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

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Math Expert
Joined: 02 Sep 2009
Posts: 56275
(3^(1/2) - 2^(1/2))^2  [#permalink]

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07 Aug 2018, 04:14
00:00

Difficulty:

15% (low)

Question Stats:

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

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

A. $$1 - 2 \sqrt{6}$$

B. $$1 - \sqrt{6}$$

C. $$5 - 2 \sqrt{6}$$

D. $$5 - 2 \sqrt{3}$$

E. $$1$$

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

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

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