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# Inequalities trick

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16 Mar 2010, 10:11
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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.
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22 Oct 2010, 06:33
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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 [ 7.9 KiB | Viewed 40244 times ]

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.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Forum Moderator Joined: 20 Dec 2010 Posts: 2021 Followers: 156 Kudos [?]: 1522 [17] , given: 376 Re: Inequalities trick [#permalink] ### Show Tags 11 Mar 2011, 06:49 17 This post received KUDOS 9 This post was BOOKMARKED vjsharma25 wrote: VeritasPrepKarishma wrote: vjsharma25 wrote: How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors? Check out my post above for explanation. 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? I always struggle with this as well!!! There is a trick Bunuel suggested; (x+2)(x-1)(x-7) < 0 Here the roots are; -2,1,7 Arrange them in ascending order; -2,1,7; These are three points where the wave will alternate. The ranges are; x<-2 -2<x<1 1<x<7 x>7 Take a big value of x; say 1000; you see the inequality will be positive for that. (1000+2)(1000-1)(1000-7) is +ve. Thus the last range(x>7) is on the positive side. Graph is +ve after 7. Between 1 and 7-> -ve between -2 and 1-> +ve Before -2 -> -ve Since the inequality has the less than sign; consider only the -ve side of the graph; 1<x<7 or x<-2 is the complete range of x that satisfies the inequality. _________________ Manager Joined: 29 Sep 2008 Posts: 147 Followers: 3 Kudos [?]: 80 [12] , given: 1 Re: Inequalities trick [#permalink] ### Show Tags 22 Oct 2010, 11:45 12 This post received KUDOS 11 This post was BOOKMARKED if = sign is included with < then <= will be there in solution like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7 in case when factors are divided then the numerator will contain = sign like for (x + 2)(x - 1)/(x -4)(x - 7) < =0 the solution will be -2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite correct me if i am wrong Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6742 Location: Pune, India Followers: 1873 Kudos [?]: 11525 [8] , given: 219 Re: Inequalities trick [#permalink] ### Show Tags 11 Mar 2011, 19:57 8 This post received KUDOS Expert's post 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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10 Aug 2011, 17:01
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WoW - This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help.

1) CORE CONCEPT
@gurpreetsingh -
Suppose you have the inequality

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

Arrange the NUMBERS in ascending order from left to right. a<b<c<d
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.

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.

Note: Make sure that the factors are of the form (ax - b), not (b - ax)...

example -
(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.

2) Variation - ODD/EVEN POWER
@ulm/Karishma -
if we have even powers like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".
This will be same as (x-b)

We can ignore squares BUT SHOULD consider ODD powers
example -
2.a
(x-a)^3(x-b)<0 is the same as (x-a)(x-b) <0
2.b
(x - a)(x - b)/(x - c)(x - d) < 0 ==> (x - a)(x - b)(x-c)^-1(x-d)^-1 <0
is the same as (x - a)(x - b)(x - c)(x - d) < 0

3) Variation <= in FRACTION
@mrinal2100 -
if = sign is included with < then <= will be there in solution
like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7

BUT if it is a fraction the denominator in the solution will not have = SIGN

example -
3.a
(x + 2)(x - 1)/(x -4)(x - 7) < =0
the solution will be -2 <= x <= 1 or 4< x < 7
we cant make 4<=x<=7 as it will make the solution infinite

4) Variation - ROOTS
@Karishma -
As for roots, you have to keep in mind that given $$\sqrt{x}$$, x cannot be negative.

$$\sqrt{x}$$ < 10
implies 0 < $$\sqrt{x}$$ < 10
Squaring, 0 < x < 100
Root questions are specific. You have to be careful. If you have a particular question in mind, send it.

Refer - inequalities-and-roots-118619.html#p959939
Some more useful tips for ROOTS....I am too lazy to consolidate

<5> THESIS -
@gmat1220 -
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.

I will save this future references....

Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post

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Just came across this useful discussion.

