Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 24 Mar 2013
Posts: 28

Re: Inequalities trick [#permalink]
Show Tags
04 May 2015, 21:55
In addition to my previous post, I'd like to understand:
If the inequality is: f(x) = (x+a)(xb)^2(x+c)(xd)^3 < 0 We are still going to work with the same number line from left to right starting with "+" and factors a,b,c,d placed on it in ascending order? Since the number of factors remain an even number (4 total). Am I on the right track?



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7996
Location: Pune, India

Re: Inequalities trick [#permalink]
Show Tags
04 May 2015, 22:20
rohitd80 wrote: Hi Guys, I'm in the process of absorbing the fundamentals and neat tricks provided by the experts here. This inequality trick is phenomenal. Would it be safe to conclude that  for odd number of factors or roots, we start with "" (cosine waveform) and for even number of roots we start with "+" (sine wave)?
As explained by Gurpreet & Karishma: If f(x) has three factors then the graph will have  +  + If f(x) has four factors then the graph will have +  +  +
Similarly: If f(x) has two factors then the graph will have +  + If f(x) has seven factors then the graph will have  +  +  +  + ?
Thanks! Yes, you can look at it like that too. You should always have the positive sign at the extreme right region. Do remember though that the factors must be of the form (x  a), (x  b) etc and not of the form (a  x).
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7996
Location: Pune, India

Re: Inequalities trick [#permalink]
Show Tags
04 May 2015, 22:25
rohitd80 wrote: In addition to my previous post, I'd like to understand:
If the inequality is: f(x) = (x+a)(xb)^2(x+c)(xd)^3 < 0 We are still going to work with the same number line from left to right starting with "+" and factors a,b,c,d placed on it in ascending order? Since the number of factors remain an even number (4 total). Am I on the right track? No, you will ignore the factor with the even power since that factor will never be negative. You should go through all three of these posts to understand exactly why this happens: http://www.veritasprep.com/blog/2012/06 ... efactors/http://www.veritasprep.com/blog/2012/07 ... nsparti/http://www.veritasprep.com/blog/2012/07 ... spartii/
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



eGMAT Representative
Joined: 04 Jan 2015
Posts: 878

Re: Inequalities trick [#permalink]
Show Tags
04 May 2015, 22:42
rohitd80 wrote: Hi Guys, I'm in the process of absorbing the fundamentals and neat tricks provided by the experts here. This inequality trick is phenomenal. Would it be safe to conclude that  for odd number of factors or roots, we start with "" (cosine waveform) and for even number of roots we start with "+" (sine wave)?
As explained by Gurpreet & Karishma: If f(x) has three factors then the graph will have  +  + If f(x) has four factors then the graph will have +  +  +
Similarly: If f(x) has two factors then the graph will have +  + If f(x) has seven factors then the graph will have  +  +  +  + ?
Thanks! Dear rohitd80, That's a good observation! However, remember that your inference will hold true when: 1. the factors of f(x) are of the form (xa), (xb)... and so on. (i.e., not of the form (ax)(bx).... If they are of this form, then again you need to make proper adjustments based on the number of factors). AND2. Also, the powers of all the factors are odd. Even if one of the factors has an even power, your patterns (+  +, +++, etc.) will fail to hold true. Below is an example. If \(f(x) = (x  1)^2 * (x  3)\)and we need to find the regions where f(x) > 0, we first draw the wavy line starting from the top right corner. It crosses the "xaxis" at x=3 and enters the negative region. However, at x=1, since there are two roots at the same place, it bounces back into the negative region. As you can see, f(x) > 0 only when x > 3 Notice that if a student doesn't pay attention to the even power of roots, then he/she is prone to draw an incorrect wavy line and therefore arrive at an incorrect answer. You can refer to this post to understand how to draw wavy lines when multiple roots are present. inequalitiestrick9148280.html#p1465609Hope this helps. Regards, Krishna
_________________
 '4 out of Top 5' Instructors on gmatclub  70 point improvement guarantee  www.egmat.com



Intern
Joined: 24 Mar 2013
Posts: 28

Re: Inequalities trick [#permalink]
Show Tags
05 May 2015, 18:17
Thanks a lot Karishma and Krishna for your responses... The links really helped me to understand this technique to the T! I made it complicated by starting with my "" or "+" from the left when it's ideal to start with a "+" from the right most region (much more convenient than worrying about even or odd number of total factors).
Thanks!



