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Inequalities trick

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Re: Inequalities trick [#permalink] New post 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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Re: Inequalities trick [#permalink] New post 09 Feb 2013, 11:41
Can some one please show the application of this method using a Gmat problem.

Thanks in advance,
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Re: Inequalities trick [#permalink] New post 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..

This method is actually called WAVY CURVE METHOD .For more info you can google it.
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Re: Inequalities trick [#permalink] New post 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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Re: Inequalities trick [#permalink] New post 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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Re: Inequalities trick [#permalink] New post 15 Sep 2013, 20:12
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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.

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Re: Inequalities trick [#permalink] New post 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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Re: Inequalities trick [#permalink] New post 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.
Answer (C)

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Re: Inequalities trick [#permalink] New post 11 Nov 2013, 06:51
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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.

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Re: Inequalities trick [#permalink] New post 11 Nov 2013, 06:51
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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/

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Re: Inequalities trick [#permalink] New post 11 Nov 2013, 22:45
Thank you Karishma and Bunuel!!
Its crystal clear now!!
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Re: Inequalities trick [#permalink] New post 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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Re: Inequalities trick [#permalink] New post 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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Re: Inequalities trick [#permalink] New post 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
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Inequalities trick [#permalink] New post 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.
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Re: Inequalities trick [#permalink] New post 06 Jul 2014, 19:58
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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/

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Re: Inequalities trick   [#permalink] 06 Jul 2014, 19:58
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