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Re: Inequalities trick [#permalink]
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13 Mar 2016, 01:28
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 (xa)(xb)(xc)(xd) < 0, we can draw the points a, b, c and d on the number line. e.g. Given (x+2)(x1)(x7)(x4) < 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. Please explain how will it be different when you have <= or >=

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Re: Inequalities trick [#permalink]
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14 Mar 2016, 01:32
RohitPrakash88 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 (xa)(xb)(xc)(xd) < 0, we can draw the points a, b, c and d on the number line. e.g. Given (x+2)(x1)(x7)(x4) < 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. Please explain how will it be different when you have <= or >= \((x+2)(x1)(x7)(x4) < 0\) gives \(2 < x < 1\) or \(4 < x < 7\) \((x+2)(x1)(x7)(x4) <= 0\) gives \(2 <= x <= 1\) or \(4 <= x <= 7\) x can be equal to each transition point such that the expression will take the value 0 in that case. The range is pretty much the same as above. With division, things are a bit different. \(\frac{(x+2)(x1)}{(x7)(x4)} < 0\) gives \(2 < x < 1\) or \(4< x < 7\) \(\frac{(x+2)(x1)}{(x7)(x4)} <= 0\) gives \(2 <= x <= 1\) or \(4< x < 7\) Note that here x cannot be 4 or 7 because then you will get a 0 in the denominator which is not acceptable.
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Re: Inequalities trick [#permalink]
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28 Jul 2016, 17:48
VeritasPrepKarishma wrote: maddyboiler wrote: Excellent post. Till now we have all seen problems in the format f(x) < 0 where f(x) is written in its factors for (xa)(xb)...
what if we have something like f(x) < k "k is a constant" (xa)(xb)(xc) < k How do we solve these kind of questions? 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. Dear Ma'am! Please try to respond to my pm... really needed that help, if possible then revert on forum so that others can benefit for it too. thanks

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28 Jul 2016, 22:01
Celestial09 wrote: Dear Ma'am! Please try to respond to my pm... really needed that help, if possible then revert on forum so that others can benefit for it too. thanks Celestial09, When you send me a pm, I automatically get a notification. You don't need to separately request me to check the pm on the forum. Check your messages  I have responded to your query.
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Re: Inequalities trick [#permalink]
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30 Sep 2016, 21:49
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 12 times to get used to it.
Suppose you have the inequality
f(x) = (xa)(xb)(xc)(xd) < 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. Thank you very much my friend, that is definitely bringing back some old memories. I just got a problem wrong on a practice problem and had I remembered this I wouldn't have! I will commit this to memory, thank you again

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Re: Inequalities trick [#permalink]
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07 Oct 2016, 11:15
As per this trick, for the inequality x^3 < x^2 we will get the range as 0<x<1. But the correct answer is "Any nonzero number less than 1". Please could someone help me understand what am I missing here? ~Kudos are free. Be generous!

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Re: Inequalities trick [#permalink]
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08 Oct 2016, 17:47
Pretty quick way of solving.

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Re: Inequalities trick [#permalink]
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24 Oct 2016, 20:02
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 (xa)(xb)(xc)(xd) < 0, we can draw the points a, b, c and d on the number line. e.g. Given (x+2)(x1)(x7)(x4) < 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. Karishma, Pls tell me it is always be that signs will change like ++ or it is possible the case when it will be the same repeats. for example, +++ or ++ thank you

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Re: Inequalities trick [#permalink]
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alesia257 wrote: Karishma, Pls tell me it is always be that signs will change like ++ or it is possible the case when it will be the same repeats. for example, +++ or ++
thank you Signs will always change in the pattern discussed above. Understand the reason why this is so  check out these posts: https://www.veritasprep.com/blog/2012/0 ... efactors/https://www.veritasprep.com/blog/2012/0 ... nsparti/https://www.veritasprep.com/blog/2012/0 ... spartii/
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25 Oct 2016, 05:18
VeritasPrepKarishma wrote: alesia257 wrote: Karishma, Pls tell me it is always be that signs will change like ++ or it is possible the case when it will be the same repeats. for example, +++ or ++
thank you Signs will always change in the pattern discussed above. Understand the reason why this is so  check out these posts: https://www.veritasprep.com/blog/2012/0 ... efactors/https://www.veritasprep.com/blog/2012/0 ... nsparti/https://www.veritasprep.com/blog/2012/0 ... spartii/Thank you for your answer. So I can put plus in the right segment and then just put ++ without trying numbers,arent I ?

