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Re: 1/(2 - \sqrt3) = ? [#permalink]
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chengliu wrote:
\(\frac{1}{2 - \sqrt{3}} = ?\)

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


Rationalize the fraction by multiplying both the numerator and the denominator by \((2 + \sqrt{3})\). The purpose of rationalization is to eliminate irrational expressions in the denominator and simplify the expression:

\(\frac{2 + \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \)

\(=\frac{2 + \sqrt{3}}{4 - 3} = \)

\(=2 + \sqrt{3}\).


Answer: B­

This algebraic manipulation is called rationalisation and is performed to eliminate irrational expression in the denominator. For this particular case we are doing rationalisation by applying the following rule: \((a-b)(a+b)=a^2-b^2\). The concept of rationalisation of a fraction used to solve this problem is often tested on the GMAT.

Questions relied on this technique:
https://gmatclub.com/forum/if-x-0-then-106291.html
https://gmatclub.com/forum/if-n-is-posi ... 31236.html
https://gmatclub.com/forum/consider-a-q ... 31083.html
https://gmatclub.com/forum/in-the-diagr ... 39282.html
https://gmatclub.com/forum/in-the-diagr ... 29962.html
https://gmatclub.com/forum/the-perimete ... 27049.html
https://gmatclub.com/forum/which-of-the ... 98531.html
https://gmatclub.com/forum/if-x-is-posi ... 63491.html

Hope it helps.­
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Re: 1/(2 - \sqrt3) = ? [#permalink]
EMPOWERgmatRichC wrote:
Hi Sandra2U,

The GMAT will often test you on ideas/concepts that you know, but in ways that you're not used to thinking about.

This question is built on 2 concepts:
1) Re-writing fractions
2) Quadratics (and how to "get rid of" radicals)

When re-writing a fraction, we're allowed to MULTIPLY or DIVIDE both parts of a fraction (as long as we do it to both parts).

For example...

2/4 = 1/2

1/3 = 5/15

With Quadratics, there are 3 that you should have memorized for Test Day. Here, we're dealing with this one....

(X-Y)(X+Y) = X^2 - Y^2

So, if the first parenthesis is (2−√3), the second parenthesis should be (2+√3).

From here, the math is just as chengliu describes.

As you continue to practice, get in the habit of thinking about what the question REMINDS you of - chances are that those exact concepts/rules are what you're supposed to use to answer the question.

GMAT assassins aren't born, they're made,
Rich


Wouldn't this lead to 4 - sqrt(3) ? (using Foil)
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Re: 1/(2 - \sqrt3) = ? [#permalink]
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Hi chrtpmdr,

If you FOIL... (2−√3)(2+√3).... then you will end up with 4 - 3 = 1

In the context of the question that is asked, if you multiply both the numerator and denominator by (2+√3).... then you will end up with (2+√3)/1.

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Re: 1/(2 - \sqrt3) = ? [#permalink]
chengliu wrote:
\(\frac{1}{2 - \sqrt{3}} = ?\)

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


Yea the answer is B, after multiplication you get the denominator 4-3 which is 1 so it is \(2 + \sqrt{3}\)
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Re: 1/(2 - \sqrt3) = ? [#permalink]
EMPOWERgmatRichC wrote:
Hi chrtpmdr,

If you FOIL... (2−√3)(2+√3).... then you will end up with 4 - 3 = 1

In the context of the question that is asked, if you multiply both the numerator and denominator by (2+√3).... then you will end up with (2+√3)/1.

GMAT assassins aren't born, they're made,
Rich


Thank you for the fast answer - it is totally clear to me I was just a bit exhausted yesterday already at this point in my studies and therefore did not see the obvious.
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Re: 1/(2 - \sqrt3) = ? [#permalink]
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chengliu wrote:
\(\frac{1}{2 - \sqrt{3}} = ?\)

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


We rationalize the fraction by multiplying both the numerator and denominator by the conjugate (2 + √3), and we have:

(2 + √3) / ((2 - √3)(2 + √3))

(2 + √3) / (2^2 - (√3)^2)

(2 + √3) / (4 - 3) = 2 + √3

Answer: B
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Re: 1/(2 - \sqrt3) = ? [#permalink]
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We need to simplify \(\frac{1}{2 - \sqrt{3}}\)

Multiplying both the numerator and the denominator by \(2 + \sqrt{3}\) we get

\(\frac{2 + \sqrt{3}}{(2 + \sqrt{3}) * (2 - \sqrt{3})}\)

Using (a+b) * (a-b) = \(a^2 - b^2\) we get

\(\frac{2 + \sqrt{3}}{(2^2 - \sqrt{3}^2)}\) = \(\frac{2 + \sqrt{3}}{4 - 3}\) = \(2 + \sqrt{3}\)

So, Answer will be B
Hope it helps!
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