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

Answer: D.

Please read and apply when posting such questions: writing-mathematical-symbols-in-posts-72468.html
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\(\sqrt{2\sqrt{63}+\frac{2}{(8+3\sqrt{7})}\frac{(8-3\sqrt{7})}{(8-3\sqrt{7})}\)

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

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

\(\sqrt{16}\)

\(4\)

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!

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.

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.

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.

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.

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

Answer = D

\((2\sqrt{63}+\frac{2}{8+3\sqrt{7}})^\frac{1}{2}\)
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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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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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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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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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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I don't think it's a question of opinion - it's either true or false that you can reliably estimate on roots questions on the real GMAT, and we can decide by looking at real GMAT questions. The question below is from an old version of GMATPrep, and there is no easy way to estimate to get an answer:

if-the-sum-of-the-square-roots-of-two-integers-is-43004.html
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All such question can be easily solved by approximation so in case readers haven't tried it before, I would urge them to think the approximate values and answer such questions. Here is the explanation that you will probably fall in love with

We have to calculate the value of √[2√63 + 2/(8+3√7)]

√63 is very close to √64 = 8
therefore,
√63 = 8 (approx.)

√7 is very close to √9 = 3
therefore,
√7 = 3 (approx.)

Now, √[2√63 + 2/(8+3√7)] = √[2*8 + 2/(8+3*3)] = √[16 + 2/(17)] = √[16] = 4 (approx)

A. 8 + 3√7 is Much Greater than 4 so INCORRECT
B. 4 + 3√7 is Much Greater than 4 so INCORRECT
C. 8 is Much Greater than 4 so INCORRECT
D. 4 CORRECT
E. √7 = 2.8 approx so INCORRECT
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HiBunuel: Can you please post similar questions to practice.

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