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sqrt3 + sqrt 2
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10 Mar 2021, 06:42
10 Mar 2021, 06:42
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Question Stats:
15% (00:51) correct
84% (01:34) wrong based on 19 sessions
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\((\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})-(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})\)
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Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Re: sqrt3 + sqrt 2
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10 Mar 2021, 07:42
10 Mar 2021, 07:42
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(a-b)(a+b) = \(a^2 + b^2\)
\((\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})-(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})\)
(3-2) - (7-5) = 1 - 2 = -1
\((\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})-(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})\)
(3-2) - (7-5) = 1 - 2 = -1
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Re: sqrt3 + sqrt 2
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10 Mar 2021, 07:47
10 Mar 2021, 07:47
Carcass wrote:
\((\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})-(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})\)
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\((\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})-(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})\)
Using property: \(a^2 - b^2 = (a + b)(a - b)\)
\([(\sqrt{3})^2 - (\sqrt{2})^2] - [(\sqrt{7})^2 - (\sqrt{5})^2]\)
\((3 - 2) - (7 - 5)\)
\(1 - 2 = -1\)
