Is |qp + q^2| > qp?

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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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.

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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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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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...

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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Hi Chetan,

Have a doubt, if we solve the inequality after squaring just to remove the modulus in any case,
we end with,

q^3(p+2q) > 0; this means q > 0 and p+2q >0 || p+2q <0 and q <0; if this is true, then how can q > 0 sufficient ? What am I missing here?

Hi....
You have gone wrong by squaring both sides..
You can square an INEQUALITY, only if both sides are POSITIVE..
Here you do not know if pq>0
Example..
|-3+2|>-3....1>-3
Square both sides..
1^2>(-3)^2......1>9?????

So be sure both sides are POSITIVE
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Hello chetan2u,

Please assist me with this question.

My approach:

Is |qp + q^2| > qp?

if qp + q^2 >0 then
=> qp + q^2 > qp
=> q^2>0
=> |q|>0; basically q not = 0

if qp + q^2 <0 then
=> qp+q^2<-qp
=> 2qp+q^2<0
=> q(2p+q)<0

so we need to find is "q not = 0" and "q(2p+q)<0"

stmt 1: q>0 -->
then is 2p+q<0
=>is q<-2p
but we can't say that since we don't know sign of p; insufficient

stmt 2: q/p > 1 --> clearly insufficient since we don't know signs of p and q

stmt 1 and 2: q>0 therefore p must be +ve
from stmt 1: is q<-2p ----> since p and q are +ve so "-2p" would be -ve
Hence, the above inequality fails ----------> sufficient
My answer=C

Please correct the fault in my analysis

Regards

Hi..

On your approach..
1) you have complicated the equation.
2) the coloured portion is incomplete..q(2p+q)<0
So if Q>0, 2p+Q<0..2p<-q.... Possible p is some NEGATIVE term
If q<0, 2p+q>0.....2p>-q..... P is some POSITIVE term possible
But the equation is NOT possible when q=0.. so here too, we have to find if Q=0, otherwise q can take any value..

From both cases ONLY one thing is to be checked-" if q=0?"
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i have used the following approach-
but i think i am missing something here----
|qp+q^2| > qp
|q(p+q)|>qp

assume q is positive
|p+q|>p
as q is positive , hence p will be positive

so for the question - it will negate if any answer choice proves p is negative.

assume q is negative
|p+q|<p
as q is negative, p is positive.

stmt 1..
q>0, 1 st case....

stmt 2...
q/p>1
hence both q and p have same signs...
as p is always +ve, q will be +ve,
hence case 1 again...

Hi chetan2u,

I think the approach that gmatexam439 used (above in the chain) is correct and necessary. I am saying this by the following general understanding in modulus that:
If |x|>a,
then either x>a, if x>0
or x<-a, if x<0

Now, the question is, is |qp + q^2| > qp?
This transcends into asking:
QI: Is qp + q^2 > qp, if (qp + q) >0?
OR
QII: Is qp + q^2 < -qp, if (qp + q) <0?

However, as my logic goes, if the stmt1 or stmt2 manages to answer "either of the 2 questions", then we are done. This is essentially because, there is a "OR" between the subordinate questions Q1 and Q2.
Here, both statements easily answer QI. So, we need not proceed to QII, which I agree complicates the question and hence the answer choice is D.

But, if the main question would have been in the format:
Is |x|<a?, which would mean asking:
Is x<a ? if x>0
AND
Is x>-a ? if x<0
In such a case, I think we will have to consider both statements equally and make sure that we are able to arrive at an answer for both the subordinate questions using the 2 statements provided with that question.

Please let me know if my analysis is correct.

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 \(q\neq{0}\)

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

2)\(\frac{q}{p}>1\)..
Again ans is YES..
If q was 0, q/p would be 0..
Suff

D
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Given
is |qp + q^2| > qp ?

can be written as, |q(p + q)| > qp


Statement 1: q > 0

Plug in numbers,
q = 1, p = -1, we get LHS = 0 & RHS = -1....we get YES
q = 1, p = 0, we get LHS = 1, RHS = 0...we get YES
q = 1, p = 1, we get LHS = 2, RHS = 1...we get YES

Statement 1 is Sufficient.

Statement 2: q/p > 1, hence q & p have the same sign.

Plug in numbers
q = 2, p = 1, we get LHS = 6, RHS = 2...we get YES
q = -2, p = -1, we get LHS = 6, RHS = 2, we get YES

Statement 2 is Sufficient.


Answer D.



Thanks,
GyM
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Left side is always positive. q^2 is making left side greater than RHS when p,q>0. when p<0, RHS is negative and LHS positive.

So, D

Hi Chetan,

Could you please explain why we are not considering value of q=0;

It is not given anywhere in the original question

I think you are referring to statement II
(2) q/p > 1
Now if we consider q=0..\(\frac{q}{p}=\frac{0}{p}=0\)
But it is given that \(\frac{q}{p}>1\), so how can 0>1
this means \(q\neq{0}\)

the equation q/p>1 itself means that q is not 0

hope it helps
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can you explain how this can be done by opening the absolute sign?

one way is -- qp+q^2-qp>o , i.e. is q not equal to 0


second way is -qp-q^2> qp
simplified to 2qp + q^2<0

with this second way, how can we say with statement 1 we can prove sufficient

Bunuel chetan2u

Let's take the second statement..
2pq+p^2<0
p^2 is surely positive so 2pq<0
This means - are p AND q of different sign
Statement II tells us both are of same sign..
Sufficient
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