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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:
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
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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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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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:
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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78% (01:55) correct 
