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Inequalities trick [#permalink]
16 Mar 2010, 09:11

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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.

Re: Inequalities trick [#permalink]
22 Oct 2010, 05:33

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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 20580 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. _________________

Re: Inequalities trick [#permalink]
11 Mar 2011, 05:49

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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 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. _________________

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

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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.... 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 Mar 2011, 18:57

3

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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. _________________

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. _________________

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. _________________

Re: Inequalities trick [#permalink]
18 Oct 2012, 04:17

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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. _________________

Re: Inequalities trick [#permalink]
20 Dec 2012, 20:04

1

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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.

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. _________________

I can define this as (1+2x).x^3(1-2x). now I have roots -1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of -1/2 and 0 I will get value as 0.

So How I will define them in graph and what range I will consider for this inequality.

Thanks

The factors must be of the form (x - a)(x - b) .... etc

Re: Inequalities trick [#permalink]
19 Mar 2010, 11:59

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.

Re: Inequalities trick [#permalink]
17 Oct 2010, 15:14

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 1-2 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? _________________

Re: Inequalities trick [#permalink]
17 Oct 2010, 16:19

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 1-2 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) = (x-a)(x-b)(x-c)(x-d) < 0 you will consider the curve with -ve inside it.. check the attached image.

f(x) = (x-a)(x-b)(x-c)(x-d) > 0 consider the +ve of the curve _________________