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# Is |qp + q^2| > qp?

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Math Expert
Joined: 02 Sep 2009
Posts: 43785
Is |qp + q^2| > qp? [#permalink]

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13 Jan 2017, 08:07
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38% (01:35) correct 62% (01:27) wrong based on 94 sessions

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Is |qp + q^2| > qp?

(1) q > 0
(2) q/p > 1
[Reveal] Spoiler: OA

_________________
Math Expert
Joined: 02 Aug 2009
Posts: 5647
Is |qp + q^2| > qp? [#permalink]

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13 Jan 2017, 08:55
5
KUDOS
Expert's post
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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Joined: 02 Nov 2013
Posts: 95
Location: India
Re: Is |qp + q^2| > qp? [#permalink]

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13 Jan 2017, 09:57
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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Location: Malaysia
Is |qp + q^2| > qp? [#permalink]

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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.

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Last edited by hazelnut on 13 Jan 2017, 21:24, edited 4 times in total.
Math Expert
Joined: 02 Aug 2009
Posts: 5647
Re: Is |qp + q^2| > qp? [#permalink]

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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.

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..
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

BANGALORE/-

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Joined: 01 Jun 2015
Posts: 14
Re: Is |qp + q^2| > qp? [#permalink]

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

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

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....

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Joined: 14 Nov 2016
Posts: 1273
Location: Malaysia
Is |qp + q^2| > qp? [#permalink]

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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.

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 $$2p + q > 0$$ .
_________________

"Be challenged at EVERY MOMENT."

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Re: Is |qp + q^2| > qp? [#permalink]

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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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