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KarishmaB
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If |x| < x^2, which of the following must be true ? Updated on: 27 Nov 2023, 23:57
Originally posted by KarishmaB on 27 May 2021, 23:12.
Last edited by KarishmaB on 27 Nov 2023, 23:57, edited 1 time in total.
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shenwenlim
When x < 0, how can -x < x^2? When you are taking the absolute value of a negative number, isn't the outcome always positive? In this case, when x < 0, shouldn't |-x| = x ? and not -x?

Bunuel VeritasKarishma
ScottTargetTestPrep JeffTargetTestPrep
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Yes, it is a common problem people face.
How can |x| be negative i.e. -x? Because we see the minus sign of -x, our brain tricks us into believing that it is a negative number. But when x itself is a negative number, -x is a POSITIVE number.
So if x = -5,
-x = 5 (positive)

So when x itself is negative, |x| = |-x| = -x (all are positive terms)

Also check:
https://anaprep.com/algebra-the-why-beh ... questions/
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If |x| < x^2, which of the following must be true ? 20 Jun 2021, 06:45
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adiagr


\(x^2(x^2 - 1)> 0\)

x^2 is always positive.

So the condition is

\(x^2 - 1> 0\)

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Can anyone please clarify my doubt here?

I generally solve such questions using number line after I have simplified to get individual values of x.

I fail to understand how x^2 has been eliminated here. What's troubling me here is that I denoted 'zero' on the number line.

Please help!
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If |x| < x^2, which of the following must be true ? 20 Jun 2021, 07:38
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BhaveshGMAT
adiagr


\(x^2(x^2 - 1)> 0\)

x^2 is always positive.

So the condition is

\(x^2 - 1> 0\)


Can anyone please clarify my doubt here?

I generally solve such questions using number line after I have simplified to get individual values of x.

I fail to understand how x^2 has been eliminated here. What's troubling me here is that I denoted 'zero' on the number line.
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In the original question, we know |x| < x^2. So it's impossible that x = 0, because x = 0 doesn't satisfy that inequality.

There are a lot of ways to solve the original question, but if you did end up with the inequality

\(x^2(x^2 - 1)> 0\)

then since x^2 is always a positive number (we know x is not zero), we can safely divide both sides of the inequality by x^2. Since x^2 must be positive, we don't need to worry about whether we might be dividing by a negative number, so we don't need to worry about whether we should reverse the inequality. And once you divide by x^2, you learn x^2 - 1 > 0, or x^2 > 1.

I don't think that approach is the best one to take on questions like this, but I'll assume other approaches were already explained earlier in the thread (I haven't read all the solutions).
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If |x| < x^2, which of the following must be true ? 05 Aug 2021, 15:57
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shenwenlim
When x < 0, how can -x < x^2? When you are taking the absolute value of a negative number, isn't the outcome always positive? In this case, when x < 0, shouldn't |-x| = x ? and not -x?

Bunuel VeritasKarishma
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Solution:

If x < 0, then -x is positive and that is why |x| is equal to -x. Let’s use some actual values to make it clear; for instance let x = -2. Now |-2| = 2 and 2 is equal to -x (because -x = -(-2) = 2).
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If |x| < x^2, which of the following must be true ? 25 Nov 2021, 06:15
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Skywalker18
If |x| < x², which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x² > 1
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I find that these questions can be solved quickly (and accurately) by testing values.

For example, if |x| < x², then x could equal 2 (since |2| < 2²)
This means we can eliminate choices B and D since they state that x cannot equal 2.

Similarly, if |x| < x², then x could equal -2 (since |-2| < (-2)²)
This means we can eliminate choices A and C since they state that x cannot equal -2.

By the process of elimination, the correct answer is E
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If |x| < x^2, which of the following must be true ? 30 Dec 2021, 00:01
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Given that |x| < x²

Let's take two cases to open the absolute value

Case 1: Whatever is inside the Absolute Value >= 0
x >= 0
=> |x| = x (Watch this video to learn more about Basics of Absolute Value)
=> x < \(x^2\)
=> \(x^2\) - x > 0
=> x*(x-1) > 0

Attachment:
x between 0 and 1.JPG
x between 0 and 1.JPG [ 17.24 KiB | Viewed 2737 times ]

=> x < 0 or x > 1 (Using Sine Wave Method: Watch this video to learn the method)
But our condition was x>=0
=> Intersection will be x > 1

Attachment:
x GT 1.JPG
x GT 1.JPG [ 14.98 KiB | Viewed 2755 times ]

Case 2: Whatever is inside the Absolute Value < 0
x < 0
=> |x| = -x (Watch this video to learn more about Basics of Absolute Value)
=> -x < \(x^2\)
=> \(x^2\) + x > 0
=> x*(x+1) > 0

Attachment:
x between -1 and 0.JPG
x between -1 and 0.JPG [ 16.8 KiB | Viewed 2754 times ]

=> x < -1 or x > 0 (Using Sine Wave Method: Watch this video to learn the method)
But our condition was x < 0
=> Intersection will be x < -1

So, x > 1 or x < -1
=> x^2 > 1

So, Answer will be E
Hope it helps!

