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JesusWalks
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Inequalities Made Easy! 08 Jun 2019, 07:06
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Bunuel


Notice that \((x+a)(x+b)(x+c)>0\) can be written as \((x-(-a))(x-(-b))(x-(-c))>0\). So, if the "roots" of (x-a)(x-b)(x-c)>0 are a, b, and c, for \((x-(-a))(x-(-b))(x-(-c))>0\) they would be -a, -b, and -c.

For example, for (x + 1)(x + 2) > 0, you put on the line -1 and -2.

Hope it helps.
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Ah perfect! Thanks!
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Inequalities Made Easy! 23 Jul 2019, 18:18
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Hi Karishma & Bunuel,

First of all, let me thank you to put together such an amazing explanation of Inequalities and strategies. It helped me immensely.

I have a small doubt in Step 2 mentioned in the very first post "Inequalities with Multiple Factors" of this thread:

"Starting from the rightmost section, mark the sections with alternate positive and negative signs. The inequality will be positive in the sections where you have the positive signs and it will be negative in the sections where you have the negative signs"

My understanding is that we have to check one value from each section to understand if the equation results in a Positive or Negative value in order to assign the sign to each of the section. is this correct? or we can blindly assign Positive and Negative sign to alternate sections respectively by starting with assigning Positive sign to the rightmost section?
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Hi Karishma & Bunuel,

First of all, let me thank you to put together such an amazing explanation of Inequalities and strategies. It helped me immensely.

I have a small doubt in Step 2 mentioned in the very first post "Inequalities with Multiple Factors" of this thread:

"Starting from the rightmost section, mark the sections with alternate positive and negative signs. The inequality will be positive in the sections where you have the positive signs and it will be negative in the sections where you have the negative signs"

My understanding is that we have to check one value from each section to understand if the equation results in a Positive or Negative value in order to assign the sign to each of the section. is this correct? or we can blindly assign Positive and Negative sign to alternate sections respectively by starting with assigning Positive sign to the rightmost section?
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You can blindly assign assuming that it's set up correctly.
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Hello Bunuel,

How did you factor this?

"Given: −2x3+17x2–30x>0−2x3+17x2–30x>0

x(−2x2+17x–30)>0x(−2x2+17x–30)>0 (taking x common)

x(2x–5)(6–x)>0x(2x–5)(6–x)>0 (factoring the quadratic)"

Thanks
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Inequalities Made Easy! 26 May 2020, 11:18
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Bunuel
Or Just Use Inequalities!
BY KARISHMA, VERITAS PREP

If you are wondering about the absurd title of this post, just take a look at the above post's title. It will make much more sense thereafter. This post is a continuation of last week’s post where we discussed number plugging. Today, as per students’ request, we will look at the inequalities approach to the same official question. You will need to go through our inequalities post to understand the method we will use here.

Recall that, given \(a < b\), \((x – a)(x – b) < 0\) gives us the range \(a < x < b\) and \((x – a)(x – b) > 0\) gives us the range \(x < a\) or \(x > b\).

Question: If x is positive, which of the following could be the correct ordering of 1/x, 2x and x^2?
(I) x^2 < 2x < 1/x
(II) x^2 < 1/x < 2x
(III) 2x < x^2 < 1/x

(A) none
(B) I only
(C) III only
(D) I and II
(E) I, II and III

Solution: The question has three complex inequalities. We will take each in turn. Note that each inequality consists of two more inequalities. We will split the complex inequality into two simpler inequalities e.g. x^2 < 2x < 1/x gives us x^2 < 2x and 2x < 1/x. Next we will find the range of values of x which satisfy each of these two inequalities and we will see if the two ranges have an overlap i.e. whether there are any values of x which satisfy both these simpler inequalities. If there are, it means there are values of x which satisfy the entire complex inequality too. Things will become clearer once we start working on it so hold on.

Let’s look at each inequality in turn. We start with the first one:

(I) x^2 < 2x < 1/x

We split it into two inequalities:

(i) x^2 < 2x

We can rewrite x^2 < 2x as x^2 – 2x < 0 or x(x – 2) < 0.