VeritasPrepKarishma has given a very lucid explanation of how this “wavy line” method works.

I have noticed that there is still a little scope to take this discussion further. So here are my two cents on it.

I would like to highlight an important special case in the application of the Wavy Line Method

When there are multiple instances of the same root:

Try to solve the following inequality using the Wavy Line Method:

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

To know how you did, compare your wavy line with the correct one below.

Did you notice how this inequality differs from all the examples above?

Notice that two of the four terms had an integral power greater than 1.

How to draw the wavy line for such expressions?

Let me directly show you how the wavy line would look and then later on the rule behind drawing it.

Attachment:
File comment: Observe how the wave bounces back at x = 1.

bounce.png [ 10.4 KiB | Viewed 4156 times ]

Notice that the curve bounced down at the point x = 1. (At every other root, including x = 4 whose power was 3, it was simply passing through them.)

Can you figure out why the wavy line looks like this for this particular inequality?

(Hint: The wavy line for the inequality

$$(x-1)^{38}(x-2)^{57}(x-3)^{15}(x-4)^{27} < 0$$

Is also the same as above)

Come on! Give it a try.

If you got it right, you’ll see that there are essentially only two rules while drawing a wavy line. (Remember, we’ll refer the region above the number line as positive region and the region below the number line as negative region.)

How to draw the wavy line?

1. How to start: Start from the top right most portion. Be ready to alternate (or not alternate) the region of the wave based on how many times a point is root to the given expression.

2. How to alternate: In the given expression, if the power of a term is odd, then the wave simply passes through the corresponding point (root) into the other region (to –ve region if the wave is currently in the positive region and to the +ve region if the wave is currently in the negative region). However, if the power of a term is even, then the wave bounces back into the same region.

Now look back at the above expression and analyze your wavy line. Were you (intuitively) using the above mentioned rules while drawing your wavy line?

Solution

Once you get your wavy line right, solving an inequality becomes very easy. For instance, for the above inequality, since we need to look for the space where the above expression would be less than zero, look for the areas in the wavy line where the curve is below the number line.

So the correct solution set would simply be {3 < x < 4} U {{x < 2} – {1}}

In words, it is the Union of two regions region1 between x = 3 and x = 4 and region2 which is x < 2, excluding the point x = 1.

Food for Thought

Now, try to answer the following questions:

1. Why did we exclude the point x = 1 from the solution set of the last example? (Easy Question)
2. Why do the above mentioned rules (especially rule #2) work? What is/are the principle(s) working behind the curtains?

Foot Note: Although the post is meant to deal with inequality expressions containing multiple roots, the above rules to draw the wavy line are generic and are applicable in all cases.

- Krishna
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Last edited by EgmatQuantExpert on 10 Jan 2015, 21:41, edited 1 time in total.
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07 Nov 2015, 23:34
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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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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: 6742 Location: Pune, India Followers: 1873 Kudos [?]: 11525 [2] , given: 219 Re: Inequalities trick [#permalink] ### Show Tags 22 Oct 2010, 12:46 2 This post received KUDOS Expert's post 1 This post was BOOKMARKED mrinal2100: Kudos to you for excellent thinking! _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Stiv wrote:
VeritasPrepKarishma wrote:
mrinal2100: Kudos to you for excellent thinking!

Correct me if I'm wrong.
If the lower part of the equation$$\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\leq x \leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation.

x cannot be equal to 4 or 7 because if x = 4 or x = 7, the denominator will be 0 and the expression will not be defined.
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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: 6742 Location: Pune, India Followers: 1873 Kudos [?]: 11525 [2] , given: 219 Re: Inequalities trick [#permalink] ### Show Tags 24 Sep 2015, 23:07 2 This post received KUDOS Expert's post 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

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ttks10 wrote:
Can u plz explainn the backgoround of this & then the explanation.

Thanks

i m sorry i dont have any background for it, you just re-read it again and try to implement whenever you get such question and I will help you out in any issue.

sidhu4u wrote:
I have applied this trick and it seemed to be quite useful.

