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brentbrent
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chix475ntu
Answer should be B.

Is it gmatclub test with the wrong answer? hmmmm :shock:


Whoops.
Answer is B. I have to edit that, gmat club isn't wrong =-)
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Bunuel
brentbrent
Hi all

\(\frac{1}{2+\sqrt{3}}=?\)

A. \(\sqrt{3}-2\)
B. \(2-\sqrt{3}\)
C. \(\sqrt{2}+\sqrt{3}\)
D. \(2+\sqrt{3}\)
E. \(\sqrt{3}+4\)

D
Solution:\((2-\sqrt{3})*(2+\sqrt{3})=4-3=1\)

Is there a fast way to recognize which one is correct here? Or can some one elaborate on the solution?

Thanks!

Answer can not be D, it should be B. The question is about the application of formula: \(a^2-b^2=(a-b)(a+b)\). Basically what we want to do is to make denominator 1, as no answer choice is in the form of fraction.

How can we do that?

Multiply \(\frac{1}{2+\sqrt{3}}\) by \(\frac{2-\sqrt{3}}{2-\sqrt{3}}\), which is 1, so that won't affect the value of our fraction. We'll get: \(\frac{1}{2+\sqrt{3}}*\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{2^2-\sqrt{3}^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}\)

Answer: B.
Thanks Bunnel. I didn't even think about that formula..sheesh.
cheers
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Answer is B

Multiply numerator & denominator by 2-\sqrt{3}.
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20 sec question and answer must be B.

multiply numerator and denominator by 2-\sqrt{3}

Cheers!
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How do you go from 2-[square_root]3/4-3 to the final answer?
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Crushit
How do you go from 2-[square_root]3/4-3 to the final answer?

just simplify it

2 - sqrt(3) / 4 - 3

= 2 - sqrt(3)/1

= 2 - sqrt(3) which is the answer
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brentbrent
Hi all

\(\frac{1}{2+\sqrt{3}}=?\)

A. \(\sqrt{3}-2\)
B. \(2-\sqrt{3}\)
C. \(\sqrt{2}+\sqrt{3}\)
D. \(2+\sqrt{3}\)
E. \(\sqrt{3}+4\)

B
Solution:\((2-\sqrt{3})*(2+\sqrt{3})=4-3=1\)

Is there a fast way to recognize which one is correct here? Or can some one elaborate on the solution?

Thanks!

edit: corrected answer choice. whoops.



Try this method instead....

we knw sqaure root of 3 is 1.732

So the equation becomes 1/(2+1.732)
that gives 1/(3.732) or may be rounded off to 1/4 or 0.25

Now working with options only B gives 2-1.732 that would give answer as 0.25.


Hope this helps.

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