We know the range of x for such inequalities can be easily found using the curve on the number line. This will give us 0 < x < 2.

(ii) 2x < 1/x

It can be rewritten as x^2 – 1/2 < 0 (Note that since x must be positive, we can easily multiply both sides of the inequality with x)

This gives us the range -1/?2 < x < 1/?2 (which is 0 < x < 1/?2 since x must be positive).

Is there a region of overlap in these two ranges i.e. can both inequalities hold simultaneously for some values of x? Yes, they can hold for 0 < x < 1/?2. Hence, x^2 < 2x < 1/x will be true for the range 0 < x < 1/?2. So this could be the correct ordering. Let’s go on to the next complex inequality.

(II) x^2 < 1/x < 2x

Again, let’s break up the inequality into two parts:

(i) x^2 < 1/x

x^1 < 1/x is rewritten as x^3 – 1 < 0 which gives us x < 1.

(ii) 1/x < 2x

1/x < 2x is rewritten as x^2 – 1/2 > 0 which gives us x < -1/?2 (not possible since x must be positive) or x > 1/?2

Can both x < 1 and x > 1/?2 hold simultaneously? Sure! For 1/?2 < x < 1, both inequalities will hold and hence x^2 < 1/x < 2x will be true. So this could be the correct ordering too.

(III) 2x < x^2 < 1/x

The inequalities here are:

(i) 2x < x^2

2x < x^2 can be rewritten as x(x – 2) > 0 which gives us x < 0 (not possible) or x > 2.

(ii) x^2 < 1/x

x^2 < 1/x gives us x^3 – 1 < 0 i.e. x < 1

Can x be less than 1 and greater than 2 simultaneously? No. Therefore, 2x < x^2 < 1/x cannot be the correct ordering.

Answer (D)

Is this method simpler?
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does anyone understand why this method works? and also bunel evaluated each a<b<c as a<b and b<c why not evaluate a<c?
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Inequalities Made Easy! 21 Oct 2020, 05:40
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Hello,

Thanks for the amazing analysis on the inequalities. I have a little question.
The questions discussed are generally inequalities with only 1 variable x and its different functions.
What is the standard approach for the questions including different variables and their functions.
Is plugging values the only solution there?

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Plugging values is almost never necessary. Check out 21:03 here:
https://youtu.be/Ob4KOxMtHf4

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Inequalities Made Easy! 21 Oct 2020, 13:42
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@avigutmann,
Hello sir

I did refer to the video. I believe I should be specific regarding the question so that you can understand my problem statement better.

For example: If a < b < c , which of the following must be true?

(I) a < b^2
(II) b − a < c
(III) a^2 < b^2 < c^2

A. None
B. I only
C. II only
D. III only
E. II and III

----
Different variables and different functions. What should be the standard approach here (other than plugging values)?
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You just have to draw them on a number line and do some thinking...
(I) we know a<b. What happens to b when it is squared? Will it move to the left? To the right? Stay in the same spot?
Depends on its position relative to -1, 0, and 1.
(II) b-a represents the (positive) distance between a and b. How does the value of c compare to the distance between a and b? Do we know anything about that distance? No...
(III) Okay so if we take three random numbers on the number line, would squaring them keep their original order, or would it potentially mess up the order? Again, depends on where they sit relative to -1, 0, and 1, because squaring a number will have very different effects in those different ranges.
Please let me know your thoughts fitshot.
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AviGutman
That's just awesome! You just widened my approach to these questions. Surely gonna help. Appreciate it!
However, I am feeling that this approach might have got some limited applicability or will consume more time on some complex questions (involving complex functions). I would love to be proved wrong :)

Can you help me with the same approach for this ques please:
If x,y,z>0 and x>y>z, which of the following could be true:
1: sqroot(x)>sqroot(y)>sqroot(z)
2: sqroot(z)>sqroot(y)>sqroot(x)
3: sqroot(x)>z^1/3> sqroot(y)
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Inequalities Made Easy! 12 Jul 2021, 12:56
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Bunuel
Some Inequalities, Mods and Sets
BY KARISHMA, VERITAS PREP

Here, let’s look at a question that involves inequalities and modulus and is best understood using the concept of sets. It is not a difficult question but it is still very tricky. You could easily get it right the first time around but if you get it wrong, it could take someone many trials before he/she is able to convince you of the right answer. Even after I write a whole post on it, I wouldn’t be surprised if I see “but I still don’t get it” in the comments below!

