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605-655 Level|   Roots|      
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samark
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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= Updated on: 20 Jul 2022, 02:51
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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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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73% (02:11) correct 27% (02:50) wrong based on 1075 sessions

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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
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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= 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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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= 12 Aug 2012, 09:56
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Thanks Bunuel -- You Are The Man!! :)
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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= 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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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= 04 Apr 2015, 07:52
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LM
\(\sqrt{3*\sqrt{80}+\frac{3}{9+4*\sqrt{5}}}=?\)

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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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= 05 May 2016, 01:17
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samark
\(\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}} =\)

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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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= 12 Nov 2017, 08:30
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LM
\(\sqrt{3*\sqrt{80}+\frac{3}{9+4*\sqrt{5}}}=?\)

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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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= 22 Oct 2018, 11:27
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samark
\(\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}} =\)

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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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= 13 Jan 2019, 16:09
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samark
\(\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}} =\)

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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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= 13 Jan 2019, 23:39
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jfranciscocuencag
samark
\(\sqrt{3\sqrt{80}+\frac{3}{9+4\sqrt{5}}} =\)

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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3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= 24 Sep 2024, 09:09
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