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(2 + 7^1/2) (4 − 7^1/2) (−2x) =

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Sajjad1994
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(2 + 7^1/2) (4 − 7^1/2) (−2x) = [#permalink] New post   09 Jul 2019, 03:58
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A
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C
D
E

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Question Stats:

95% (01:26) correct 5% (01:51) wrong based on 76 sessions
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\((2 + \sqrt{7}) (4 − \sqrt{7}) (−2x) =\)

(A) \(78x − 4x \sqrt{7}\)

(B) \(\sqrt{7}x\)

(C) \(−2x − 4x \sqrt{7}\)

(D) \(–2x\)

(E) \(4x \sqrt{7}\)
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C
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KarishmaB
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(2 + 7^1/2) (4 − 7^1/2) (−2x) = [#permalink] New post   09 Jul 2019, 06:33
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SajjadAhmad wrote:
\((2 + \sqrt{7}) (4 − \sqrt{7}) (−2x) =\)

(A) \(78x − 4x \sqrt{7}\)

(B) \(\sqrt{7}x\)

(C) \(−2x − 4x \sqrt{7}\)

(D) \(–2x\)

(E) \(4x \sqrt{7}\)


\((2 + \sqrt{7}) (4 − \sqrt{7}) (−2x) = (1 + 2\sqrt{7})(-2x)\)

\(= -2x -4x\sqrt{7}\)
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Kinshook
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(2 + 7^1/2) (4 − 7^1/2) (−2x) = [#permalink] New post   Updated on: 09 Jul 2019, 08:01
SajjadAhmad wrote:
\((2 + \sqrt{7}) (4 − \sqrt{7}) (−2x) =\)

(A) \(78x − 4x \sqrt{7}\)

(B) \(\sqrt{7}x\)

(C) \(−2x − 4x \sqrt{7}\)

(D) \(–2x\)

(E) \(4x \sqrt{7}\)


\((2 + \sqrt{7}) (4 − \sqrt{7}) (−2x) =\)
\(=(8 +4 \sqrt{7} -2\sqrt{7}-7)(-2x)\)
\(=(1+2\sqrt{7})(-2x)\)
\(=-2x-4x\sqrt{7}\)

IMO C

Originally posted by Kinshook on 09 Jul 2019, 07:46.
Last edited by Kinshook on 09 Jul 2019, 08:01, edited 1 time in total.
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Re: (2 + 7^1/2) (4 − 7^1/2) (−2x) = [#permalink] New post   09 Jul 2019, 07:59
Kinshook There is a typo in your solution. The answer is C, which is what you got, not D.
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Kinshook
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Re: (2 + 7^1/2) (4 − 7^1/2) (−2x) = [#permalink] New post   09 Jul 2019, 08:02
bebs wrote:
Kinshook There is a typo in your solution. The answer is C, which is what you got, not D.


OK. Thanks for noticing.
Corrected
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Re: (2 + 7^1/2) (4 − 7^1/2) (−2x) = [#permalink] New post   28 Jul 2019, 10:22
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SajjadAhmad wrote:
\((2 + \sqrt{7}) (4 − \sqrt{7}) (−2x) =\)

(A) \(78x − 4x \sqrt{7}\)

(B) \(\sqrt{7}x\)

(C) \(−2x − 4x \sqrt{7}\)

(D) \(–2x\)

(E) \(4x \sqrt{7}\)


Simplifying, we have:

(8 - 2√7 + 4√7 - 7)(-2x)

(1 + 2√7)(-2x)

-2x -4x√7

Answer: C
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Re: (2 + 7^1/2) (4 − 7^1/2) (−2x) = [#permalink]
28 Jul 2019, 10:22
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