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

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Joined: 09 Sep 2013
Posts: 1
\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?  [#permalink]

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Updated on: 15 Mar 2014, 05:05
12
00:00

Difficulty:

85% (hard)

Question Stats:

49% (01:51) correct 51% (01:59) wrong based on 340 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, 12:45.
Last edited by Bunuel on 15 Mar 2014, 05:05, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Joined: 02 Sep 2009
Posts: 52344

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15 Mar 2014, 05:04
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, $$\sqrt{7 + \sqrt{48}} - \sqrt{3} = (2+\sqrt{3})-\sqrt{3}=2$$.

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

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16 Mar 2014, 19:22
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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16 Mar 2014, 19:59
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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17 Jan 2015, 17:08
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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17 Jan 2015, 17:09
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, $$\sqrt{7 + \sqrt{48}} - \sqrt{3} = (2+\sqrt{3})-\sqrt{3}=2$$.

can anyone show me how we reduced( 2+sqrt3)^2-sqrt3 to become just (2+sqrt3)-sqrt3
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Posts: 52344
Re: \sqrt{7 + \sqrt{48}} - \sqrt{3} = ?  [#permalink]

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18 Jan 2015, 02:02
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, $$\sqrt{7 + \sqrt{48}} - \sqrt{3} = (2+\sqrt{3})-\sqrt{3}=2$$.

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

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18 Jan 2015, 02:17
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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22 Mar 2015, 08:32
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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22 Mar 2015, 11:00
√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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25 Dec 2017, 03:11
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: \sqrt{7 + \sqrt{48}} - \sqrt{3} = ? &nbs [#permalink] 25 Dec 2017, 03:11
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