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[4(2)^1/2]^2 =

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Bunuel
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[4(2)^1/2]^2 = [#permalink] New post   17 Mar 2022, 04:30
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\((4\sqrt{2})^2 =\)

A. 8\(\sqrt{2}\)

B. 8

C. 16\(\sqrt{2}\)

D. 32

E. 64
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D
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Charli08
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[4(2)^1/2]^2 = [#permalink] New post   Updated on: 20 Mar 2022, 11:04
(4√2)^2 = 4^2 * √2^2 = 16 * 2 = 32

==> D

Originally posted by Charli08 on 19 Mar 2022, 09:04.
Last edited by Charli08 on 20 Mar 2022, 11:04, edited 1 time in total.
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Re: [4(2)^1/2]^2 = [#permalink] New post   19 Mar 2022, 09:22
Bunuel wrote:
\((4\sqrt{2})^2 =\)

A. 8\(\sqrt{2}\)

B. 8

C. 16\(\sqrt{2}\)

D. 32

E. 64


(4/\sqrt{2})^2 = [(2^2)/\sqrt{2}]^2 = 2^[(2*2)-(0.5*2)] = 2^3 = 8

IMO B
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Re: [4(2)^1/2]^2 = [#permalink] New post   19 Mar 2022, 10:09
Bunuel wrote:
\((4\sqrt{2})^2 =\)

A. 8\(\sqrt{2}\)

B. 8

C. 16\(\sqrt{2}\)

D. 32

E. 64

\((4\sqrt{2})^2 =\)

Or, \(4^2*\sqrt{2}^2\)

Or, \(16*2=32\), Answer must be (D)
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GyanaDash
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Re: [4(2)^1/2]^2 = [#permalink] New post   19 Mar 2022, 10:37
GyanaDash wrote:
Bunuel wrote:
\((4\sqrt{2})^2 =\)

A. 8\(\sqrt{2}\)

B. 8

C. 16\(\sqrt{2}\)

D. 32

E. 64


(4/\sqrt{2})^2 = [(2^2)/\sqrt{2}]^2 = 2^[(2*2)-(0.5*2)] = 2^3 = 8

IMO B



Don't understand how I read this question wrong. Apologies.

Its (4*sqrt(s))^2, so its 4^2 * (sqrt(2))^2 = 16*2 = 32.

So, asnwer is D.
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Re: [4(2)^1/2]^2 = [#permalink] New post   20 Mar 2022, 01:57
Charli08 wrote:
(4√2)^2 = (2^2 * 2^1/2)^2 = (2^3/2)^2 = 2^3 = 8

==> B


2 + 1/2 = 5/2 not 3/2
So it will be 2^5 = 32

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Re: [4(2)^1/2]^2 = [#permalink]
20 Mar 2022, 01:57
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