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(3+sqrt5)(3+sqrt5)
[#permalink]
07 May 2024, 01:45
07 May 2024, 01:45
Expert Reply
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Question Stats:
83% (00:38) correct
16% (00:45) wrong based on 18 sessions
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Quantity A |
Quantity B |
\([3+\sqrt{5}][3+\sqrt{5}] \) |
20 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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Re: (3+sqrt5)(3+sqrt5)
[#permalink]
08 May 2024, 10:19
08 May 2024, 10:19
1
Using the "FOIL" method of multiplying,
\([3+\sqrt{5}][3+\sqrt{5}]\)
\(= (3*3) + (3*\sqrt{5}) + (3*\sqrt{5}) + (\sqrt{5}*\sqrt{5})\)
\(= 9 + 6\sqrt{5} + 5\)
\(= 14 + 6\sqrt{5}\)
Now we can compare \(14 + 6\sqrt{5}\) to 20.
Simplify to compare \(6\sqrt{5}\) to 6.
Since \(6\sqrt{5} > 6\), Quantity A is greater.
\([3+\sqrt{5}][3+\sqrt{5}]\)
\(= (3*3) + (3*\sqrt{5}) + (3*\sqrt{5}) + (\sqrt{5}*\sqrt{5})\)
\(= 9 + 6\sqrt{5} + 5\)
\(= 14 + 6\sqrt{5}\)
Now we can compare \(14 + 6\sqrt{5}\) to 20.
Simplify to compare \(6\sqrt{5}\) to 6.
Since \(6\sqrt{5} > 6\), Quantity A is greater.
gmatclubot
Re: (3+sqrt5)(3+sqrt5) [#permalink]
08 May 2024, 10:19
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