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√[2√63 + 2/(8+3√7)] =

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\sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}}  [#permalink]

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New post 21 Jan 2010, 06:31
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\(\sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}}\) =

A) \(8+3\sqrt{7}\)
B) \(4+3\sqrt{7}\)
C) 8
D) 4
E) \(\sqrt{7}\)
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Re: square root problem  [#permalink]

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New post 26 Sep 2010, 12:07
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5
udaymathapati wrote:
√[2√63 + 2/(8+3√7)] =
A. 8 + 3√7
B. 4 + 3√7
C. 8
D. 4
E. √7


\(\sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}}=\sqrt{6\sqrt{7}+\frac{2}{8+3\sqrt{7}}*\frac{8-3\sqrt{7}}{8-3\sqrt{7}}}=\sqrt{6\sqrt{7}+\frac{2(8-3\sqrt{7})}{64-63}}=\sqrt{6\sqrt{7}+16-6\sqrt{7}}=\sqrt{16}=4\).

Answer: D.

Please read and apply when posting such questions: writing-mathematical-symbols-in-posts-72468.html
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√[2√63 + 2/(8+3√7)] =  [#permalink]

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New post 26 Sep 2010, 11:31
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√[2√63 + 2/(8+3√7)] =

A. 8 + 3√7
B. 4 + 3√7
C. 8
D. 4
E. √7
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Re: \sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}}  [#permalink]

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New post 21 Jan 2010, 09:22
1
4
rahulms wrote:
\(\sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}\) =

A) \(8+3\sqrt{7}\)
B) \(4+3\sqrt{7}\)
C) 8
D) 4
E) \(\sqrt{7}\)


\(\sqrt{2\sqrt{63}+\frac{2}{(8+3\sqrt{7})}\frac{(8-3\sqrt{7})}{(8-3\sqrt{7})}\)

\(\sqrt{{2\{3\sqrt{7}})+\frac{2 (8-3\sqrt{7})}{(64-63)}}\)

\(\sqrt{{2\{3\sqrt{7}})+{2 (8)-(2)(3\sqrt{7})}\)

\(\sqrt{16}\)

\(4\)
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√[2√63 + 2/(8+3√7)] =  [#permalink]

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New post 06 Jul 2010, 11:08
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Re: √[2√63 + 2/(8+3√7)] =  [#permalink]

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New post 06 Jul 2010, 13:19
2
This is a fairly simple problem but one that requires combination of a few concepts. Here's my simplification:

You first identify that \(\sqrt{63}= 2*3\sqrt{7}\) and then multiply the numerator and denominator of the second term with \(8-3\sqrt{7}\) to rationalize it.

Rationalizing the second term yields the following result:

\(\frac{16-6\sqrt{7}}{64-63}\) = \(\frac{16-6\sqrt{7}}{1}\) = \(16-6\sqrt{7}\)

The first term is \(2*3\sqrt{7}\) = \(6\sqrt{7}\)

When you put them together, the expression inside the overall square root symbol becomes:

\(6\sqrt{7}\) - \(6\sqrt{7} + 16 = 16\)

The \(6\sqrt{7}\) cancels out leaving you with \(\sqrt{16} = 4\)

Thus, answer is D. Hope this helps!
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Re: √[2√63 + 2/(8+3√7)] =  [#permalink]

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New post 06 Jul 2010, 22:39
I dont understand,,

why do you multiply with \(8-3\sqrt{7}\) but not \(8+3\sqrt{7}\)..

And how did you get \(\frac{16-6\sqrt{7}}{64-63}\)

thanks...
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Re: √[2√63 + 2/(8+3√7)] =  [#permalink]

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New post 07 Jul 2010, 06:45
1
gmatJP wrote:
I dont understand,,

why do you multiply with \(8-3\sqrt{7}\) but not \(8+3\sqrt{7}\)..

And how did you get \(\frac{16-6\sqrt{7}}{64-63}\)

thanks...


It's called rationalization gmatJP.

When you multiply \(8+3\sqrt{7}\) with \(8+3\sqrt{7}\) you get \((8+3\sqrt{7})^2\) which can only be simplified as \(64+ 2*8*3\sqrt{7}+63\) on the denominator which will not help us much in the process of simplification. The rationale behind multiplying it with \(8-3\sqrt{7}\) is because it'll be of the form \((a+b)(a-b) = a^2-b^2\) which gives us \((8^2)-(3\sqrt{7})^2\) = \(64 - (3)^2*(\sqrt{7})^2\) = \(64-63 = 1\) which literally makes the denominator inconsequential to the simplification process and helps us cancel terms.

