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\sqrt5+\sqrt7
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15 May 2018, 15:56
15 May 2018, 15:56
1
Expert Reply
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Question Stats:
84% (00:29) correct
16% (00:18) wrong based on 50 sessions
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\((\sqrt{5}+\sqrt{7})^2=\)
A. 12
B. \(12 + 2\sqrt{3}\)
C. \(12 + 4\sqrt{3}\)
D. \(12 + \sqrt{35}\)
E. \(12 + 2\sqrt{35}\)
A. 12
B. \(12 + 2\sqrt{3}\)
C. \(12 + 4\sqrt{3}\)
D. \(12 + \sqrt{35}\)
E. \(12 + 2\sqrt{35}\)
Retired Moderator
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Re: exponential problem
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13 May 2019, 04:48
13 May 2019, 04:48
1
sandy wrote:
\((\sqrt{5}+\sqrt{7})^2=\)
A. 12
B. \(12 + 2\sqrt{3}\)
C. \(12 + 4\sqrt{3}\)
D. \(12 + \sqrt{35}\)
E. \(12 + 2\sqrt{35}\)
A. 12
B. \(12 + 2\sqrt{3}\)
C. \(12 + 4\sqrt{3}\)
D. \(12 + \sqrt{35}\)
E. \(12 + 2\sqrt{35}\)
Use FOIL to expand the product
Note: \((\sqrt{x})(\sqrt{y})=\sqrt{xy}\)
So, \((\sqrt{5}+\sqrt{7})^2=(\sqrt{5}+\sqrt{7})(\sqrt{5}+\sqrt{7})\)
\(=\sqrt{25}+\sqrt{35}+\sqrt{35}+\sqrt{49}\)
\(=5+2\sqrt{35}+7\)
\(=12+2\sqrt{35}\)
Answer: E
Cheers,
Brent
gmatclubot
Re: exponential problem [#permalink]
13 May 2019, 04:48
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