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Re: root[3*root{80}+3/(9+4*root{5})]=? [#permalink]
12 Aug 2012, 10:29

This can be solved by taking the denominator as 9+4\sqrt{5} and then taking this as the common factor. We get the the ans as C. _________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

Re: root[3*root{80}+3/(9+4*root{5})]=? [#permalink]
16 Aug 2012, 22:16

This kind of problems always seemed to me very scary and requires a lot of calculation, but later i realised that GMAT never asks something that you need to calculate a lot, so one needs to look for some pattern or similar numbers/sets. In our case, we look at denominator 9+\sqrt{5} and 3\sqrt{80}, so 80 is 2^4*5, which means 4\sqrt{5}, from here we feel that numerator and denominator could be reduced. The rest is just calculations. In my opinion the most crucial part is this one. _________________

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Re: Que from MGMAT - [#permalink]
07 Apr 2013, 10:07

Lets analyze the first part 3\sqrt{80} = 3\sqrt{5*16} = 3*4\sqrt{5} The second term: Denominator (9+4\sqrt{5})*(9-4\sqrt{5})=9^2-4^2*5=1 Rule: (x+y)(x-y)=x^2-y^2 The second term: Numerator 3*(9-4\sqrt{5})=27-12\sqrt{5} Now putting all in one: \sqrt{(}12\sqrt{5}+27-12\sqrt{5})= \sqrt{27}=\sqrt{3*3^2}=3\sqrt{3}

Hope it's clear now _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: Que from MGMAT - [#permalink]
14 Apr 2013, 07:16

Zarrolou wrote:

Lets analyze the first part 3\sqrt{80} = 3\sqrt{5*16} = 3*4\sqrt{5} The second term: Denominator (9+4\sqrt{5})*(9-4\sqrt{5})=9^2-4^2*5=1 Rule: (x+y)(x-y)=x^2-y^2 The second term: Numerator 3*(9-4\sqrt{5})=27-12\sqrt{5} Now putting all in one: \sqrt{(}12\sqrt{5}+27-12\sqrt{5})= \sqrt{27}=\sqrt{3*3^2}=3\sqrt{3}

Hope it's clear now

thank you!! .. that 3 is so small that i took cube root 80...