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by umaa » Tue Dec 23, 2008 9:52 pm
Explanation pls..
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by cramya » Tue Dec 23, 2008 10:17 pm
SQRT(4) = 2

4THROOT(4) = (2^2)1/4 = SQRT(2) = 1.414

3.414 just with evaluating 2 roots Eliminate A,B

Now its either c,d or e

cuberoot(4) ~= 1.6 (started with 1.5 and then did 1.6-> by trial)

Therefore greater than 4
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by Stuart@KaplanGMAT » Tue Dec 23, 2008 10:30 pm
We can quickly estimate.

root4 is 2

cuberoot 4 is between 1 and 2

quardicroot 4 is between 1 and 2.

So, the final answer is > 2 + 1 + 1, i.e > 4.. choose (E).

As an aside, for all n>1 and x>1, the nth root of x>1. In other words, even the billiionth root of 4 is > 1.
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by cramya » Tue Dec 23, 2008 10:32 pm
As an aside, for all n>1 and x>1, the nth root of x>1. In other words, even the billiionth root of 4 is > 1.
Thanks Stuart!
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by vishubn » Tue Dec 23, 2008 10:34 pm
cramya wrote:
As an aside, for all n>1 and x>1, the nth root of x>1. In other words, even the billiionth root of 4 is > 1.
Thanks Stuart!
I swear !! that was nice way to put it !! :D

vishu

thanks stuart
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by ronniecoleman » Tue Dec 23, 2008 11:14 pm
square Root 4 = 2
quad root 4 = root 2 = 1.414
cube root 4 = 2^2/3 ( 2/3 > 1/2 ) hence > 1.414

so X > 2+ 1.414 + 1.414
X> 4.8
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