Intern
Joined: 18 Aug 2012
Posts: 4

Inequalities trick [#permalink]
Show Tags
27 Aug 2015, 03:15
Can somebody please explain for me why this one doesnt work that way?
Which of the following is a value of x for which x^11  x^9 > 0 ?
A. 2 B. 1 C. 1/2 D. 1/2 E. 1
x11−x9>0 > x9(x2−1)>0 > (x+1)x9(x−1)>0 > roots are 1, 0 and 1 > −1<x<0 or x>1. Only C fits.
Answer: C.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7996
Location: Pune, India

Re: Inequalities trick [#permalink]
Show Tags
27 Aug 2015, 04:50
vihavivi wrote: Can somebody please explain for me why this one doesnt work that way?
Which of the following is a value of x for which x^11  x^9 > 0 ?
A. 2 B. 1 C. 1/2 D. 1/2 E. 1
x11−x9>0 > x9(x2−1)>0 > (x+1)x9(x−1)>0 > roots are 1, 0 and 1 > −1<x<0 or x>1. Only C fits.
Answer: C. But it does. Everything you have done is correct (ignoring the formatting issues).
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Intern
Joined: 18 Aug 2012
Posts: 4

Re: Inequalities trick [#permalink]
Show Tags
27 Aug 2015, 09:03
VeritasPrepKarishma wrote: vihavivi wrote: Can somebody please explain for me why this one doesnt work that way?
Which of the following is a value of x for which x^11  x^9 > 0 ?
A. 2 B. 1 C. 1/2 D. 1/2 E. 1
x11−x9>0 > x9(x2−1)>0 > (x+1)x9(x−1)>0 > roots are 1, 0 and 1 > −1<x<0 or x>1. Only C fits.
Answer: C. But it does. Everything you have done is correct (ignoring the formatting issues). I knew the answer makes sense, but if I apply the inequalities trick into this, with 3 roots 1, 0, and 1 and for the f(x) >0, shouldn't it be x <1 and 0 <x<1? Thanks



Manager
Joined: 04 May 2015
Posts: 73
Concentration: Strategy, Operations
WE: Operations (Military & Defense)

Re: Inequalities trick [#permalink]
Show Tags
27 Aug 2015, 17:33
2
This post received KUDOS
vihavivi wrote: VeritasPrepKarishma wrote: vihavivi wrote: Can somebody please explain for me why this one doesnt work that way?
Which of the following is a value of x for which x^11  x^9 > 0 ?
A. 2 B. 1 C. 1/2 D. 1/2 E. 1
x11−x9>0 > x9(x2−1)>0 > (x+1)x9(x−1)>0 > roots are 1, 0 and 1 > −1<x<0 or x>1. Only C fits.
Answer: C. But it does. Everything you have done is correct (ignoring the formatting issues). I knew the answer makes sense, but if I apply the inequalities trick into this, with 3 roots 1, 0, and 1 and for the f(x) >0, shouldn't it be x <1 and 0 <x<1? Thanks Hi Vihavivi, I think you have your inequalities curve flipped upside down. When you have the roots plotted on your line, it should always open up positive on the right and everything alternates from there. There was a previous post on the first page by fluke where he suggest plugging in large numbers to intuitively highlight what's going on. In your case being \(x^9(x+1)(x1)\) just choose say x=1000... In this case all three components will be positiive...\((1000^x)(10001)(1000+1)\) will surely yield a positive answer. Use this as a reminder that in these cases the inequalities line will open up positive to the right. Have a look at my freehand below. If you find this useful maybe throw me my first Kudos???
Attachments
File comment: Show's this equations "inequalities trick" line
IMG_20150828_3395.jpg [ 50.19 KiB  Viewed 1737 times ]
_________________
If you found my post useful, please consider throwing me a Kudos... Every bit helps



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7996
Location: Pune, India

Re: Inequalities trick [#permalink]
Show Tags
27 Aug 2015, 20:56
vihavivi wrote: VeritasPrepKarishma wrote: vihavivi wrote: Can somebody please explain for me why this one doesnt work that way?
Which of the following is a value of x for which x^11  x^9 > 0 ?
A. 2 B. 1 C. 1/2 D. 1/2 E. 1
x11−x9>0 > x9(x2−1)>0 > (x+1)x9(x−1)>0 > roots are 1, 0 and 1 > −1<x<0 or x>1. Only C fits.
Answer: C. But it does. Everything you have done is correct (ignoring the formatting issues). I knew the answer makes sense, but if I apply the inequalities trick into this, with 3 roots 1, 0, and 1 and for the f(x) >0, shouldn't it be x <1 and 0 <x<1? Thanks Remember, the rightmost region is always the positive region. That is how you start assigning  from the right. Of course, keep in mind that the inequality must be of the form (xa)*(xb)... and not of the form (a  x)*(xb)... (x+1)x^9(x−1)>0 > roots are 1, 0 and 1 So to the right of 1 will be positive and between 1 and 0, it will be positive. So x > 1 or 1 < x < 0
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Director
Status: I don't stop when I'm Tired,I stop when I'm done
Joined: 11 May 2014
Posts: 561
Location: Bangladesh
Concentration: Finance, Leadership
GPA: 2.81
WE: Business Development (Real Estate)

