Inequalities trick : GMAT Quantitative Section - Page 4
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# Inequalities trick

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20 Dec 2012, 20:04
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VeritasPrepKarishma wrote:
The entire concept is based on positive/negative factors which means <0 or >0 is a must. If the question is not in this format, you need to bring it to this format by taking the constant to the left hand side.

e.g.
(x + 2)(x + 3) < 2
x^2 + 5x + 6 - 2 < 0
x^2 + 5x + 4 < 0
(x+4)(x+1) < 0

Now use the concept.

Yes this is probable but it might not be possible always to group them. So in case you are unsure just follow the number plugging approach. But most of the times this trick would be very handy.
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09 Feb 2013, 11:41
Can some one please show the application of this method using a Gmat problem.

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14 Jun 2013, 04:26
hey guys...yes the above method is very useful in determining ranges. = < > signs and saves a lot of time.

There are many more points associated with it.I guess only a few have been mentioned..

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09 Sep 2013, 13:42
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0

Just arrange them in order as shown in the picture and draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

Don't forget to arrange then in ascending order from left to right. a<b<c<d

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.

Hi Gurpreet,
Thanks for the wonderful method.
I am trying to understand it so that i can apply it in tests.
Can you help me in applying this method to the below expression to find range of x.
x^3 – 4x^5 < 0?

I am getting the roots as -1/2, 0, 1/2 and when i plot them using this method, putting + in the rightmost region, I am not getting correct result. Not sure where i am going wrong. Can you pls help.
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09 Sep 2013, 22:35
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karannanda wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

Suppose you have the inequality

f(x) = (x-a)(x-b)(x-c)(x-d) < 0

Just arrange them in order as shown in the picture and draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

Don't forget to arrange then in ascending order from left to right. a<b<c<d

So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.

For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.

Hi Gurpreet,
Thanks for the wonderful method.
I am trying to understand it so that i can apply it in tests.
Can you help me in applying this method to the below expression to find range of x.
x^3 – 4x^5 < 0?

I am getting the roots as -1/2, 0, 1/2 and when i plot them using this method, putting + in the rightmost region, I am not getting correct result. Not sure where i am going wrong. Can you pls help.

Before you apply the method, ensure that the factors are of the form (x - a)(x - b) etc

$$x^3 - 4x^5 < 0$$

$$x^3 ( 1 - 4x^2) < 0$$

$$x^3(1 - 2x) (1 + 2x) < 0$$

$$4x^3(x - 1/2)(x + 1/2) > 0$$ (Notice the flipped sign. We multiplied both sides by -1 to convert 1/2 - x to x - 1/2)

Now the transition points are 0, -1/2 and 1/2 so put + in the rightmost region.
The solution will be x > 1/2 or -1/2 < x< 0.

Check out these posts discussing such complications:
http://www.veritasprep.com/blog/2012/06 ... e-factors/
http://www.veritasprep.com/blog/2012/07 ... ns-part-i/
http://www.veritasprep.com/blog/2012/07 ... s-part-ii/
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7121 Location: Pune, India Followers: 2133 Kudos [?]: 13638 [0], given: 222 Re: Inequalities trick [#permalink] ### Show Tags 15 Sep 2013, 20:12 VeritasPrepKarishma wrote: vjsharma25 wrote: I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0 Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2. Now let me add another factor: (x+8)(x+2)(x-1)(x-7) Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes. So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region. Note: Make sure that the factors are of the form (ax - b), not (b - ax)... e.g. (x+2)(x-1)(7 - x)<0 Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1') Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-' Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0. Responding to a pm: Quote: i have a doubt in the highlighted region. U have said that it will be always +ve but in a bunel post he has asked to substitute the extreme values and if the f(x) is -ve then the right most of the inequality will be -ve. Please clarify me on this. Different people use different methods of solving problems. Both of us are correct. But you cannot mix up the methods. When you follow one, you have to follow that through and through. When I say that the rightmost region will always be positive, it is after I make appropriate changes. Right below the highlighted portion, notice the note given: Note: Make sure that the factors are of the form (ax - b), not (b - ax)... I convert all factors to (ax - b) form. Now the rightmost region is positive by default. Bunuel prefers to keep the factors as it is and check for the rightmost region. What you would like to follow is your personal choice. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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11 Nov 2013, 05:14
Can you please explain the above mentioned concept in relation to the following question?
Is a > O?
(1) a^3 - 0 < 0
(2) 1- a^2 > 0

Can you please explain the scenario when (x-a)(x-B)(x-C)(x-d)>0?
Sorry, but finding it difficult to understand.
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11 Nov 2013, 06:43
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Responding to a pm:

Is a > O?
(1) a^3 - a < 0
(2) 1- a^2 > 0

(1) a^3 - a < 0
a(a+1)(a - 1) < 0

Points of transition are -1, 0 and 1. Make the wave.
The expression is negative when 0 < a < 1 or a < -1.
a could be positive or negative. Not sufficient.