Nice to hear this....good luck.
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26 May 2011, 20:55
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chethanjs wrote:
mrinal2100 wrote:
if = sign is included with < then <= will be there in solution
like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7

in case when factors are divided then the numerator will contain = sign

like for (x + 2)(x - 1)/(x -4)(x - 7) < =0
the solution will be -2 <= x <= 1 or 4< x < 7
we cant make 4<=x<=7 as it will make the solution infinite

correct me if i am wrong

Can you please tell me why the solution gets infinite for 4<=x<=7 ?

Thanks.

(x -4)(x - 7) is in denominator.
Making x=4 or 7 would make the denominator 0 and the entire function undefined.

Thus, the range of x can't be either 4 or 7.
4<=x<=7 would be wrong.
4<x<7 is correct because now we removed "=" sign.
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Asher wrote:
gurpreetsingh wrote:
ulm wrote:
if we have smth like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".

yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a

This way of solving inequalities actually makes it soo much easier. Thanks gurpreetsingh and karishma

However, i am confused about how to solve inequalities such as: (x-a)^2(x-b) and also ones with root value.

When you have (x-a)^2(x-b) < 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.

Similarly for (x-a)^2(x-b) > 0, x > b

As for roots, you have to keep in mind that given $$\sqrt{x}$$, x cannot be negative.

$$\sqrt{x}$$ < 10
implies 0 < $$\sqrt{x}$$ < 10
Squaring, 0 < x < 100
Root questions are specific. You have to be careful. If you have a particular question in mind, send it.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 16 Feb 2012 Posts: 237 Concentration: Finance, Economics Followers: 7 Kudos [?]: 219 [1] , given: 121 Re: Inequalities trick [#permalink] ### Show Tags 22 Jul 2012, 03:03 1 This post received KUDOS VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation$$\frac {(x+2)(x-1)}{(x-4)(x-7)}$$ were $$4\leq x \leq 7$$, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. _________________ Kudos if you like the post! Failing to plan is planning to fail. CEO Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2797 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Followers: 219 Kudos [?]: 1474 [1] , given: 235 Re: Inequalities trick [#permalink] ### Show Tags 18 Oct 2012, 05:17 1 This post received KUDOS 1 This post was BOOKMARKED GMATBaumgartner wrote: gurpreetsingh wrote: ulm wrote: in addition: if we have smth like (x-a)^2(x-b) we don't need to change a sign when jump over "a". yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a Hi Gurpreet, Could you elaborate what exactly you meant here in highlighted text ? Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above. If the powers are even then the inequality won't be affected. eg if u have to find the range of values of x satisfying (x-a)^2 *(x-b)(x-c) >0 just use (x-b)*(x-c) >0 because x-a raised to the power 2 will not affect the inequality sign. But just make sure x=a is taken care off , as it would make the inequality zero. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6742 Location: Pune, India Followers: 1873 Kudos [?]: 11525 [1] , given: 219 Re: Inequalities trick [#permalink] ### Show Tags 18 Oct 2012, 10:27 1 This post received KUDOS Expert's post GMATBaumgartner wrote: Hi Gurpreet, Could you elaborate what exactly you meant here in highlighted text ? Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above. In addition, you can check out this post: http://www.veritasprep.com/blog/2012/07 ... s-part-ii/ I have discussed how to handle powers in it. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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02 Dec 2012, 07:25
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GMATGURU1 wrote:
Going ahead in this inequalities area of GMAT, can some one has the problem numbers of inequalities in OG 11 and OG 12?

Cheers,
Danny

Search for hundreds of question with solutions by tags: viewforumtags.php

DS questions on inequalities: search.php?search_id=tag&tag_id=184
PS questions on inequalities: search.php?search_id=tag&tag_id=189
Hardest DS inequality questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.
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Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
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20 Dec 2012, 21:04
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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] 20 Dec 2012, 21:04

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