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\sqrta^2+39=8
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28 Sep 2019, 09:53
28 Sep 2019, 09:53
Expert Reply
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Question Stats:
29% (00:27) correct
70% (00:25) wrong based on 24 sessions
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\(\sqrt{a^2+39}=8\)
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.
Kudos for the right answer and explanation
Quantity A |
Quantity B |
a |
4 |
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.
Kudos for the right answer and explanation
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: \sqrta^2+39=8
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29 Sep 2019, 05:20
29 Sep 2019, 05:20
1
Carcass wrote:
\(\sqrt{a^2+39}=8\)
Quantity A |
Quantity B |
a |
4 |
Take: \(\sqrt{a^2+39}=8\)
Square both sides to get: \(a^2+39=64\)
Subtract 39 from both sides to get: \(a^2=25\)
So, EITHER \(a=5\) OR \(a=-5\)
If \(a=5\), then Quantity A is greater
If \(a=-5\), then Quantity B is greater
Answer: D
Cheers,
Brent
gmatclubot
Re: \sqrta^2+39=8 [#permalink]
29 Sep 2019, 05:20
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