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\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?

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\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?  [#permalink]

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New post Updated on: 15 Mar 2014, 06:05
12
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

49% (01:50) correct 51% (01:58) wrong based on 331 sessions

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\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)

A. 1
B. 1.7
C. 2
D. 2.4
E. 3

M11-18

Originally posted by faiint on 14 Mar 2014, 13:45.
Last edited by Bunuel on 15 Mar 2014, 06:05, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: GMATCLUB M11-18  [#permalink]

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New post 15 Mar 2014, 06:04
10
7
faiint wrote:
sqrt{7 + \sqrt{48}} - \sqrt{3} = ?

How come we can't do {sqrt{7 + \sqrt{48}} }^2 = {\sqrt{3}}^2 to solve?


Can you please elaborate what you mean? Thank you.

\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)

A. 1
B. 1.7
C. 2
D. 2.4
E. 3

The expression \(7 + \sqrt{48}\) can be rewritten as \(4 + 4\sqrt{3}+3\) which is the same as \((2+\sqrt{3})^2\). Therefore, \(\sqrt{7 + \sqrt{48}} - \sqrt{3} = (2+\sqrt{3})-\sqrt{3}=2\).

Answer: C.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 2, 3, 7, 8, 9 and 10. Thank you.
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Re: \sqrt{7 + \sqrt{48}} - \sqrt{3} = ?  [#permalink]

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New post 16 Mar 2014, 20:22
2
You may solve in this way:

Let \(\sqrt{ 7+ \sqrt{48}} - \sqrt{3} = x\)

\(\sqrt{ 7+ \sqrt{48}} = x + \sqrt{3}\)

Squaring both sides

\(7 + \sqrt{48} = 3 + 2\sqrt{3} x + x^2\)

\(4 + \sqrt{48} = 2\sqrt{3}x + x^2\)

\(x^2 + 2\sqrt{3}x - 43 - 4 = 0\)

\(x^2 - 4 + 2\sqrt{3}(x-2) = 0\)

\((x+2) (x-2) + 2\sqrt{3} (x-2) = 0\)

\((x-2) (x + 2 + 2\sqrt{3}) = 0\)

From the above equation, either x-2 = 0 or \(x + 2 + 2\sqrt{3} = 0\)

x = 2 OR \(x = -(2+2\sqrt{3})\)

From the options, x has a positive value, hence Answer = 2 = C

This is a long method; I would prefer Bunuel's method :)
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Re: \sqrt{7 + \sqrt{48}} - \sqrt{3} = ?  [#permalink]

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New post 16 Mar 2014, 20:59
1
If some1 is really fast in calculation then they might try this method as well.

\(\sqrt{ 7+ \sqrt{48}} - \sqrt{3}\) = \(\sqrt{ 14} - \sqrt{3}\) ( approximating 48 as 49)

= 3.7xxx - 1.732

which is close to 2.

but looking at the question it is better to do it the way Bunuel did it as GMAT doesn't want us to be good at calculations but at observation.
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Re: \sqrt{7 + \sqrt{48}} - \sqrt{3} = ?  [#permalink]

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New post 17 Jan 2015, 18:08
can anyone show me how we reduced( 2+sqrt3)^2-sqrt3 to become just (2+sqrt3)-sqrt3
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Re: \sqrt{7 + \sqrt{48}} - \sqrt{3} = ?  [#permalink]

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New post 17 Jan 2015, 18:09
Bunuel wrote:
faiint wrote:
sqrt{7 + \sqrt{48}} - \sqrt{3} = ?

How come we can't do {sqrt{7 + \sqrt{48}} }^2 = {\sqrt{3}}^2 to solve?


Can you please elaborate what you mean? Thank you.

\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)

A. 1
B. 1.7
C. 2
D. 2.4
E. 3

The expression \(7 + \sqrt{48}\) can be rewritten as \(4 + 4\sqrt{3}+3\) which is the same as \((2+\sqrt{3})^2\). Therefore, \(\sqrt{7 + \sqrt{48}} - \sqrt{3} = (2+\sqrt{3})-\sqrt{3}=2\).

