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Inequalities trick [#permalink]
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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 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.
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Re: Inequalities trick [#permalink]
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16 Mar 2010, 10:46
Can u plz explainn the backgoround of this & then the explanation.
Thanks

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Re: Inequalities trick [#permalink]
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19 Mar 2010, 12:48
I have applied this trick and it seemed to be quite useful.
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Re: Inequalities trick [#permalink]
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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 reread 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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Re: Inequalities trick [#permalink]
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17 Oct 2010, 16:14
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.
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)
I don't understand this part alone. Can you please explain?
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Re: Inequalities trick [#permalink]
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17 Oct 2010, 17:19
Dreamy 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.
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)
I don't understand this part alone. Can you please explain? Suppose you have the inequality f(x) = (xa)(xb)(xc)(xd) < 0 you will consider the curve with ve inside it.. check the attached image. f(x) = (xa)(xb)(xc)(xd) > 0 consider the +ve of the curve
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Re: Inequalities trick [#permalink]
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18 Oct 2010, 09:23
hey i remember having used this in my Boards ?(School exams) . Thanks Gurpreet . This makes life and its inequalities a lil easier . +1.

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Re: Inequalities trick [#permalink]
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19 Oct 2010, 04:09
in addition: if we have smth like (xa)^2(xb) we don't need to change a sign when jump over "a".

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Re: Inequalities trick [#permalink]
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19 Oct 2010, 12:56
ulm wrote: in addition: if we have smth like (xa)^2(xb) 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
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Re: Inequalities trick [#permalink]
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21 Oct 2010, 10:52
gurpreetsingh wrote: Dreamy 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.
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)
I don't understand this part alone. Can you please explain? Suppose you have the inequality f(x) = (xa)(xb)(xc)(xd) < 0 you will consider the curve with ve inside it.. check the attached image. f(x) = (xa)(xb)(xc)(xd) > 0 consider the +ve of the curve Thank you gurpreet. I understand it now. You need fix the initial post. The second f(x) is also having < 0.
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Re: Inequalities trick [#permalink]
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21 Oct 2010, 10:57
Thanks typo fixed.
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Re: Inequalities trick [#permalink]
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21 Oct 2010, 23:12
Quite interesting. Do you have any sample questions we can use this method on?
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Re: Inequalities trick [#permalink]
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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 (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 [ 7.9 KiB  Viewed 52154 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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Re: Inequalities trick [#permalink]
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if = sign is included with < then <= will be there in solution like for (x+2)(x1)(x7)(x4) <=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

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Re: Inequalities trick [#permalink]
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mrinal2100: Kudos to you for excellent thinking!
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Re: Inequalities trick [#permalink]
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31 Jan 2011, 03:28
this is good thinking , i will use it in solving problems.
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Re: Inequalities trick [#permalink]
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11 Mar 2011, 05:40
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.
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Re: Inequalities trick [#permalink]
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11 Mar 2011, 06:29
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 Yaxis for four values and starts from ve Yaxis for three values. What if the equation you mentioned is (x+2)(x1)(x7)<0,will the last two ranges be excluded or the graph will also change?

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Re: Inequalities trick [#permalink]
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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 Yaxis for four values and starts from ve Yaxis for three values. What if the equation you mentioned is (x+2)(x1)(x7)<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)(x1)(x7) < 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)(10001)(10007) 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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Re: Inequalities trick [#permalink]
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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.

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