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# D01-13

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Intern
Joined: 23 Apr 2018
Posts: 15

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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.
Intern
Joined: 28 Jun 2018
Posts: 1

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 53831

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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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Intern
Joined: 13 Jul 2018
Posts: 3

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06 Aug 2018, 09:27
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 21 Aug 2018
Posts: 2

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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.
Math Expert
Joined: 02 Sep 2009
Posts: 53831

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20 Nov 2018, 21:35
razatr wrote:
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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Intern
Joined: 02 Jun 2018
Posts: 2

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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?
Manager
Joined: 22 Jun 2017
Posts: 178
Location: Argentina
Schools: HBS, Stanford, Wharton

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18 Feb 2019, 09:12
I think this is a high-quality question and I agree with explanation.
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The HARDER you work, the LUCKIER you get.
Re D01-13   [#permalink] 18 Feb 2019, 09:12

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