(2) 1- a^2 > 0
a^2 - 1 < 0 (multiplied the inequality by -1 which flipped the sign)
(a-1)(a+1) < 0

Points of transition are -1 and 1. Make the wave.
The expression is negative when -1 < a < 1.
a could be positive or negative. Not sufficient.

Using both together, we see that only 0 < a < 1 is possible for both inequalities to hold. In this case a must be positive. Sufficient.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 36545 Followers: 7076 Kudos [?]: 93078 [1] , given: 10542 Re: Inequalities trick [#permalink] ### Show Tags 11 Nov 2013, 06:51 1 This post received KUDOS Expert's post anujpadia wrote: Can you please explain the above mentioned concept in relation to the following question? Is a > O? (1) a^3 - 0 < 0 (2) 1- a^2 > 0 Can you please explain the scenario when (x-a)(x-B)(x-C)(x-d)>0? Sorry, but finding it difficult to understand. Check alternative solutions here: is-a-0-1-a-3-a-0-2-1-a-86749.html Hope this helps. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7121 Location: Pune, India Followers: 2133 Kudos [?]: 13638 [0], given: 222 Re: Inequalities trick [#permalink] ### Show Tags 11 Nov 2013, 06:51 anujpadia wrote: Can you please explain the above mentioned concept in relation to the following question? Is a > O? (1) a^3 - 0 < 0 (2) 1- a^2 > 0 Can you please explain the scenario when (x-a)(x-B)(x-C)(x-d)>0? Sorry, but finding it difficult to understand. If you have doubts in any particular steps given above, feel free to ask. Also, check out my posts on this method: http://www.veritasprep.com/blog/2012/06 ... e-factors/ http://www.veritasprep.com/blog/2012/07 ... ns-part-i/ http://www.veritasprep.com/blog/2012/07 ... s-part-ii/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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11 Nov 2013, 22:45
Thank you Karishma and Bunuel!!
Its crystal clear now!!
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22 Apr 2014, 23:42
VeritasPrepKarishma wrote:
Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.

If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.
e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.

Attachment:
doc.jpg

This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.

When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.

When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.

Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7

It should be obvious that it will also work in cases where factors are divided.
e.g. (x - a)(x - b)/(x - c)(x - d) < 0
(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.

Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.

Hi Karishma,

I have a query. I have following question

x^3 - 4x^5 < 0

I can define this as (1+2x).x^3(1-2x). now I have roots -1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of -1/2 and 0 I will get value as 0.

So How I will define them in graph and what range I will consider for this inequality.

Thanks
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23 Apr 2014, 03:43
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PathFinder007 wrote:

I have a query. I have following question

x^3 - 4x^5 < 0

I can define this as (1+2x).x^3(1-2x). now I have roots -1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of -1/2 and 0 I will get value as 0.

So How I will define them in graph and what range I will consider for this inequality.

Thanks

The factors must be of the form (x - a)(x - b) .... etc

x^3 - 4x^5 < 0
x^3 * (1 - 4x^2) < 0
x^3 * (1 - 2x) * (1 + 2x) < 0
x^3 * (2x - 1) * (2x + 1) > 0 (Note the sign flip because 1-2x was changed to 2x - 1)
x^3 * 2(x - 1/2) *2(x + 1/2) > 0

So transition points are 0, 1/2 and -1/2.

____________ - 1/2 _____ 0 ______1/2 _________

This is what it looks like on the number line.
The rightmost region is positive. We want the positive regions in the inequality.

So the desired range of x is given by x > 1/2 or -1/2 < x< 0

For more on this method, check these posts:

http://www.veritasprep.com/blog/2012/06 ... e-factors/
http://www.veritasprep.com/blog/2012/07 ... ns-part-i/
http://www.veritasprep.com/blog/2012/07 ... s-part-ii/

