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Is pq < 4 ? (1) p^2 + q^2 < 8 (2) p - q < 1

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Is pq < 4 ? (1) p^2 + q^2 < 8 (2) p - q < 1  [#permalink]

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New post 11 Nov 2019, 04:46
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Is pq < 4 ? (1) p^2 + q^2 < 8 (2) p - q < 1  [#permalink]

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New post 11 Nov 2019, 07:41
Is \(pq < 4\) ?


(1) \(p^2 + q^2 < 8\)
. \(p^2 + q^2=(p-q)^2+2pq\) so \(p^2 + q^2 < 8\) MEANS \((p-q)^2+2pq<8\) ..
The least value of \((p-q)^2\) is 0 when p=q, so MAXIMUM value of 2pq will be \(0+2pq<8...pq<4.\)..TRUE

(2) \(p - q < 1\)
If p is 8 and q is 7.5...pq>4If p=2 and q is 1.5...pq<4

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Re: Is pq < 4 ? (1) p^2 + q^2 < 8 (2) p - q < 1  [#permalink]

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New post 11 Nov 2019, 08:58
Bunuel wrote:
Is \(pq < 4\) ?


(1) \(p^2 + q^2 < 8\)

(2) \(p - q < 1\)




statement 1: sufficient : we know p and q mat take take any value so p can root 7.8 and q can be root .111111111 and make total less than 8
now root 7.8 *root .111111 will give value <4 similarly if we check negatives p and q each has to be less than -2 as putting -2 will give a total more than 8. pq has to less than 4

statement 2 :p<1+q no boundaries so insufficient

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Re: Is pq < 4 ? (1) p^2 + q^2 < 8 (2) p - q < 1   [#permalink] 11 Nov 2019, 08:58
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