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kyatin
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Is 4^{\frac12}+4^{\frac13}+4^{\frac14} greater than 5? I'm 02 Apr 2008, 23:34
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That was a cruel joke walker :-D
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walker
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Is 4^{\frac12}+4^{\frac13}+4^{\frac14} greater than 5? I'm 02 Apr 2008, 23:55
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kyatin
That was a cruel joke walker :-D
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:-D
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Is 4^{\frac12}+4^{\frac13}+4^{\frac14} greater than 5? I'm 17 Apr 2008, 17:47
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walker
kyatin
That was a cruel joke walker :-D
:-D
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Walker go to off topic... I have question for you on riding.
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Schools: Chicago (Booth) - Class of 2011
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Is 4^{\frac12}+4^{\frac13}+4^{\frac14} greater than 5? I'm 17 Apr 2008, 20:53
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dominion
Walker go to off topic... I have question for you on riding.
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Ok. Do you also like to ride? :cool
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Is 4^{\frac12}+4^{\frac13}+4^{\frac14} greater than 5? I'm 20 Apr 2008, 13:28
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walker
Is \(4^{\frac12}+4^{\frac13}+4^{\frac14}\) greater than 5?

I'm interested in the fastest approach. So, don't forget to start your timer :wink:
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I've tried to get it down to a point where you'd need to round only once. I think that's very important under given conditions. Actually, it was due to rounding that guys ended up with different answers.
assume \(4^{\frac12}+4^{\frac13}+4^{\frac14}>5\)
-->
\(4^{\frac13}+4^{\frac14}>5-2\)
\(2^{\frac23}+2^{\frac12}>3\)
\(2^{\frac23}>3-2^{\frac12}\)
take it to the power of 3
\(2^2>(3-2^{\frac12})(3-2^{\frac12})(3-2^{\frac12})\)
\(4>(9-3*2*2^{\frac12}+2)(3-2^{\frac12})\)
\(4>(11-6*2^{\frac12})(3-2^{\frac12})\)
\(4>(33-18*2^{\frac12}-11*2^{\frac12}+12)\)
\(4>(45-29*2^{\frac12})\)

and this is the point when you have to be cautious rounding \(sqrt(2)\)
if you take 1.4 you will get the wrong answer. If, however, you multiply 1.414 by 29 you will see that the inequality holds true under given assumption.
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Is 4^{\frac12}+4^{\frac13}+4^{\frac14} greater than 5? I'm 20 Apr 2008, 21:12
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NightAlum, good point!
(+1)
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Is 4^{\frac12}+4^{\frac13}+4^{\frac14} greater than 5? I'm 20 Apr 2008, 21:52
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NightAlum
walker
Is \(4^{\frac12}+4^{\frac13}+4^{\frac14}\) greater than 5?

I'm interested in the fastest approach. So, don't forget to start your timer :wink:

I've tried to get it down to a point where you'd need to round only once. I think that's very important under given conditions. Actually, it was due to rounding that guys ended up with different answers.
assume \(4^{\frac12}+4^{\frac13}+4^{\frac14}>5\)
-->
\(4^{\frac13}+4^{\frac14}>5-2\)
\(2^{\frac23}+2^{\frac12}>3\)
\(2^{\frac23}>3-2^{\frac12}\)
take it to the power of 3
\(2^2>(3-2^{\frac12})(3-2^{\frac12})(3-2^{\frac12})\)
\(4>(9-3*2*2^{\frac12}+2)(3-2^{\frac12})\)
\(4>(11-6*2^{\frac12})(3-2^{\frac12})\)
\(4>(33-18*2^{\frac12}-11*2^{\frac12}+12)\)
\(4>(45-29*2^{\frac12})\)

and this is the point when you have to be cautious rounding \(sqrt(2)\)
if you take 1.4 you will get the wrong answer. If, however, you multiply 1.414 by 29 you will see that the inequality holds true under given assumption.
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definitely your approach is nice one but is it the fastest way to do so?
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Is 4^{\frac12}+4^{\frac13}+4^{\frac14} greater than 5? I'm 20 Apr 2008, 21:53
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Have a great trip, Walker. Like you I visited the US for the first time only last year and envy you in a sense that Id love to repeat that first experience in NYC. Have a great time there!

2 Fistail: Thanks. I'm not sure.
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Is 4^{\frac12}+4^{\frac13}+4^{\frac14} greater than 5? I'm 20 Apr 2008, 22:57
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NightAlum
Have a great trip, Walker. Like you I visited the US for the first time only last year and envy you in a sense that Id love to repeat that first experience in NYC. Have a great time there!
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Thanks! I'm waiting for my flight on Thursday :bouncer2



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