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root[3*root{80}+3/(9+4*root{5})]=?

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root[3*root{80}+3/(9+4*root{5})]=?  [#permalink]

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New post 26 Jan 2012, 05:30
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\(\sqrt{3*\sqrt{80}+\frac{3}{9+4*\sqrt{5}}}=?\)

A) \(\sqrt{3*\sqrt{5}}\)

B) 3

C) \(3*\sqrt{3}\)

D) \(3+2*\sqrt{5}\)

E) \(9+4*\sqrt{5}\)
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Re: Sqaure Root  [#permalink]

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New post 26 Jan 2012, 05:59
5
4
LM wrote:
\(\sqrt{3*\sqrt{80}+\frac{3}{9+4*\sqrt{5}}}=?\)

A) \(\sqrt{3*\sqrt{5}}\)

B) 3

C) \(3*\sqrt{3}\)

D) \(3+2*\sqrt{5}\)

E) \(9+4*\sqrt{5}\)


\(\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}=\sqrt{12\sqrt{5}+\frac{3}{9+4\sqrt{5}}*\frac{9-4\sqrt{5}}{9-4\sqrt{5}}}=\sqrt{12\sqrt{5}+\frac{3(9-4\sqrt{5})}{81-80}}=\sqrt{12\sqrt{5}+27-12\sqrt{5}}=\sqrt{27}=3\sqrt{3}\).

Answer: C.
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Re: root[3*root{80}+3/(9+4*root{5})]=?  [#permalink]

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New post 12 Aug 2012, 08:56
Thanks Bunuel -- You Are The Man!! :)
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Re: root[3*root{80}+3/(9+4*root{5})]=?  [#permalink]

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New post 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.
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Re: root[3*root{80}+3/(9+4*root{5})]=?  [#permalink]

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New post 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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New post 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
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root[3*root{80}+3/(9+4*root{5})]=?  [#permalink]

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New post 02 Jan 2015, 00:12
\(\sqrt{3*\sqrt{80}+\frac{3}{9+4*\sqrt{5}}}\)

\(= \sqrt{3 * (4\sqrt{5}+\frac{1}{9+4*\sqrt{5}})}\)

\(= \sqrt{3 * \frac{36\sqrt{5} + 80 + 1}{9+4*\sqrt{5}}}\)

\(= \sqrt{3 * 9 * \frac{9+4*\sqrt{5}}{9+4*\sqrt{5}}}\)

\(= 3 \sqrt{3}\)
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Re: root[3*root{80}+3/(9+4*root{5})]=?  [#permalink]

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New post 04 Apr 2015, 06:52
2
1
LM wrote:
\(\sqrt{3*\sqrt{80}+\frac{3}{9+4*\sqrt{5}}}=?\)

A) \(\sqrt{3*\sqrt{5}}\)

B) 3

C) \(3*\sqrt{3}\)

D) \(3+2*\sqrt{5}\)

E) \(9+4*\sqrt{5}\)


Though, I am quite comfortable with the method mentioned by Bunuel, I found an alternative way by The Economist.

\(9 + 4*\sqrt{5}\) will be approx. equal to 9 + 4 x 2 = 17. Hence 3/ 17 will be quite less to contribute towards the value of expression.

\(3\sqrt{80}\) is approx. 3 x 9 = 27. now \(\sqrt{27}\) will be something more than 5.

Now coming to options:

(A)\(\sqrt{3*\sqrt{5}}\) is approx \(\sqrt{6}\) which is quite less than 5. Rejected.
(B) Rejected.
(C) 3 x 1.732 = 5.1 , which is in our desired range.
(D) 3 + 4 =7. Rejected
(E) 9 + 4 x 2 = 17. Rejected.

Hence the correct ans is (C).

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

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New post 27 Oct 2016, 18:41
Multiplying by the conjugate is the trick for this problem.

[3/(9+4sqrt(5))][(9-4sqrt(5)/9-4sqrt(5)] = 27-12sqrt(5)

sqrt(80) = 4sqrt(5)

Plug into equation 12sqrt(5) cancels out and we are left with sqrt(27) = 3sqrt(3)
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Re: root[3*root{80}+3/(9+4*root{5})]=?  [#permalink]

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New post 12 Nov 2017, 07:30
LM wrote:
\(\sqrt{3*\sqrt{80}+\frac{3}{9+4*\sqrt{5}}}=?\)

A) \(\sqrt{3*\sqrt{5}}\)

B) 3

C) \(3*\sqrt{3}\)

D) \(3+2*\sqrt{5}\)

E) \(9+4*\sqrt{5}\)


Let’s first simplify 3√80:

3√80 = 3 x √16 x √5 = 12√5

Next, let’s simplify 3/(9 + 4√5) by rationalizing the denominator:

3/(9 + 4√5) x (9 - 4√5)/(9 - 4√5) = (27 - 12√5)/(81 - 80) = 27 - 12√5

Finally, we have:

√(12√5 + 27 - 12√5) = √27 = √9 x √3 = 3√3

Answer: C
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Re: root[3*root{80}+3/(9+4*root{5})]=? &nbs [#permalink] 12 Nov 2017, 07:30
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