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M13-23

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Manager
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P
Joined: 17 May 2017
Posts: 146

Kudos [?]: 117 [0], given: 226

GPA: 3
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Re: M13-23 [#permalink]

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New post 11 Sep 2017, 23:11
still confused if we take x= 5 then a can be both positive and negative so how is the first statement sufficient?

Kudos [?]: 117 [0], given: 226

Manager
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G
Joined: 01 Sep 2016
Posts: 208

Kudos [?]: 185 [0], given: 33

GMAT 1: 690 Q49 V35
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Re: M13-23 [#permalink]

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New post 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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we shall fight on the landing grounds,
we shall fight in the fields and in the streets,
we shall fight in the hills;
we shall never surrender!

Kudos [?]: 185 [0], given: 33

Intern
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B
Joined: 10 Mar 2017
Posts: 9

Kudos [?]: 0 [0], given: 41

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Re M13-23 [#permalink]

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New post 22 Sep 2017, 23:17
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.

Kudos [?]: 0 [0], given: 41

Re M13-23   [#permalink] 22 Sep 2017, 23:17

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