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(3^(1/2) - 2^(1/2))^2

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(3^(1/2) - 2^(1/2))^2  [#permalink]

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New post 07 Aug 2018, 04:14
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

76% (00:35) correct 24% (00:46) wrong based on 38 sessions

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Re: (3^(1/2) - 2^(1/2))^2  [#permalink]

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New post 07 Aug 2018, 04:30
Bunuel wrote:
\((\sqrt3 - \sqrt2)^2\)


A. \(1 - 2 \sqrt{6}\)

B. \(1 - \sqrt{6}\)

C. \(5 - 2 \sqrt{6}\)

D. \(5 - 2 \sqrt{3}\)

E. \(1\)



Given

\((\sqrt3 - \sqrt2)^2\)


applying \((a-b)^2\). we will surely get \(5 - 2\sqrt{}6\)

The best answer is C.
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Re: (3^(1/2) - 2^(1/2))^2  [#permalink]

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New post 07 Aug 2018, 06:49
+1 for C.

= (√3−√2)^2
= √3^2 - 2*√3*√2 + √2^2
= 3 - 2√6 + 2
= 5 - 2√6

Hence, C.
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Re: (3^(1/2) - 2^(1/2))^2  [#permalink]

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New post 07 Aug 2018, 07:10
Bunuel wrote:
\((\sqrt3 - \sqrt2)^2\)


A. \(1 - 2 \sqrt{6}\)

B. \(1 - \sqrt{6}\)

C. \(5 - 2 \sqrt{6}\)

D. \(5 - 2 \sqrt{3}\)

E. \(1\)


\({\sqrt{3} - \sqrt{2}}^2\)

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

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

= \(5 - 2\sqrt{6}\), Answer must be (C)
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Re: (3^(1/2) - 2^(1/2))^2  [#permalink]

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New post 08 Aug 2018, 07:03
Bunuel wrote:
\((\sqrt3 - \sqrt2)^2\)


A. \(1 - 2 \sqrt{6}\)

B. \(1 - \sqrt{6}\)

C. \(5 - 2 \sqrt{6}\)

D. \(5 - 2 \sqrt{3}\)

E. \(1\)


I faced similar question type in my GMAT in first 10 questions.

\((\sqrt3 - \sqrt2)^2\)=\(3+2 -2 \sqrt{6}\)= \(5 -2 \sqrt{6}\)

Answer: C
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Re: (3^(1/2) - 2^(1/2))^2   [#permalink] 08 Aug 2018, 07:03
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