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Re: Inequalities trick [#permalink]
11 Mar 2011, 18:57

4

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. _________________

Re: Inequalities trick [#permalink]
11 Mar 2011, 20:48

VeritasPrepKarishma wrote:

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.

I think this is the key point,which has cleared my doubt about this theory That's why I was wondering how to assign the first sign on the number line graph.

Thanks Karishma for showing patience to resolve this for me.

Re: Inequalities trick [#permalink]
02 Jun 2011, 10:38

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?

if the equation being (x+2)(x-1)(x-7)<0 then soln will be for 1<x<7 bcoz in this range outcome of this equation is negative which is required _________________

Re: Inequalities trick [#permalink]
08 Aug 2011, 10:59

Expert's post

Asher 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

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. _________________

Re: Inequalities trick [#permalink]
08 Aug 2011, 21:22

Quote:

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

Thanks Karishma for the explanation. Hope you wouldn't mind clarifying a few more doubts.

Firstly, in the above case, since x>b could we say that everything with be positive.

would the graph look something like this: positive.. b..postive.. a.. positive

On the other hand if (x-a)^2(x-b) < 0, x < b, (x-a)^2 would be positive and for (x-b) if x<b the the left side would be negative.

would the graph look something like this: negative.. b..postive.. a.. postive

Am i right?

If it is not too much of a trouble, could you please show the graphical representation.

problems with \sqrt{x}.. this is all i could find (googled actually ):

1. √(-x+4) ≤ √(x) 2. x^\sqrt{x} =< (\sqrt{x})^x

{P.S.: i tried to insert the graphical representation that i came up with, but i am a bit technically challenged in this area it seems} _________________

Re: Inequalities trick [#permalink]
09 Aug 2011, 02:28

Expert's post

Asher wrote:

Firstly, in the above case, since x>b could we say that everything with be positive.

would the graph look something like this: positive.. b..postive.. a.. positive

On the other hand if (x-a)^2(x-b) < 0, x < b, (x-a)^2 would be positive and for (x-b) if x<b the the left side would be negative.

would the graph look something like this: negative.. b..postive.. a.. postive

So when you have (x - a)^2(x - b) < 0, you ignore x = a and just plot x = b. It is positive in the rightmost region and negative on the left. So the graph looks like this:

negative ... b ... positive

Am i right?

If it is not too much of a trouble, could you please show the graphical representation.

problems with \sqrt{x}.. this is all i could find (googled actually ):

1. √(-x+4) ≤ √(x) 2. x^\sqrt{x} =< (\sqrt{x})^x

{P.S.: i tried to insert the graphical representation that i came up with, but i am a bit technically challenged in this area it seems}

Squared terms are ignored. You do not put them in the graph. They are always positive so they do not change the sign of the expression. e.g. (x-4)^2(x - 9)(x+11) < 0 We do not plot x = 4 here, only x = -11 and x = 9. We start with the rightmost section as positive. So it looks something like this:

positive... -11 ... negative ... 9 ... positive

Since we need the region where x is negative, we get -11 < x < 9. Basically, the squared term is like a positive number in that it doesn't affect the sign of the expression.

I would be happy to solve inequalities questions related to roots but please put them in a separate post and pm the link to me. That way, everybody can try them. _________________

Re: Inequalities trick [#permalink]
09 Aug 2011, 07:16

VeritasPrepKarishma wrote:

Asher wrote:

Firstly, in the above case, since x>b could we say that everything with be positive.

would the graph look something like this: positive.. b..postive.. a.. positive

On the other hand if (x-a)^2(x-b) < 0, x < b, (x-a)^2 would be positive and for (x-b) if x<b the the left side would be negative.

would the graph look something like this: negative.. b..postive.. a.. postive

So when you have (x - a)^2(x - b) < 0, you ignore x = a and just plot x = b. It is positive in the rightmost region and negative on the left. So the graph looks like this:

negative ... b ... positive

Am i right?

If it is not too much of a trouble, could you please show the graphical representation.

problems with \sqrt{x}.. this is all i could find (googled actually ):

1. √(-x+4) ≤ √(x) 2. x^\sqrt{x} =< (\sqrt{x})^x

{P.S.: i tried to insert the graphical representation that i came up with, but i am a bit technically challenged in this area it seems}

Squared terms are ignored. You do not put them in the graph. They are always positive so they do not change the sign of the expression. e.g. (x-4)^2(x - 9)(x+11) < 0 We do not plot x = 4 here, only x = -11 and x = 9. We start with the rightmost section as positive. So it looks something like this:

positive... -11 ... negative ... 9 ... positive

Since we need the region where x is negative, we get -11 < x < 9. Basically, the squared term is like a positive number in that it doesn't affect the sign of the expression.

