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root[root{96}+2/(5+2*root{6})] lies between

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Director
Joined: 03 Sep 2006
Posts: 840

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26 Jan 2012, 06:26
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80% (01:04) correct 20% (01:34) wrong based on 577 sessions

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$$\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}$$ lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6
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Joined: 02 Sep 2009
Posts: 46143

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26 Jan 2012, 06:39
4
2
LM wrote:
$$\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}$$ lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6

Multiply both nominator and denominator of $$\frac{2}{5+2*\sqrt{6}}$$ by $${5-2*\sqrt{6}}$$ and apply $$(a+b)(a-b)=a^2-b^2$$: $$\frac{2*(5-2*\sqrt{6})}{(5+2*\sqrt{6})(5-2*\sqrt{6})}=\frac{2*(5-2*\sqrt{6})}{25-24}=2*(5-2*\sqrt{6})$$.

Now, $$\sqrt{6}$$ is a little bit more than 2 (~2.5), hence $$2*(5-2*\sqrt{6})\approx{2(5-2*2.5)}=0$$;

So we have: $$\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}\approx{\sqrt{\sqrt{96}+0}=\sqrt{\sqrt{96}}$$ --> $$\sqrt{96}$$ is more than 9 but less than 10 (~9.5), hence $$\sqrt{\sqrt{96}}\approx{\sqrt{9.5}}$$, which is more than 3 but less than 4.

Or another way, using the same approximations as above:
$$\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}\approx{\sqrt{9.5+\frac{2}{5+2*2.5}}}=\sqrt{9.5+\frac{1}{5}}\approx{\sqrt{10}}\approx{3.something}$$

Hope it's clear.
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05 Jun 2013, 04:30
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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05 Jun 2013, 07:27
5
LM wrote:
$$\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}$$ lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6

$$\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}=\sqrt{4\sqrt{6}+\frac{2}{5+2*\sqrt{6}}}=\sqrt{2(2*\sqrt{6}+\frac{1}{5+2*\sqrt{6}})}=\sqrt{2(\frac{25+10\sqrt{6}}{5+2*\sqrt{6}})}=\sqrt{2*5(\frac{5+2\sqrt{6}}{5+2*\sqrt{6}})}=\sqrt{2*5}$$

a bit more than $$3$$,C
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05 Jun 2013, 08:15
3
Here is how I solved this:
1) Follow what Bunuel did up to a point
Bunuel wrote:
LM wrote:
$$\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}$$ lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6

Multiply both nominator and denominator of $$\frac{2}{5+2*\sqrt{6}}$$ by $${5-2*\sqrt{6}}$$ and apply $$(a+b)(a-b)=a^2-b^2$$: $$\frac{2*(5-2*\sqrt{6})}{(5+2*\sqrt{6})(5-2*\sqrt{6})}=\frac{2*(5-2*\sqrt{6})}{25-24}=2*(5-2*\sqrt{6})$$.

2) multiply out $$2(5-2\sqrt{6}) = 10-4\sqrt{6}$$
3) $$\sqrt{96} = 4\sqrt{6}$$
4) $$\sqrt{4\sqrt{6}+10-4\sqrt{6}} = \sqrt{10}$$
5) 3² = 9 and 4² = 16, so $$\sqrt{10}$$ is between 3 and 4
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06 Jun 2013, 06:48
1
LM wrote:
$$\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}$$ lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6

In approximation questions we can approximate, so:
$$\sqrt{96}$$ is between 9 and 10, $$\sqrt{6}$$ is between 2 and 3, and the expression $$\frac{2}{5+2*\sqrt{6}}$$ is about 1/5 (0.20) which means it does not affect the final answer too much in our case. Since the first number is bigger than 9 and smaller than 10 its square root will be more than 3 but less than 4.

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Joined: 23 May 2013
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06 Jun 2013, 07:30
ziko wrote:
LM wrote:
$$\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}$$ lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6

In approximation questions we can approximate, so:
$$\sqrt{96}$$ is between 9 and 10, $$\sqrt{6}$$ is between 2 and 3, and the expression $$\frac{2}{5+2*\sqrt{6}}$$ is about 1/5 (0.20) which means it does not affect the final answer too much in our case. Since the first number is bigger than 9 and smaller than 10 its square root will be more than 3 but less than 4.

No need to solve for the second part. moment we get to know that sqrt of 96 will be approx 9 and sqrt of 9 is 3. So ans should be between 3 and 4
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16 Dec 2013, 04:49
LM wrote:
$$\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}$$ lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6

I used ballparking. root96 rounded up to root100 is 10. root6 rounded down to root4 is 2. so I have 10+ 2/9 which is a little bigger than root9 which is 3.

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Re: root[root{96}+2/(√+25+2∗6√−−−−−−−−−−−−√ 5+2*root{6})] lies between [#permalink]

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01 Oct 2015, 23:58
NOTE: When ever we have an expression which contains a square root in the denominator, always rationalize the denominator.
Example: $$a/b+\sqrt{c}$$
We should multiply both the numerator and the denominator by $$b - \sqrt{c}$$

Coming to the question, first step is to rationalize $$2/(5 + 2*\sqrt{6})$$
Multiplying both numerator and denominator by $$5 - 2*\sqrt{6}$$

The expression becomes:

$$(4 \sqrt{6} + 2*( 5 - 2*\sqrt{6)}/(25 - 24))$$^(1/2)
$$(4 *\sqrt{6} + 10 - 4* \sqrt{6})$$^(1/2)
$$\sqrt{10} = 3. xx$$

Option C
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Joined: 28 May 2017
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30 Aug 2017, 04:32
No need for any complex calculations.
( 10- + 2/(5+2*2+) )^1/2
= ( 92/9+ )^1/2
= 10-/3+
= between 3 and 4
Manager
Joined: 05 Dec 2016
Posts: 114

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17 Mar 2018, 06:10
1
LM wrote:
$$\sqrt{\sqrt{96}+\frac{2}{5+2*\sqrt{6}}}$$ lies between

A) 1 & 2
B) 2 & 3
C) 3 & 4
D) 4 & 5
E) 5 & 6

The fastest way for my is this : we have$$\sqrt{96}=\sqrt{6*8*2}=4*\sqrt{6}$$ and $$\frac{2}{(5+2*\sqrt{6})}$$ =$$\frac{2*(5-2*sqr(6))}{5^2-2^2*sqr(6)^2}$$ = $$10-4*\sqrt{6}$$

so we have $$\sqrt{10}$$ which must lie between the closest perfect squares $$\sqrt{9}<\sqrt{10}<\sqrt{12}$$ therefore answer is C
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Re: root[root{96}+2/(5+2*root{6})] lies between   [#permalink] 17 Mar 2018, 06:10
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