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Is |qp + q^2| > qp?

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Is |qp + q^2| > qp? [#permalink]

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New post 13 Jan 2017, 08:07
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Is |qp + q^2| > qp? [#permalink]

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Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0
(2) q/p > 1



Hi,

On the first look itself, the equation will hold true for every value except when left side is 0.
And the left side will be 0, only when q is 0..
So is |qp+q^2|>qp actually MEANS is q NOT EQUAL to 0

Let's see the statements..
1) q>0..
Ans is YES, q is not equal to 0..
Suff

2) q/P>1..
Again ans is YES..
If q was 0, q/P=0..
Suff

D
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Re: Is |qp + q^2| > qp? [#permalink]

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New post 13 Jan 2017, 09:57
My answer is D.
Taking any value of q or p, absolute value will be greater than the product of the two value here.
Option 1: As q is positive, either value of p negative or positive will true for |qp + q^2| > qp
Option 2: As q>p, Either positive value of both, both negative value or q positive and p negative results in |qp + q^2| > qp.
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Is |qp + q^2| > qp? [#permalink]

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New post 13 Jan 2017, 19:29
Bunuel wrote:
Is \(|qp + q^{2}| > qp\) ?

(1) \(q > 0\)
(2) \(\frac{q}{p}> 1\)


Square both side \(|qp + q^{2}| > qp\)
\((qp+q^2)^{2}\) - \((qp)^{2}> 0\)
\((qp)^{2}+2q^{2}(qp) + (q^{2})^{2}-(qp)^{2}> 0\)
\(2q^{2}(qp) + (q^2)^{2}> 0\)
\(2(qp)+q^{2} > 0\)
\(q (2p+q) > 0\)

(1) \(q > 0\), q must be greater than zero in order for \(q (2p+q) > 0\) to hold. Sufficient

(2) \(\frac{2(qp)+q^{2}}{qp} > 0\), \(2 + \frac{q}{p} > 0\) , \(\frac{q}{p}> 1\) . Therefore, it is sufficient to solve the equation.

Answer : D
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Last edited by hazelnut on 13 Jan 2017, 21:24, edited 4 times in total.
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Re: Is |qp + q^2| > qp? [#permalink]

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New post 13 Jan 2017, 20:52
ziyuenlau wrote:
Bunuel wrote:
Is \(|qp + q^{2}| > qp\) ?

(1) \(q > 0\)
(2) \(\frac{q}{p}> 1\)


Square both side \(|qp + q^{2}| > qp\)
\((qp+q^2)^{2}\) - \((qp)^{2}> 0\)
\((qp)^{2}+2q^{2}(qp) + (q^{2})^{2}-(qp)^{2}> 0\)
\(2q^{2}(qp) + (q^2)^{2}> 0\)
\(2(qp)+q^{2} > 0\)

(1) \(q > 0\), q must be greater than zero in order for \(2(qp)+q^{2}> 0\) to hold. Sufficient

(2) \(\frac{2(qp)+q^{2}}{qp} > 0\), \(2 + \frac{q}{p} > 0\) . \(\frac{q}{p}> 1\) . Therefore, it is sufficient to solve the equation.

Answer : D


Hi
Relook into the coloured portion above..
Say q is 2 and p is -8..
2pq+q^2=2*2*-8+2^2= -32+4=-28, this is not GREATER than 0..
On other hand if both q and p are negative, 2pq+q^2>0..
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Re: Is |qp + q^2| > qp? [#permalink]

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New post 13 Jan 2017, 20:54
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Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0
(2) q/p > 1


Solution: Answer is D..!!

Take any value of q&p except zero , the situation would hold true..
Irrespective of whether q>p or p>q..Anything would hold true..
Example Let q=2 And p=3
|6+4|=6
Anything would hold true except the condition which i mentioned in the BOLD....

So answer is D...
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Is |qp + q^2| > qp? [#permalink]

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New post 13 Jan 2017, 21:27
chetan2u wrote:
ziyuenlau wrote:
Bunuel wrote:
Is \(|qp + q^{2}| > qp\) ?

(1) \(q > 0\)
(2) \(\frac{q}{p}> 1\)


Square both side \(|qp + q^{2}| > qp\)
\((qp+q^2)^{2}\) - \((qp)^{2}> 0\)
\((qp)^{2}+2q^{2}(qp) + (q^{2})^{2}-(qp)^{2}> 0\)
\(2q^{2}(qp) + (q^2)^{2}> 0\)
\(2(qp)+q^{2} > 0\)

(1) \(q > 0\), q must be greater than zero in order for \(2(qp)+q^{2}> 0\) to hold. Sufficient

(2) \(\frac{2(qp)+q^{2}}{qp} > 0\), \(2 + \frac{q}{p} > 0\) . \(\frac{q}{p}> 1\) . Therefore, it is sufficient to solve the equation.

Answer : D


Hi
Relook into the coloured portion above..
Say q is 2 and p is -8..
2pq+q^2=2*2*-8+2^2= -32+4=-28, this is not GREATER than 0..
On other hand if both q and p are negative, 2pq+q^2>0..


chetan2u, I have factorised the equation. \(2 (qp)+q^{2} > 0\) , \(q (2p+q) > 0\) . Therefore, \(q > 0\) and \(2p + q > 0\) .
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“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”

"Each stage of the journey is crucial to attaining new heights of knowledge."

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Re: Is |qp + q^2| > qp? [#permalink]

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New post 10 Feb 2018, 06:08
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Re: Is |qp + q^2| > qp?   [#permalink] 10 Feb 2018, 06:08
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