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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, 12:59
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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, 20:23.
Last edited by ENGRTOMBA2018 on 13 Jan 2016, 12:59, 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, 20:45
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samark wrote:
Attachment:
solve.jpg


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

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

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New post 05 May 2016, 00:17
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, 11:40
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, 10:27
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}}}= &nbs [#permalink] 22 Oct 2018, 10:27
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