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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.

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 46505 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.
_________________

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.
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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.
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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.... Please add anything that you feel will help. Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post _________________

Labor cost for typing this post >= Labor cost for pushing the Kudos Button http://gmatclub.com/forum/kudos-what-are-they-and-why-we-have-them-94812.html

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 5021 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 – region1between 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.

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 984 times ]

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.

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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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 \(x^9(x+1)(x-1)\) just choose say x=1000... In this case all three components will be positiive...\((1000^x)(1000-1)(1000+1)\) 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 1197 times ]

_________________

If you found my post useful, please consider throwing me a Kudos... Every bit helps

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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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.

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

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.

could someone please explain.

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.
_________________

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.
_________________

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.
_________________

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