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# Which of the following is equal to 1/[(3^(1/2)-2^(1/2))^1/2]

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Which of the following is equal to 1/[(3^(1/2)-2^(1/2))^1/2] [#permalink]  04 Aug 2010, 02:56
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Which of the following is equal to $$\frac{1}{(\sqrt{3}-\sqrt{2})^2$$.

A. 1

B. 5

C. $$\sqrt{6}$$

D. $$5 - \sqrt{6}$$

E. $$5 + 2\sqrt{6}$$
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Re: Which of the following is equal to 1/[(3^(1/2)-2^(1/2))^1/2] [#permalink]  04 Aug 2010, 03:07
Hi,

[1/ (sqrt3-sqrt2)]^2 = [1/ (3+2 - 2sqrt6)] = (5 + 2sqrt6)/ (25-24) = 5 + 2sqrt6

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Jack
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Re: Which of the following is equal to 1/[(3^(1/2)-2^(1/2))^1/2] [#permalink]  04 Aug 2010, 03:10
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kwhitejr wrote:
Can anyone demonstrate the following?

Which of the following is equal to [1/(\sqrt{3} - \sqrt{2})]^2 ?

A. 1
B. 5
C. \sqrt{6}
D. 5 - \sqrt{6}
E. 5 + 2\sqrt{6}

Two things to know to deal with this problem (helps to rationalize):

1. $$(a-b)^2=a^2-2ab+b^2$$ and 2. $$(a-b)(a+b)=a^2-b^2$$.

$$\frac{1}{(\sqrt{3}-\sqrt{2})^2}=\frac{1}{3-2\sqrt{6}+2}=\frac{1}{5-2\sqrt{6}}$$. Now multiply both nominator and denominator by $$5+2\sqrt{6}$$ --> $$\frac{5+2\sqrt{6}}{(5-2\sqrt{6})(5+2\sqrt{6})}=\frac{5+2\sqrt{6}}{25-24}=5+2\sqrt{6}$$.

Hope it's clear.

 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

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Re: Which of the following is equal to 1/[(3^(1/2)-2^(1/2))^1/2] [#permalink]  22 May 2014, 19:22
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Multiply both numerator & denominator by $$(\sqrt{3} + \sqrt{2})^2$$

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

$$= \frac{3 + 2\sqrt{6} + 2 }{((\sqrt{3} - \sqrt{2}) (\sqrt{3} + \sqrt{2}))^2}$$

$$= \frac{5 + 2\sqrt{6}}{ (3 - 2)^2}$$

$$= 5 + 2\sqrt{6}$$

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Re: Which of the following is equal to 1/[(3^(1/2)-2^(1/2))^1/2]   [#permalink] 22 May 2014, 19:22
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