So, now when we multiply the numerator of the second term (2) with \(8-3\sqrt{7}\) we get \(2*(8-3\sqrt{7})\) = \(16-6\sqrt{7}\)

So we get \(\frac{16-6\sqrt{7}}{64-63}\) as the final product. Hope this is clear.
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Re: √[2√63 + 2/(8+3√7)] =  [#permalink]

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New post 07 Jul 2010, 12:25
Hussain15 wrote:
√[2√63 + 2/(8+3√7)] =

A. 8 + 3√7
B. 4 + 3√7
C. 8
D. 4
E. √7


There is another way to do this without rationalizing it, which I'm not so sure the GMAT even expects you to do.

√[2√63 + 2/(8+3√7)

Multiply the terms: 2√63 + 2/(8+3√7) -> (16√63+6√(63*7)+2)/(8+3√7). Now simplify.

(16*3√7+6*3*7+2)/(8+3√7) -> (16*3√7+128)/(8+3√7)

Now, notice if we take out the 16 from the top we get -> 16*(3√7+8)/(8+3√7) --> and the "messy portion gets cancelled out and we are left with 16. Don't forget that we still have a √ left

√16 = 4 and thus D is our answer.
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Re: square root problem  [#permalink]

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New post 26 Sep 2010, 12:15
In case you have forgotten how to work these, I found a good site on conjugates:

http://www.regentsprep.org/Regents/math ... lesson.htm Look at situation 2 and 3.
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Re: Square root  [#permalink]

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New post 25 Oct 2010, 03:46
In these type of question u multiply the nominator and denominator with conjugate...Conjugate of 8 +3sqrt(7) is 8 -3sqrt(7)

sqrt[2sqrt(63) + 2 { 8 - 3sqrt(7)}/{64 - 63}]
=sqrt[2sqrt(63) +16 - 2sqrt(63)] =4

Answer is 4.

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Re: square root problem  [#permalink]

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New post 11 Feb 2011, 08:41
√[2√63 + 2/(8+3√7)]


48√7 + 2 * 3 * 3 * √7 * √7 + 2
√[-----------------------------------------]
8 + 3√7

48√7 + 128
√[---------------]
8 + 3√7


16(3√7 + 8)
√[----------------]
8 + 3√7

√[16] = 4
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Re: square root problem  [#permalink]

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New post 15 Jul 2012, 15:06
udaymathapati wrote:
√[2√63 + 2/(8+3√7)] =
A. 8 + 3√7
B. 4 + 3√7
C. 8
D. 4
E. √7


Here is my approach that I think work good for problems like these and on the actual test:

\(\sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}}\).

Since \(\sqrt{64}=8\)
then take: \(\sqrt{2*8}\)

And since this part will be very small
\(\frac{2}{8+3\sqrt{7}}\).

I just take: \(\sqrt{2*8+ something\)

And get \(\sqrt{16}=4\)



Answer: D.
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Re: √[2√63 + 2/(8+3√7)] =  [#permalink]

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New post 21 Oct 2014, 23:55
\(\sqrt{2\sqrt{63}+\frac{2}{8+3\sqrt{7}}} = \sqrt{2\sqrt{63}+\frac{2}{8+\sqrt{63}}} = \sqrt{2\sqrt{63}+\frac{2}{8+\sqrt{63}} * \frac{8-\sqrt{63}}{8-\sqrt{63}}} = \sqrt{2\sqrt{63}+16 - 2\sqrt{63}} = \sqrt{16} = 4\)

Answer = D
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Roots problem  [#permalink]

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New post 18 Jun 2015, 19:18
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\((2\sqrt{63}+\frac{2}{8+3\sqrt{7}})^\frac{1}{2}\)


A) \(8+3\sqrt{7}\)
B) \(4+3\sqrt{7}\)
C) 8
D) 4
E) \(\sqrt{7}\)
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Re: Roots problem  [#permalink]

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New post 18 Jun 2015, 19:35
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anurag356 wrote:
\((2\sqrt{63}+\frac{2}{8+3\sqrt{7}})^\frac{1}{2}\)


A) \(8+3\sqrt{7}\)
B) \(4+3\sqrt{7}\)
C) 8
D) 4
E) \(\sqrt{7}\)



