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# M13-23

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Manager
Joined: 17 May 2017
Posts: 119
Schools: ESSEC '20
GPA: 3

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11 Sep 2017, 23:11
1
still confused if we take x= 5 then a can be both positive and negative so how is the first statement sufficient?
Manager
Joined: 01 Sep 2016
Posts: 185
GMAT 1: 690 Q49 V35

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17 Sep 2017, 04:56
Bunuel : Can you help me with a simple question:

Why haven't you used discriminant method to solve Option B as well. If you solve it using D method, you will get A>0, what am I missing here please?
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Intern
Joined: 10 Mar 2017
Posts: 4
GMAT 1: 660 Q49 V32

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22 Sep 2017, 23:17
1
I think this is a high-quality question and I agree with explanation. Hi Bunuel,
Is the following approach for statement-1 wrong?
x^2-2x+a is positive for all values of x,
Hence it should hold its ground even when x=0.
So, when x=0,
x^2-2x+a is reduced to a
Hence a has to be greater than 0.
Manager
Joined: 26 Dec 2017
Posts: 148

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13 Jan 2018, 04:44
4
1
Hi Bunnel ,

I too have same doubt that why imaginary roots discriminant method is not used in S2 if used you will get a>0 but also actually it is not true as a=0 also satisfies the equation which is the reason for insufficiency.

So m main doubt is Can I know whether discriminant method is suffice(S1) or not (S2) for checking whether statements suffice or not
Intern
Joined: 02 Feb 2018
Posts: 31

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01 Dec 2018, 08:53
Addition to (2): A can also be negative when x =0
Senior Manager
Joined: 03 Sep 2018
Posts: 252
Location: Netherlands
GMAT 1: 710 Q48 V40
GMAT 2: 780 Q50 V49
GMAT 3: 760 Q49 V44
GPA: 4

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01 Nov 2019, 04:38
I am wondering whether we can do the following.

$$x^2-2x+a>0 \implies a>x(2-x) \implies a>0$$ OR $$a>2$$

Is that legal Bunuel?
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Re: M13-23   [#permalink] 01 Nov 2019, 04:38

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# M13-23

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