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alesia257 wrote: VeritasPrepKarishma wrote: alesia257 wrote: Karishma, Pls tell me it is always be that signs will change like ++ or it is possible the case when it will be the same repeats. for example, +++ or ++
thank you Signs will always change in the pattern discussed above. Understand the reason why this is so  check out these posts: https://www.veritasprep.com/blog/2012/0 ... efactors/https://www.veritasprep.com/blog/2012/0 ... nsparti/https://www.veritasprep.com/blog/2012/0 ... spartii/Thank you for your answer. So I can put plus in the right segment and then just put ++ without trying numbers,arent I ? Yes, absolutely but after you bring the factors in the form (ax +/ b)
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VeritasPrepKarishma wrote: 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. \((x4)^2(x  9)(x+11) < 0\) 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. Responding to a pm: Quote: I have a question on this particular thing, wouldn't the range for <0 exclude 4? I understand that it does not change the sign of the graph but it does = 0: \((x4)^2(x  9)(x+11) < 0\) if x=4, then the whole thing goes to 0 and it would not be inside the 'valid' range.
Am I correct?
We are given that \((x4)^2(x  9)(x+11)\) is less than 0. We need to find the range of values that x can take in that case. Note that the expression IS LESS THAN 0. This means that it cannot be 0. So x cannot be 4 because that will make the expression 0. So we don't plot 4 on the number line. (x4)^2 is positive only and hence doesn't affect our signs.
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Re: Inequalities trick [#permalink]
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@p00rv@ wrote: As per this trick, for the inequality x^3 < x^2 we will get the range as 0<x<1. But the correct answer is "Any nonzero number less than 1". Please could someone help me understand what am I missing here? ~Kudos are free. Be generous! x^3<x^2 => x^3x^2<0 (We can get x^2 on to LHS because we now that x^2 is nonnegative and hence we need not change the inequality sign) x^2(x1)<0 x1<0 x<1 Hope this helps And yeah, ~Kudos are free. Be generous!
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12 Nov 2016, 01:12
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 Answer = 11 Check solution as attached
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Re: Inequalities trick [#permalink]
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25 Dec 2016, 10:38
Hi, I tried solving the following question algebraically and had a doubt:
Is X negative? (i) (x^3)  (x^5) < 0 (ii) (x^2)  1 < 0
after simplifying the first statement, I got (x^3) (1+x) (1x) < 0. Now if we plot the three roots (1,0,1) on the number line and start mark the regions as  +  + (since f(x) has three factors), it should result in x<1 and 0<x<1. But when I plugged numbers to confirm, the ranges satisfying the statement are 1<x<0 and x>1. Please help me identify where I'm going wrong.

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Re: Inequalities trick [#permalink]
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02 Apr 2017, 02:43
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 12 times to get used to it.
Suppose you have the inequality
f(x) = (xa)(xb)(xc)(xd) < 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. Thanks a lot for this post , it's very helpful. Would be great if you clarify the below query: Given: If f(x) has three factors then the graph will have  +  + If f(x) has four factors then the graph will have +  +  + Question: What if we have 2 factors? do we need to start out interpretation as +  + and for 5 factors, do we need to have it in this way?  +  +  + Thanks in advance, Uma

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Re: Inequalities trick [#permalink]
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03 Apr 2017, 05:34
umabharatigudipalli 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 12 times to get used to it.
Suppose you have the inequality
f(x) = (xa)(xb)(xc)(xd) < 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. Thanks a lot for this post , it's very helpful. Would be great if you clarify the below query: Given: If f(x) has three factors then the graph will have  +  + If f(x) has four factors then the graph will have +  +  + Question: What if we have 2 factors? do we need to start out interpretation as +  + and for 5 factors, do we need to have it in this way?  +  +  + Thanks in advance, Uma Yes, that is correct. Usually, n factors will divide the number line into (n+1) regions. You start out by giving + to the rightmost region (provided all your factors are of the form (ax + b) or (ax  b)) and then alternating the signs on the left. It will be good if you take a look at this post which gives links to explanations on why this happens: https://gmatclub.com/forum/inequalities ... l#p1753431
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Hi karishma, abhimahna, mikemcgarry, Bunuel, Skywalker18, and other experts I am not able to understand the graph. How do we know which one is negative/positive? How does this graph help us in solving inequalities? Thanks
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Shiv2016 wrote: Hi karishma, abhimahna, mikemcgarry, Bunuel, Skywalker18, and other experts I am not able to understand the graph. How do we know which one is negative/positive? How does this graph help us in solving inequalities? Thanks Hi Shiv2016 , The concept is very easy. 1. Find out the zero points. 2. Arrange them in ascending order. 3. Draw them on a number line. 4. Take the right most as +ve and proceed towards left taking alternate signs. 5. If the inequality is of > form, then take all +ve ranges. 6. if the inequality is of lesser form, then take all ve ranges. E.g. (x2) (x3 ) > 0 Equality form is greater than(>). Zero points = 2 and 3 Draw them on number line. You will get 3 ranges. x<2; 2<x<3; and x >2. Here, right most will be +ve, or x>2 will be +ve. then 2<x<3 will be ve then x<2 will be +ve. Since inequality is of (>) form, we will take all the ranges which have +ve sign. Hence, the answer will be x>2 and x<2. I hope it makes sense.
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