Anyway, enough of introduction! Let’s get to the question now.

Question: If \(\frac{x}{|x|} < x\), which of the following must be true about x?
(A) x > 1
(B) x > -1
(C) |x|< 1
(D) |x| = 1
(E) |x|^2 > 1

Solution:
First thing we do is tackle the mod. We know that |x| is just the absolute value of x.

So, \(\frac{x}{|x|}\) can take only 2 values: 1 or -1
If x is positive, \(\frac{x}{|x|} = 1\) e.g. if \(x = 4\), then \(\frac{4}{|4|} = 1\)
If x is negative, \(\frac{x}{|x|} = -1\) e.g. if \(x = -4\), then \(\frac{-4}{|-4|} = \frac{-4}{4} = -1\)
x cannot be 0 because we cannot have 0 in the denominator of an expression.

Now let’s work on the inequality.

\(x >\frac{x}{|x|}\) implies \(x > 1\) if x is positive or \(x > -1\) if x is negative.

Hence, for this inequality to hold, either \(x > 1\) (when x is positive) or \(-1 < x< 0\) (when x is negative)

x can take many values e.g. -1/3, -4/5, 2, 5, 10 etc.

Now think – which of the following MUST BE TRUE about every value that x can take?
(A) x > 1
or
(B) x > -1

I hope that you agree that x > 1 doesn’t hold for every possible value of x whereas x > -1 holds for every possible value of x. Mind you, every value greater than -1 need not be a possible value of x.

This concept might need some more work. Let me explain with another example.

Forget this question for a minute. Say instead you have this question:

Example 1: x > 2 and x < 7. What integral values can x take?
I guess most of you will come up with 3, 4, 5, 6. That’s correct. I can represent this on the number line.



The top arrow shows x > 2 and the bottom arrow shows x < 7. You see that the overlapping area includes 3, 4, 5 and 6. That is the region that satisfies both the inequalities.

Now consider this:

Example 2: x > 2 or x > 5. What integral values can x take?
Let’s draw that number line again.



So is the solution again the overlapping numbers i.e. all integers greater than 5? No. This question is different. x is greater than 2 OR greater than 5. This means that if x satisfies at least one of these conditions, it is included in your answer. Think of sets. AND means it should be in both the sets (i.e. the overlapping part) as was the case in example 1. OR means it should be in at least one of the sets. Hence, which values can x take? All integral values starting from 3 onwards i.e. 3, 4, 5, 6, 7, 8, 9 …

Now go back to this question. The solution is a one liner.
\(\frac{x}{|x|}\) is either 1 or -1.
So \(x > 1\) or \(x > -1\)
So which values can x take? All values included in at least one of the sets. Therefore, x > -1.

Note that the confusion lies only between the first two options. The other three options are rejected outright.
(C) |x|< 1 implies -1 < x < 1. Definitely doesn’t hold.
(D) |x| = 1 implies x = 1 or -1. Definitely doesn’t hold.
(E) |x|^2 > 1 implies either x < -1 or x > 1. Definitely doesn’t hold.

So what do you say? Are you convinced that the answer is (B)? This question is discussed HERE.

