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Re: Inequalities trick [#permalink]
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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?

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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).
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Re: Inequalities trick [#permalink]
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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/
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Re: Inequalities trick [#permalink]
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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
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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!

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Inequalities trick [#permalink]
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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.

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Re: Inequalities trick [#permalink]
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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).
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Re: Inequalities trick [#permalink]
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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

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Re: Inequalities trick [#permalink]
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27 Aug 2015, 17:33
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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???
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Re: Inequalities trick [#permalink]
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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
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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
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Re: Inequalities trick [#permalink]
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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?

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24 Sep 2015, 07:18

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24 Sep 2015, 08:01
Thanks bunuel,
But couldnt find answer to my question there. Yet, I reasked the question.. )

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

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Re: Inequalities trick [#permalink]
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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.

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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!
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Re: Inequalities trick [#permalink]
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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...

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Re: Inequalities trick [#permalink]
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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.

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