2)+2^(1/2)] aka [sqrt(2)-sqrt(3)]/[sqrt(3)+sqrt(2)] if i tried to rationalize it, i might try this, but it's been a long time, and it's late: [sqrt(2)-sqrt(3)]/[sqrt(3)+sqrt(2)] * [sqrt(2)-sqrt(3)]/[sqrt(2)-sqrt(3)] [[sqrt(2)-sqrt(3)]^2]/{[sqrt(2)-sqrt(3)] * [sqrt(2)+sqrt(3)]} [[sqrt(2)-sqrt(3)]^2]/[2-3] ? - [sqrt(2)-sqrt(3)]^2 (which, btw, is NOT "-2-3" => "1") in other words, you're right, you probably shouldn't have been the only person that could do anything with that expression. You're making it more difficult for yourself. The equation given is: 8^1/6 - ( 9^1/4 / 3^1/2 ) + 2^1/2 This comes out to: sqrt(2) - ( sqrt(3) / sqrt(3) ) + sqrt(2) Which simplifies to: 2*sqrt(2) - 1
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Posted: Thu Jan 31, 2008 4:04 pm
Zaxoth Arturos Doomschneider in theory, yes. it's fairly simple, but the numbers end up messy and, depending on the class you're in, there are a lot of people who forget how to do fraction exponents. how's this: [(8^(1/3))^(1/2)-(9^(1/2)^(1/2))]/[(3^(1/2)+2^(1/2))] [2^(1/2)-3^(1/2)]/[3^(1/2)+2^(1/2)] aka [sqrt(2)-sqrt(3)]/[sqrt(3)+sqrt(2)] if i tried to rationalize it, i might try this, but it's been a long time, and it's late: [sqrt(2)-sqrt(3)]/[sqrt(3)+sqrt(2)] * [sqrt(2)-sqrt(3)]/[sqrt(2)-sqrt(3)] [[sqrt(2)-sqrt(3)]^2]/{[sqrt(2)-sqrt(3)] * [sqrt(2)+sqrt(3)]} [[sqrt(2)-sqrt(3)]^2]/[2-3] ? - [sqrt(2)-sqrt(3)]^2 (which, btw, is NOT "-2-3" => "1") in other words, you're right, you probably shouldn't have been the only person that could do anything with that expression. You're making it more difficult for yourself. The equation given is: 8^1/6 - ( 9^1/4 / 3^1/2 ) + 2^1/2 This comes out to: sqrt(2) - ( sqrt(3) / sqrt(3) ) + sqrt(2) Which simplifies to: 2*sqrt(2) - 1 ok, wait, she is almost right, but this is my fault for not making the problem clearer, I've edited the problem, it's correct now. edit: wait, sorry, Doomschneider, you are, in fact, wrong. You rationalized the numerator incorrectly
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Posted: Thu Jan 31, 2008 4:13 pm
slayerlx ok, wait, she is mostly right, but this is my fault for not making the problem clearer, I've edited the problem, it's correct now. Simple enough, even now, to be sure. Does the teacher require square roots to not be in the denominator? If so, the answer comes to: 2*sqrt(6) - 5
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Posted: Thu Jan 31, 2008 4:35 pm
Zaxoth Arturos slayerlx ok, wait, she is mostly right, but this is my fault for not making the problem clearer, I've edited the problem, it's correct now. Simple enough, even now, to be sure. Does the teacher require square roots to not be in the denominator? If so, the answer comes to: 2*sqrt(6) - 5 correct! and yes, she does want us to always rationalize the denominator
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Posted: Wed May 28, 2008 8:41 pm
O: i dun get it
im in the 7th grade... am i supposed to know this?
i stink at math T.T
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Posted: Sun Jun 22, 2008 4:20 pm
slayerlx simplify: (8^1/6 - 9^1/4) / (3^1/2 + 2^1/2) Not only am I the only one to get it right in my class, but apparently out of all the classes. But, it's seems sorta easy. Does anyone else?? this is where the rules of operations come in handy biggrin EPMDAS biggrin exponents,paranthesis,multiply/divide,add/subtract
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Posted: Sun Aug 10, 2008 2:36 pm
I thought order of op. was PEMDAS.
I know my Calc teachers never made us rationalize the denominator, only my algebra II teacher did.
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Posted: Mon Sep 22, 2008 11:09 am
Nyahoi O: i dun get it im in the 7th grade... am i supposed to know this? i stink at math T.T Nah don't worry about it dude, your at a lower lvl of math, this guy is talking about higher high school lvl math, and your only in middle school.
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Posted: Fri May 08, 2009 2:49 pm
SS4 Gogeta Forever slayerlx simplify: (8^1/6 - 9^1/4) / (3^1/2 + 2^1/2) Not only am I the only one to get it right in my class, but apparently out of all the classes. But, it's seems sorta easy. Does anyone else?? this is where the rules of operations come in handy biggrin EPMDAS biggrin exponents,paranthesis,multiply/divide,add/subtract our teacher says PEMDAS not epmsad
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