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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}=

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samark
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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= [#permalink] New post   Updated on: 20 Jul 2022, 02:51
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\(\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}} =\)

(A) \(2\sqrt{3\sqrt{5}}\)

(B) \(3\)

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

(D) \(9+4\sqrt{5}\)

(E) \(3+2\sqrt{5}\)
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C

Originally posted by samark on 31 Oct 2010, 21:23.
Last edited by Bunuel on 20 Jul 2022, 02:51, edited 3 times in total.
Reformatted the question and added the OA.
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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= [#permalink] New post   31 Oct 2010, 21:45
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\(\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}} =\)

(A) \(2\sqrt{3\sqrt{5}}\)

(B) \(3\)

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

(D) \(9+4\sqrt{5}\)

(E) \(3+2\sqrt{5}\)


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

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

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

    \(=\sqrt{12\sqrt{5}+27-12\sqrt{5}}=\)

    \(=\sqrt{27}=\)

    \(=3\sqrt{3}\).

Answer: C.
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Re: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= [#permalink] New post   12 Aug 2012, 09:56
Thanks Bunuel -- You Are The Man!! :)
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Re: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= [#permalink] New post   07 Apr 2013, 11:07
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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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Re: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= [#permalink] New post   04 Apr 2015, 07:52
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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: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= [#permalink] New post   05 May 2016, 01:17
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samark wrote:
\(\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}} =\)

A) \(2\sqrt{3\sqrt{5}}\)
B) \(3\)
C) \(3\sqrt{3}\)
D) \(9+4\sqrt{5}\)
E) \(3+2\sqrt{5}\)


You can simplify this question by changing \(\sqrt{80}\) to \(\sqrt{81}\)

\(4\sqrt{5}\) is actually \(\sqrt{80}\), change it also...

\(\sqrt{81}\) is 9

so approximately the equation produce \(\sqrt{27}\)

= \(3\sqrt{3}\)
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Re: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= [#permalink] New post   12 Nov 2017, 08:30
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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: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= [#permalink] New post   22 Oct 2018, 11:27
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samark wrote:
\(\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}} =\)

A) \(2\sqrt{3\sqrt{5}}\)
B) \(3\)
C) \(3\sqrt{3}\)
D) \(9+4\sqrt{5}\)
E) \(3+2\sqrt{5}\)

\(?\,\,\,:\,\,\,{\rm{expression}}\)

\(\sqrt {80} \,\, = \,\,\underleftrightarrow {\sqrt {8 \cdot 10} = \sqrt {{2^4} \cdot 5} } = 4\sqrt 5 \,\)

\(\frac{1}{{9 + 4\sqrt 5 }} = \frac{1}{{9 + 4\sqrt 5 }} \cdot \frac{{9 - 4\sqrt 5 }}{{9 - 4\sqrt 5 }} = \frac{{9 - 4\sqrt 5 }}{{81 - 16 \cdot 5}} = 9 - 4\sqrt 5\)

\(3\left( {\sqrt {80} + \frac{1}{{9 + 4\sqrt 5 }}} \right)\,\, = \,\,3\,\left( 9 \right)\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,?\,\, = \,\,\sqrt {3 \cdot 9} = 3\sqrt 3\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= [#permalink] New post   13 Jan 2019, 16:09
samark wrote:
\(\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}} =\)

A) \(2\sqrt{3\sqrt{5}}\)
B) \(3\)
C) \(3\sqrt{3}\)
D) \(9+4\sqrt{5}\)
E) \(3+2\sqrt{5}\)


Hello math experts!

I think this might be a silly question but Why can't we square the whole term in order to get fid of the square root?

I'm a bit confused.

Thank you in advance!
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Re: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= [#permalink] New post   13 Jan 2019, 23:39
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jfranciscocuencag wrote:
samark wrote:
\(\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}} =\)

A) \(2\sqrt{3\sqrt{5}}\)
B) \(3\)
C) \(3\sqrt{3}\)
D) \(9+4\sqrt{5}\)
E) \(3+2\sqrt{5}\)


Hello math experts!

I think this might be a silly question but Why can't we square the whole term in order to get fid of the square root?

I'm a bit confused.

Thank you in advance!


Squaring to get rid of the square root is useful technique in some cases but here it does not give anything super helpful. We'd get \(3\sqrt{80}+\frac{3}{9+4\sqrt{5}}\). How is this any better to calculate then what we had before? Plus we should not forget un-squaring back to get the correct answer.


Check the following question to see when squaring might be a good call: https://gmatclub.com/forum/new-tough-an ... l#p1029216
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Re: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= [#permalink] New post   30 Aug 2023, 06:01
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Re: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= [#permalink]
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