The links will give you the theory behind this method in detail.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 27 Jan 2014 Posts: 12 Followers: 0 Kudos [?]: 1 [0], given: 19 Re: Inequalities trick [#permalink] ### Show Tags 13 Jun 2014, 04:26 VeritasPrepKarishma wrote: Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain. If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line. e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown. Attachment: doc.jpg This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive. When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive. When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section. Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7 It should be obvious that it will also work in cases where factors are divided. e.g. (x - a)(x - b)/(x - c)(x - d) < 0 (x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above. Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion. Thanks a lot for the wonderful explanation... kudos +1 Manager Joined: 17 Jul 2013 Posts: 110 Followers: 0 Kudos [?]: 6 [0], given: 67 Inequalities trick [#permalink] ### Show Tags 03 Jul 2014, 23:36 VeritasPrepKarishma wrote: Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain. If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line. e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown. Attachment: doc.jpg This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive. When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive. When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section. Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7 It should be obvious that it will also work in cases where factors are divided. e.g. (x - a)(x - b)/(x - c)(x - d) < 0 (x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above. Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion. can we do modulus questions with the help of this method? Also please suggest when we decide to take the + ve curve values or -ve values. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7121 Location: Pune, India Followers: 2133 Kudos [?]: 13638 [0], given: 222 Re: Inequalities trick [#permalink] ### Show Tags 06 Jul 2014, 19:58 Expert's post 1 This post was BOOKMARKED GmatDestroyer2013 wrote: VeritasPrepKarishma wrote: Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain. If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line. e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown. Attachment: doc.jpg This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive. When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive. When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section. Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7 It should be obvious that it will also work in cases where factors are divided. e.g. (x - a)(x - b)/(x - c)(x - d) < 0 (x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above. Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion. can we do modulus questions with the help of this method? Also please suggest when we decide to take the + ve curve values or -ve values. This method is only for inequalities with factors such that they are of the form (x-a). You should check out the links of the 3 posts I have given in my post above. They discuss the theory in detail and tell you why the regions are positive or negative. Also, they tell you how to handle complications such as factors of the form (a - x) or (ax + b) etc. When you have all factors of the form (x-a) (or when you convert them to this form), the rightmost region is always positive. To solve absolute value questions you have a few different methods that you can use. http://www.veritasprep.com/blog/2014/06 ... -the-gmat/ http://www.veritasprep.com/blog/2011/01 ... edore-did/ http://www.veritasprep.com/blog/2011/01 ... h-to-mods/ http://www.veritasprep.com/blog/2011/01 ... s-part-ii/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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22 Sep 2014, 21:31
VeritasPrepKarishma wrote:
Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.

If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.
e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.

Attachment:
doc.jpg

This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.

When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.

When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.

Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7

It should be obvious that it will also work in cases where factors are divided.
e.g. (x - a)(x - b)/(x - c)(x - d) < 0
(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.

Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.

Hi Karishma

Can you explain the logic when the expression has a denominator or when factors are divided. I did read the solution given below, but can you please explain the logic?
Or if you can share a link in which you have explained expressions with denominators, I would be grateful
Thanks
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22 Sep 2014, 21:48
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alphonsa wrote:

Hi Karishma

Can you explain the logic when the expression has a denominator or when factors are divided. I did read the solution given below, but can you please explain the logic?
Or if you can share a link in which you have explained expressions with denominators, I would be grateful
Thanks

Here: http://www.veritasprep.com/blog/2012/07 ... s-part-ii/
Last question here handles factors in the denominator.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Current Student Joined: 25 Feb 2014 Posts: 49 GMAT 1: 730 Q50 V39 GPA: 3.83 Followers: 0 Kudos [?]: 4 [0], given: 12 Re: Inequalities trick [#permalink] ### Show Tags 16 Oct 2014, 11:10 gurpreetsingh wrote: I learnt this trick while I was in school and yesterday while solving one question I recalled. Its good if you guys use it 1-2 times to get used to it. Suppose you have the inequality f(x) = (x-a)(x-b)(x-c)(x-d) < 0 Just arrange them in order as shown in the picture and draw curve starting from + from right. now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful. Don't forget to arrange then in ascending order from left to right. a<b<c<d So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d) and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x) If f(x) has three factors then the graph will have - + - + If f(x) has four factors then the graph will have + - + - + If you can not figure out how and why, just remember it. Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis. For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively. Hi Could you please explain this question to me with this logic? The explanation at (range-for-variable-x-in-a-given-inequality-109468.html ) starts with - + - + logic instead of + - + - as advised above.... Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7121 Location: Pune, India Followers: 2133 Kudos [?]: 13638 [0], given: 222 Re: Inequalities trick [#permalink] ### Show Tags 17 Oct 2014, 09:21 Expert's post 1 This post was BOOKMARKED mayankpant wrote: gurpreetsingh wrote: I learnt this trick while I was in school and yesterday while solving one question I recalled. Its good if you guys use it 1-2 times to get used to it. Suppose you have the inequality f(x) = (x-a)(x-b)(x-c)(x-d) < 0 Just arrange them in order as shown in the picture and draw curve starting from + from right. now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful. Don't forget to arrange then in ascending order from left to right. a<b<c<d So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d) and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x) If f(x) has three factors then the graph will have - + - + If f(x) has four factors then the graph will have + - + - + If you can not figure out how and why, just remember it. Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis. For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively. Hi Could you please explain this question to me with this logic? The explanation at (range-for-variable-x-in-a-given-inequality-109468.html ) starts with - + - + logic instead of + - + - as advised above.... There are two methods to go about it. Either decide whether the rightmost region is positive or negative depending on your factors or ALWAYS take rightmost region to be positive and adjust your factors accordingly i.e. bring all factors to the form (x-a). The second method is discussed in detail in the links given here: inequalities-trick-91482-60.html#p1358533 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Inequalities trick   [#permalink] 17 Oct 2014, 09:21

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