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Can we do this by squaring both sides and through a wavy line method?
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for equation x>x/|x| Can anyone any algebraically prove that if x is positive x>1 and if x is negative x>-1;
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Hi Where can i find more resources on modulus, word problems, number properties (odd/even, factors, LCM etc.) similar to the one above shared on Inequalities? Bunuel

Thank you
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Hi Where can i find more resources on modulus, word problems, number properties (odd/even, factors, LCM etc.) similar to the one above shared on Inequalities? Bunuel

Thank you
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Check the links below:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.
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Is it fair to say that while a>b and c>d for non-negative numbers gives ac>bd, the reverse isn't necessarily true.
In other words, ac>bd does not imply that a>b and c>d or a>d and c>b?

For example, a=6,b=2,c=3,d=8:
ac = 18, bd = 16. Therefore, ac > bd

But, c<d and a<d, so therefore the "split inequalities" do not hold.

I find this strange so just want to check whether I have missed something here?
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Is the graph for final question correct, shoudl not it be -++-
Bunuel
Inequalities with Complications - Part II
BY KARISHMA, VERITAS PREP

Above we discussed a couple of variations of inequality questions with many factors. Let’s now look at some more complications that we should know how to handle.

Complication No 3: Even powers

\((x – a)^2(x – b)(x – c)(x – d) > 0\)

How would you handle even powers of factors? Say, if the question has a factor (x – a) which has an even power, would you still plot ‘a’ on the number line? No, you will just ignore that factor while making the number line! Why? This is so because this factor will never be negative. It will be either 0 (in case the inequality includes the equality sign e.g. \((x – a)^2(x – b)(x – c)(x – d) = 0\)) or it will be positive. Therefore, a factor with an even power acts just like a positive constant. Then, does it mean factors with even powers have no role to play at all? No, it doesn’t. While writing out the range, they could impact the final answer. We will discuss this in more detail using an example later on.

e.g. \((x – 4)(x – 2)(x + 8)^4 < 0\)

You don’t need to plot -8 here. Since \((x + 8)^4\) can never be negative, it doesn’t change the sign of the expression.



The expression will be negative in the range \(2 < x < 4\).

Complication No 4: Odd powers

\((x – a)^3(x – b)(x – c)(x – d) > 0\)

What will you do in case of odd powers? Notice that in the last two posts, we have handled questions where the power of all the factors is 1. 1 is an odd power. So when you have any other odd power, you will handle it the same way. You can assume that the odd power is equal to 1 and proceed as usual. This is so because the sign of (x – a) will be the same as the sign of \((x – a)^3\).

e.g. \((x – 4)(x – 2)(x + 1)^3 < 0\)



The expression will be negative in the range \(x < -1\) or \(2 < x < 4\).

Now let’s look at a question involving both these complications.

Question: Find the range of x for which the given inequality holds.
\(\frac{(2x – 5) * (6 – x)^3}{x^2} < 0\)

Solution:

I hope you agree that it doesn’t matter whether the factors are multiplied or divided. We are only concerned with the sign of the factors.

\(\frac{2(x – \frac{5}{2}) * (-1)(x – 6)^3}{x^2}< 0\) (take 2 common)

\(\frac{2(x – \frac{5}{2}) * (x – 6)^3}{(x – 0)^2} > 0\) (multiply both sides by -1)

Now let’s draw the number line. We don’t need to plot 0 since (x – 0) has an even power.



We want to find the range where the expression is positive. The required range is \(x < \frac{5}{2}\) or \(x > 6\). But we are missing something here. \(x < \frac{5}{2}\) implies that all values less than 5/2 are acceptable but note that x cannot be 0 since x^2 is in the denominator. Hence the acceptable range is \(x < \frac{5}{2}\) or \(x > 6\) but \(x \neq 0\).

When you have the equal to sign, you have to be careful about the way you choose your range.

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bishal128
Is the graph for final question correct, shoudl not it be -++-

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You forgot that x^2 is always positive, so it never changes the sign. It only means x cannot be 0. After simplifying, the only factors that affect the sign are (x - 5/2) and (x - 6)^3, which gives + - +.

So (x - 5/2)(x - 6)^3/x^2 > 0 gives x < 0, 0 < x < 5/2, or x > 6.


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