Answer: C.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 2, 3, 7, 8, 9 and 10. Thank you.



can anyone show me how we reduced( 2+sqrt3)^2-sqrt3 to become just (2+sqrt3)-sqrt3
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Re: \sqrt{7 + \sqrt{48}} - \sqrt{3} = ?  [#permalink]

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New post 18 Jan 2015, 03:02
23a2012 wrote:
Bunuel wrote:
faiint wrote:
sqrt{7 + \sqrt{48}} - \sqrt{3} = ?

How come we can't do {sqrt{7 + \sqrt{48}} }^2 = {\sqrt{3}}^2 to solve?


Can you please elaborate what you mean? Thank you.

\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)

A. 1
B. 1.7
C. 2
D. 2.4
E. 3

The expression \(7 + \sqrt{48}\) can be rewritten as \(4 + 4\sqrt{3}+3\) which is the same as \((2+\sqrt{3})^2\). Therefore, \(\sqrt{7 + \sqrt{48}} - \sqrt{3} = (2+\sqrt{3})-\sqrt{3}=2\).

Answer: C.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 2, 3, 7, 8, 9 and 10. Thank you.



can anyone show me how we reduced( 2+sqrt3)^2-sqrt3 to become just (2+sqrt3)-sqrt3


\(\sqrt{7 + \sqrt{48}} - \sqrt{3} =\sqrt{(2+\sqrt{3})^2} - \sqrt{3}= (2+\sqrt{3})-\sqrt{3}=2\)
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: \sqrt{7 + \sqrt{48}} - \sqrt{3} = ?  [#permalink]

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New post 18 Jan 2015, 03:17
faiint wrote:
\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)

A. 1
B. 1.7
C. 2
D. 2.4
E. 3

M11-18


\(sqrt{48}\) is approximately 7; \(\sqrt{7 + \sqrt{48}\) falls between 3 and 4, say 3,7d \(sqrt{3}\) is approximately 1,7; result is very close to 2. A & B too small D & E too big C = correct.
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Re: \sqrt{7 + \sqrt{48}} - \sqrt{3} = ?  [#permalink]

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New post 22 Mar 2015, 09:32
PareshGmat wrote:
You may solve in this way:

Let \(\sqrt{ 7+ \sqrt{48}} - \sqrt{3} = x\)

\(\sqrt{ 7+ \sqrt{48}} = x + \sqrt{3}\)

Squaring both sides

\(7 + \sqrt{48} = 3 + 2\sqrt{3} x + x^2\)

\(4 + \sqrt{48} = 2\sqrt{3}x + x^2\)

\(x^2 + 2\sqrt{3}x - 43 - 4 = 0\)

\(x^2 - 4 + 2\sqrt{3}(x-2) = 0\)

\((x+2) (x-2) + 2\sqrt{3} (x-2) = 0\)

\((x-2) (x + 2 + 2\sqrt{3}) = 0\)

From the above equation, either x-2 = 0 or \(x + 2 + 2\sqrt{3} = 0\)

x = 2 OR \(x = -(2+2\sqrt{3})\)

From the options, x has a positive value, hence Answer = 2 = C

This is a long method; I would prefer Bunuel's method :)



How is square root of 48 suddenly reduced to 43 ?
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Re: \sqrt{7 + \sqrt{48}} - \sqrt{3} = ?  [#permalink]

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New post 22 Mar 2015, 12:00
√of 48 =6.9 so we make it 49. √7+7 -√3 = √14 -√3= 3.7-1.5= 2 (C)
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Re: \sqrt{7 + \sqrt{48}} - \sqrt{3} = ?  [#permalink]

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New post 25 Dec 2017, 04:11
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Re: \sqrt{7 + \sqrt{48}} - \sqrt{3} = ? &nbs [#permalink] 25 Dec 2017, 04:11
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