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D01-13

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D01-13  [#permalink]

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New post 09 May 2018, 19:23
I thought this was a great question, but I got it wrong.
I managed to bring the equation to x^2(x^2-x-6) = 0, but I applied Viete's Theorem (sum of roots is -b/a) instead of just factoring again. I guess the formula only applies under strict circumstances and didn't work here because of the initial requirement that the roots be even.
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New post 29 Jun 2018, 01:23
I think this is a poor-quality question and I don't agree with the explanation. the solution of minus two is correct , you can have the 4th root of (X=-2) it is a 4th root of 16 ,and 16 it is a positive number which can use for the root.
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New post 29 Jun 2018, 01:30
Eyaltau wrote:
I think this is a poor-quality question and I don't agree with the explanation. the solution of minus two is correct , you can have the 4th root of (X=-2) it is a 4th root of 16 ,and 16 it is a positive number which can use for the root.


You are wrong. Your doubt is addressed MANY times on previous pages:

\(\sqrt{}\) denotes a function. Mathematically the square root function cannot give negative result. When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{25}=5\), NOT +5 or -5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.

OFFICIAL GUIDE:
\(\sqrt{n}\) denotes the positive number whose square is n.

Piece of advice, people here, experts/tutors, are here to help students, not to deceive them. So, if you see the same thing repeated many times by experts/tutors you should now that it's true.

P.S. Generally I would suggest to brush-up fundamental before attempting hard questions.
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New post 06 Aug 2018, 09:27
I think this is a high-quality question and I agree with explanation.
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New post 20 Nov 2018, 12:11
Hi team,

I think Answer is wrong as
-2 can also be the solution . x3+6x2>=0
if we give x=-2 ,this value is 16 which is positive.
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New post 20 Nov 2018, 21:35
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New post 23 Nov 2018, 19:15
Hello,

In the absolute values concept, sqr root(x^2) = |x|

Which means, x can be +ve or -ve and the modulus will make it +ve.

Is x on the left side assumed to be |x| in the problem we have to eliminate -2 as an option?
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Re D01-13  [#permalink]

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New post 18 Feb 2019, 09:12
I think this is a high-quality question and I agree with explanation.
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Re D01-13   [#permalink] 18 Feb 2019, 09:12

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