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M13-23

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Re: M13-23  [#permalink]

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New post 11 May 2017, 05:00
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Re: M13-23  [#permalink]

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New post 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?
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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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Re M13-23  [#permalink]

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New post 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.
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Re: M13-23  [#permalink]

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New post 01 Dec 2018, 08:53
Addition to (2): A can also be negative when x =0
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Re: M13-23  [#permalink]

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New post 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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