\((2\sqrt{63}+\frac{2}{8+3\sqrt{7}})^\frac{1}{2}\)
mutliply \(\frac{2}{8+3\sqrt{7}}\)with \([fraction]8-3\sqrt{7}\) in noth num and denominator
we get\(2(8-3\sqrt{7})=16-6\sqrt{7}\)
\(2\sqrt{63}=6\sqrt{7}\)
therefore ans is \(\sqrt{16}=4\)
ans D
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Re: Roots problem  [#permalink]

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New post 18 Jun 2015, 20:31
anurag356 wrote:
\((2\sqrt{63}+\frac{2}{8+3\sqrt{7}})^\frac{1}{2}\)


A) \(8+3\sqrt{7}\)
B) \(4+3\sqrt{7}\)
C) 8
D) 4
E) \(\sqrt{7}\)



You can solve it easily if you use approximation.

You have two terms:
\(2\sqrt{63}\)
\(\frac{2}{8+3\sqrt{7}}\)

The term \(\frac{2}{8+3\sqrt{7}}\) is 2 divided by a relatively much larger number so its value will be small (in decimals 0.1 types). That is much smaller than the other term \(2\sqrt{63}\) because \(\sqrt{63}\) is almost 8. So the combined effect of these terms is:

2* almost 8 + very small term = 16 = 4^2 (approximately)

\((4^2)^{1/2}= 4\)

Answer (D)
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Re: Roots problem  [#permalink]

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New post 18 Jun 2015, 22:06
anurag356 wrote:
\((2\sqrt{63}+\frac{2}{8+3\sqrt{7}})^\frac{1}{2}\)


A) \(8+3\sqrt{7}\)
B) \(4+3\sqrt{7}\)
C) 8
D) 4
E) \(\sqrt{7}\)


Here is a different and Much faster approach of Pure Aptitude (and less of maths)

We can solve every such question with the approximation

Now we know that \(\sqrt{63}\) is very close to\(\sqrt{64}\)=8

i.e. we can take \(\sqrt{63}\)= 7.9 or almost 8 for approximation
i.e. \(2*\sqrt{63} = 2*8\) = approx. \(16\)

Now we know that \(\sqrt{7}\) is very close to \(\sqrt{9}\)=3
i.e. we can take \(\sqrt{7}= 2.8\) or almost \(3\) for approximation

Now, \(8+3\sqrt{7} = 8+3*3 = 17\) approximately

Now, \(\frac{2}{8+3\sqrt{7}} = 2/17\) = approx. \(1/8\) = \(0.125\) approximately

i.e. \((2\sqrt{63}+\frac{2}{8+3\sqrt{7}})^\frac{1}{2}\) = \((16+0.125)^{1/2}\) = Approx. 4

Answer: Option

Except the answer choice, no other option is even close
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Re: Roots problem  [#permalink]

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New post 18 Jun 2015, 22:53
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When you see a square root in a denominator of a fraction, you need to get it out of the denominator - that's a required simplification in math. When that root is in a sum or subtraction, you need to use the difference of squares pattern, as below. Note also we can factor a perfect square from √63, so we can also simplify that root:

\(\begin{align}
\left( 2 \sqrt{63} + \frac{2}{8 + 3\sqrt{7}} \right)^{\frac{1}{2}} &= \left( 2 \sqrt{9} \sqrt{7} + \left( \frac{2}{8 + 3\sqrt{7}} \right) \left( \frac{8 - 3\sqrt{7}}{8 - 3\sqrt{7}} \right) \right)^{\frac{1}{2}} \\
&= \left( 6 \sqrt{7} + \frac{16 - 6\sqrt{7}}{8^2 - (9)(7)} \right)^{\frac{1}{2}} \\
&= \left(6 \sqrt{7} + \frac{16 - 6\sqrt{7}}{1} \right)^{\frac{1}{2}} \\
&= (16)^{\frac{1}{2}} \\
&= 4
\end{align}\)

This is identical to the approach chetan2u used above, but I post it in case the additional detail helps anyone. Estimation also works here, but I wouldn't advise relying on that strategy, since it will only work when the answer choices are sufficiently far apart.
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Re: Roots problem  [#permalink]

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New post 18 Jun 2015, 23:03
in GMAT, answer choices are sufficiently far apart in the questions of roots (atleast my little 10 years of experience say so about such questions). ;)

But yes, We all are entitled to have different opinions. (y)
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Re: Roots problem   [#permalink] 18 Jun 2015, 23:03

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