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

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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}=  [#permalink]

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New post Updated on: 13 Jan 2016, 13:59
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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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Originally posted by samark on 31 Oct 2010, 21:23.
Last edited by ENGRTOMBA2018 on 13 Jan 2016, 13:59, edited 2 times in total.
Reformatted the question and added the OA.
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Re: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}=  [#permalink]

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New post 31 Oct 2010, 21:45
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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}=  [#permalink]

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New post 05 May 2016, 01:17
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]

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New post 19 Jun 2017, 12:40
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) \(3+2\sqrt{5}\)


We can deconstruct this problem by knowing 2 basic things:

We can find the normal form of a radical expression like 3 root 80 by multiplying the square root of the coefficient in front of the radical by the number inside. So 3 root 80 is 9 times 80 720. We also need to rationalize the denominator.
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Re: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}=  [#permalink]

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New post 22 Oct 2018, 11:27
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]

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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]

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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] 13 Jan 2019, 23:39
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