Re: Inequalities trick [#permalink]
Show Tags
28 Aug 2015, 23:40
1
This post was BOOKMARKED
LalaB wrote: VeritasPrepKarishma wrote: When you have (xa)^2(xb) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x  b) should be negative i.e. x  b < 0 or x < b. .
IMHO, it should be x<b and also x is not equal to a . so, one shouldn't totally ignore the squared term. We can ignore it, if the expression is <=0 correct me, if I am wrong my question is  do we always have a sequence of + and  from rightmost to the left side. I mean is it possible to have + and then + again ? We can ignore it,if the expression is =0
_________________
Md. Abdur Rakib
Please Press +1 Kudos,If it helps Sentence CorrectionCollection of Ron Purewal's "elliptical construction/analogies" for SC Challenges



Manager
Joined: 10 Sep 2013
Posts: 66
Concentration: General Management, Finance
GPA: 3.9

Re: Inequalities trick [#permalink]
Show Tags
24 Sep 2015, 06:59
Dear All, Is x a negative number? (1) 9x > 10x (2) is positive. GMAT OG claims A is sufficient. But, what if x is a fraction number? Such as 2/90. In this case 2/90 also satisfies ST 1 .. What is the exact solution for this?



Math Expert
Joined: 02 Sep 2009
Posts: 44383

Re: Inequalities trick [#permalink]
Show Tags
24 Sep 2015, 07:18



Manager
Joined: 10 Sep 2013
Posts: 66
Concentration: General Management, Finance
GPA: 3.9

Re: Inequalities trick [#permalink]
Show Tags
24 Sep 2015, 08:01
Thanks bunuel,
But couldnt find answer to my question there. Yet, I reasked the question.. )



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7996
Location: Pune, India

Re: Inequalities trick [#permalink]
Show Tags
24 Sep 2015, 23:07
mrslee wrote: Dear All, Is x a negative number? (1) 9x > 10x (2) is positive. GMAT OG claims A is sufficient. But, what if x is a fraction number? Such as 2/90. In this case 2/90 also satisfies ST 1 .. What is the exact solution for this? 2/90 does not satisfy statement 1. No positive value satisfies statement 1. 9*2/90 > 10*2/90 18/90 > 20/90  this is false
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Manager
Joined: 10 Sep 2013
Posts: 66
Concentration: General Management, Finance
GPA: 3.9

Re: Inequalities trick [#permalink]
Show Tags
25 Sep 2015, 02:38
Thanks dear,
I noticed it very late. Yeah, that was my unknown mistake.



Intern
Joined: 24 Sep 2015
Posts: 1

Re: Inequalities trick [#permalink]
Show Tags
07 Nov 2015, 01:12
if (k – 5)(k – 1)(k – 6) < 0 or (k – 5)(k – 1)(k – 6) > 0 what will be range for both? I have tried method suggested by gurpreetsingh, but got confused when to take ">" and "<". can anyone explain? It will be more helpful if anyone can upload an image of the solution. Thanks.



Current Student
Joined: 20 Jul 2015
Posts: 98
Location: India
Concentration: Marketing, General Management
GMAT 1: 720 Q49 V40 GMAT 2: 720 Q50 V38 GMAT 3: 760 Q50 V42
GPA: 3.8
WE: Engineering (NonProfit and Government)

Re: Inequalities trick [#permalink]
Show Tags
07 Nov 2015, 23:34
4
This post received KUDOS
Manali888 wrote: if (k – 5)(k – 1)(k – 6) < 0 or (k – 5)(k – 1)(k – 6) > 0 what will be range for both? I have tried method suggested by gurpreetsingh, but got confused when to take ">" and "<". can anyone explain? It will be more helpful if anyone can upload an image of the solution. Thanks. See the attached solution. Hope its clear!
Attachments
IMG_20151108_115517448[1].jpg [ 1.56 MiB  Viewed 1530 times ]



Intern
Joined: 11 Oct 2012
Posts: 41

Re: Inequalities trick [#permalink]
Show Tags
02 Jan 2016, 22:16
gmat1220 wrote: Once algebra teacher told me  signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane. Great thinking .. this really renders the inequalities so much clearer and easy to visualize...



Manager
Joined: 24 May 2013
Posts: 85

Re: Inequalities trick [#permalink]
Show Tags
06 Mar 2016, 07:35
1
This post received KUDOS
lets assume we have a,b,c,d and e in ascending order as follows:
(xa)(xb)(xc)(xd)<0
As number of factors are EVEN so for any x<a (i.e. the starting sign) we have (ve)*(ve)*(ive)*(ive) = +ve Then alternate signs will follow.
(xa)(xb)(xc)(xd)(xe)<0
As number of factors are ODD so for any x<a (i.e. the starting sign) we have (ve)*(ve)*(ive)*(ive)*(ve) = ve Then alternate signs will follow.
The same is true for for any number of factors in any form i.e. Numerator, Denominator, Combination of both.




Re: Inequalities trick
[#permalink]
06 Mar 2016, 07:35



Go to page
Previous
1 2 3 4 5 6 7 8
Next
[ 147 posts ]



