Inequalities trick

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.



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.

Here is the video explanation of this: https://youtu.be/PWsUOe77__E
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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

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.

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

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

Just came across this useful discussion.

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.




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 – region1 between 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.

- Krishna
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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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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
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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?

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.
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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}

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.
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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.
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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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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.
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Thankyou Karishma.

Further, say if the same expression was (x+2)(1-x)/(x-4)(x-7) and still the question was for what values of x is the expression positive, then ... make it x-1 and with the same roots, have the rightmost as -ve. Then we look for the +ve intervals and check for those intervals if the expression is positive. for examples, in this case, -2<x<1 and 4<x<7 both depict positive interval but only first range satisfies the condition. Please confirm

However, if for the same equation as mentioned, say the expression was (x+2)(x-1)/(x-4)(x-7) >0 and then we were asked to give the range where this is valid, then we would also multiply the -ve sign and make is <0 and then make the range after extreme right root -ve and provide all the intervals where it is negative. Please confirm