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

If you like this Ballparking method, please press "Kudos".
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Re: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= [#permalink]
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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]
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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]
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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]
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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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]
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Re: 3\sqrt{80}+\frac{3}{9+4\sqrt{5}}}= [#permalink]
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