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sqrt of x^4+6x^2+9
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Updated on: 25 May 2020, 16:08
Updated on: 25 May 2020, 16:08
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68% (00:40) correct
31% (00:39) wrong based on 57 sessions
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Quantity A |
Quantity B |
\(\sqrt{{x^4 + 6x^2 + 9}}\) |
\(x^2+3\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
I used in solving this question choosing number strategy, my question if the \(\sqrt{16}\) will it equal 4 or -4, or only 4? How does the GRE deal with that ?
Source : Greg mat
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Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Square root
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25 May 2020, 05:17
25 May 2020, 05:17
1
Asmakan wrote:
Quantity A |
Quantity B |
\(\sqrt{{x^4 + 6x^2 + 9}}\) |
\(x^2+3\) |
I used in solving this question choosing number strategy, my question if the \(\sqrt{16}\) will it equal 4 or -4, or only 4? How does the GRE deal with that ?
Source : Greg mat
16 has two square roots (4 and -4), but the square root notation tells us to take only the POSITIVE square root.
So, \(\sqrt{16}=4\), \(\sqrt{49}=7\), \(\sqrt{100}=10\), etc
Alternatively, if \(x^2 = 16\) (i.e., no square root notation), then either \(x=4\) or \(x=-4\)
Likewise, if \(x^2 = 81\) (i.e., no square root notation), then either \(x=9\) or \(x=-9\)
Does that help?
Cheers,
Brent
Re: Square root
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26 May 2020, 03:08
26 May 2020, 03:08
GreenlightTestPrep wrote:
Asmakan wrote:
Quantity A |
Quantity B |
\(\sqrt{{x^4 + 6x^2 + 9}}\) |
\(x^2+3\) |
I used in solving this question choosing number strategy, my question if the \(\sqrt{16}\) will it equal 4 or -4, or only 4? How does the GRE deal with that?
Source: Greg mat
16 has two square roots (4 and -4), but the square root notation tells us to take only the POSITIVE square root.
So, \(\sqrt{16}=4\), \(\sqrt{49}=7\), \(\sqrt{100}=10\), etc
Alternatively, if \(x^2 = 16\) (i.e., no square root notation), then either \(x=4\) or \(x=-4\)
Likewise, if \(x^2 = 81\) (i.e., no square root notation), then either \(x=9\) or \(x=-9\)
Does that help?
Cheers,
Brent
Yes, but need more clarification, now the \(\sqrt{16}\) , 16 could be a result of 4 and -4 eventhough the notation is not -ve.
