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Is pq > 0? (1) x^2 - px + q = 0 has real and positive roots

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Is pq > 0? (1) x^2 - px + q = 0 has real and positive roots  [#permalink]

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New post 07 Apr 2019, 08:46
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Question Stats:

36% (01:49) correct 64% (02:31) wrong based on 44 sessions

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Is pq > 0?


(1) x^2 - px + q = 0 has real and positive roots

(2) p(p - q) > 0

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Is pq > 0? (1) x^2 - px + q = 0 has real and positive roots  [#permalink]

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New post 07 Jun 2019, 03:07
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Asad wrote:
Is pq > 0?


(1) x^2 - px + q = 0 has real and positive roots

(2) p(p - q) > 0


let \(x_1\) and \(x_2\) be the roots of the quadratic , given both are positive.
also sum of roots of a quadratic\(\>\) \(x_1+x_2=-(\frac{-p}{1}) \Rightarrow x_1+x_2=p\)
Product of roots\(\>\)\(x_1*x_2=q\)
since \(x_1\) and \(x_2\) both are positive hence p and q are also positive.Hence pq>0

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Re: Is pq > 0? (1) x^2 - px + q = 0 has real and positive roots  [#permalink]

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New post 07 Apr 2019, 10:20
1.
p²-4q>O
p²>4q
but cant say about q's sign.
so, insufficient

2. p²-pq>O
p²>pq
p² will always be positive.
hence sufficient.

IMO B

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Re: Is pq > 0? (1) x^2 - px + q = 0 has real and positive roots   [#permalink] 07 Apr 2019, 10:20
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