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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]
05 Jun 2013, 07:15
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]
06 Jun 2013, 05:48
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]
06 Jun 2013, 06:30
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]
16 Dec 2013, 03:49
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]
30 Sep 2015, 21:35
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Re: root[root{96}+2/(√+25+2∗6√−−−−−−−−−−−−√ 5+2*root{6})] lies between [#permalink]
01 Oct 2015, 22:58
Expert's post
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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