I would be happy to solve inequalities questions related to roots but please put them in a separate post and pm the link to me. That way, everybody can try them.

Thanks a ton Karishma, i really appreciate it. I will post pm you after posting the roots questions. _________________

Re: Inequalities trick [#permalink]
10 Aug 2011, 06:00

hey , can u please tel me the solution for this ques

a car dealership sells only sports cars and luxury cars and has atleast some of each type of car in stock at all times.if exactly 1/7 of sports car and 1/2 of luxury cars have sunroofs and there are exactly 42 cars on the lot.what is the smallest number of cars that could have roofs?

Re: Inequalities trick [#permalink]
10 Aug 2011, 16:01

4

This post received KUDOS

3

This post was BOOKMARKED

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 _________________

Re: Inequalities trick [#permalink]
11 Aug 2011, 19:33

sushantarora wrote:

hey , can u please tel me the solution for this ques

a car dealership sells only sports cars and luxury cars and has atleast some of each type of car in stock at all times.if exactly 1/7 of sports car and 1/2 of luxury cars have sunroofs and there are exactly 42 cars on the lot.what is the smallest number of cars that could have roofs?

ans -11

Interesting question, but i guess it would be better to post it as a new topic. This would ensure that more ppl see it and ans it.

But than again that's just my suggestion _________________

Re: Inequalities trick [#permalink]
11 Aug 2011, 21:57

Expert's post

sushantarora wrote:

hey , can u please tel me the solution for this ques

a car dealership sells only sports cars and luxury cars and has atleast some of each type of car in stock at all times.if exactly 1/7 of sports car and 1/2 of luxury cars have sunroofs and there are exactly 42 cars on the lot.what is the smallest number of cars that could have roofs?

ans -11

Please put questions in new posts. Put it in the same post only if it is totally related or a variation of the question that we are discussing.

Now for the solution:

There are 42 cars on the lot. 1/7 of sports cars and 1/2 of luxury cars have sunroofs. This means that 1/7 of number of sports cars and 1/2 of number of luxury cars should be integers (You cannot have 1.5 cars with sunroofs, right?) We want to minimize the sunroofs. Since 1/2 of luxury cars have sunroofs and only 1/7 of sports cars have them, it will be good to have fewer luxury cars and more sports cars. Best would be to have all sports cars. But, the question says there are some of each kind at any time. So let's say there are 2 luxury cars (since 1/2 of them should be an integer value). But 1/7 of 40 (the rest of the cars are sports cars) is not an integer number. Let's instead look for the multiple of 7 that is less than 42. The multiple of 7 that is less than 42 is 35. So we could have 35 sports cars. But then, 1/2 of 7 (since 42 - 35 = 7 are luxury cars) is not an integer. The next smaller multiple of 7 is 28. This works. 1/2 of 14 (since 42 - 28 = 14 are luxury cars) is 7. So we can have 14 luxury cars and 28 sports cars. That is the maximum number of sports cars that we can have. 1/7 of 28 sports cars = 4 cars have sunroofs 1/2 of 14 luxury cars = 7 cars have sunroofs

So at least 11 cars will have sunroofs. _________________

Re: Inequalities trick [#permalink]
06 May 2012, 23:15

Edvento 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 gurpreetsingh,

This is really great!!

Regards,

Shouvik.

By the way,

For further reference, this method is called the 'wavy curve method'.

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. _________________

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. _________________

Just for my reference, say if the equation was (x+2)(x-1)/(x-4)(x-7) and the question was for what values of x is this expression >0, then the roots will be -2,1,4,7 and by placing on the number line and making the extreme right as positive...

----(-2)----(1)----(4)---(7)----then x>7, 1<x<4 and x<-2...Please confirm.. However, is say it was >=0 then x>7, 1<=x,4 and x<=-2; given that the denominator cannot be zero. Please confirm

Just for my reference, say if the equation was (x+2)(x-1)/(x-4)(x-7) and the question was for what values of x is this expression >0, then the roots will be -2,1,4,7 and by placing on the number line and making the extreme right as positive...

----(-2)----(1)----(4)---(7)----then x>7, 1<x<4 and x<-2...Please confirm.. However, is say it was >=0 then x>7, 1<=x<4 and x<=-2; given that the denominator cannot be zero. Please confirm

Yes, you are right in both the cases.

Also, if you want to verify that the range you have got is correct, just plug in some values to see. Put x = 0, the expression is -ve. Put x = 2, the expression is positive. _________________

For my Cambridge essay I have to write down by short and long term career objectives as a part of the personal statement. Easy enough I said, done it...