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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

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.

Answer: C.

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

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

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05 Jun 2013, 08:15

2

This post received KUDOS

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 Answer is C

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

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06 Jun 2013, 06:48

1

This post received KUDOS

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.

Answer is C.
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

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

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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.

Answer is C.

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
_________________

“Confidence comes not from always being right but from not fearing to be wrong.”

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

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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.

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

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30 Sep 2015, 22:35

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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 \(5 - 2*\sqrt{6}\)

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