Watch the following video to learn the Basics of Absolute Values

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If |x| < x^2, which of the following must be true ? 09 Dec 2022, 07:31
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i tried plugging in values to eliminate options.
can someone tell me how which values i can pick for option C to eliminate it
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If |x| < x^2, which of the following must be true ? 09 Dec 2022, 07:32
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can someone tell me how which values i can pick for option C to eliminate it Bunuel KarishmaB
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If |x| < x^2, which of the following must be true ? 14 Feb 2023, 23:37
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ShaunakSawant
can someone tell me how which values i can pick for option C to eliminate it Bunuel KarishmaB
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Hello ShaunakSawant,

Let me begin by telling you that this approach of plugging in values is not scalable (would take a lot of time to check each choice) and something that I would not recommend. Had you adopted the methodical approach of solving the question, you would have known exactly what value would eliminate choice C.

That said, I will address your request right now. 😊

You wished for a value to help us eliminate option C: x > 1. If we find a case where the original condition, |x| < x\(^2\) is true but condition x > 1 is not true, then we will be done - that is, we will be able to reject choice C as need not be true.

Now, since we want x > 1 to not be true, our x must be anywhere in the range x <= 1. And remember that it must satisfy the condition given in the question: |x| < x\(^2\).

To find such numbers, let’s just take numbers from the range x <= 1 and plug them into the given inequality: |x| < x\(^2\)to see which ones work. (Again, the ones that do work will help us reject choice C.)

  1. x = 1
    |x| = 1 and x\(^2\) = 1
    |x| < x\(^2\) – NOT TRUE
  2. x = 0
    |x| = 0 and x\(^2\) = 0
    |x| < x\(^2\) – NOT TRUE
  3. x = -1
    |x| = 1 and x\(^2\) = 1
    |x| < x\(^2\) – NOT TRUE
  4. x = -2
    |x| = 2 and x\(^2\) = 4
    |x| < x\(^2\) – TRUE

So, we have found a value in the range x <= 1 that satisfies the original condition! Hence, choice C can be rejected.

Note: I would again suggest that you build methodical approaches to solve questions. This will enable you to tackle the hardest of questions with ease.
You can check the solution given by BrushMyQuant just above your query.

Hope this helps!

Best Regards,
Ashish Arora
Quant Expert, e-GMAT
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If |x| < x^2, which of the following must be true ? 18 Jun 2024, 23:48
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After squaring both the sides, we can rearrange the equation as x^2 * (x-1) * (x+1) >0
Now, we can plot the inflexion points using the wave curly method:

————————(-1)——————————(1)————————
——Positive———————-Negative——————-Positive———

Hence, x^2>1
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If |x| < x^2, which of the following must be true ? 14 May 2025, 06:54
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Can we figure it out like this?
the only case where this equation can flaw is when x = decimal or 1 / 0
as decimal ^2 < decimal

hence we know x > 1 or x < -1
the only possible case is E
Skywalker18
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1
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If |x| < x^2, which of the following must be true ? 14 May 2025, 17:05
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in what case is x >1 ever untrue?

Bunuel
Skywalker18
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

Given: \(|x|<x^2\) --> reduce by \(|x|\) (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too, as if x would be zero inequality wouldn't hold true, so \(|x|>0\)) --> \(1<|x|\) --> \(x<-1\) or \(x>1\).

So we have that \(x<-1\) or \(x>1\).


A. x > 0. Not always true.
B. x < 0. Not always true.
C. x > 1. Not always true.
D. -1 < x < 1. Not true.
E. x^2 > 1. Always true.

Answer: E.

To understand the underline concept better practice other Trickiest Inequality Questions Type: Confusing Ranges.

Hope it helps.
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If |x| < x^2, which of the following must be true ? 14 May 2025, 23:32
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benbitz
in what case is x >1 ever untrue?

Bunuel
Skywalker18
If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

Given: \(|x|<x^2\) --> reduce by \(|x|\) (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too, as if x would be zero inequality wouldn't hold true, so \(|x|>0\)) --> \(1<|x|\) --> \(x<-1\) or \(x>1\).

So we have that \(x<-1\) or \(x>1\).


A. x > 0. Not always true.
B. x < 0. Not always true.
C. x > 1. Not always true.
D. -1 < x < 1. Not true.
E. x^2 > 1. Always true.

Answer: E.

To understand the underline concept better practice other Trickiest Inequality Questions Type: Confusing Ranges.

Hope it helps.
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Since we have that \(x<-1\) or \(x>1\), C (x > 1) will be false for any value which is less than -1, for example if x is -10.
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If |x| < x^2, which of the following must be true ? 28 Sep 2025, 09:02
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Hey scott, I want to understand one thing, In case 2, you divided both sides by x, but then changed the inequality sign, Can you explain me that part, as of what I know is we change the inequalities when we multiply or divide by any -ve number. But here you divided by x.
ScottTargetTestPrep


If x = 0, then |0| will not be less than 0^2 since they both are equal to 0. Thus, we know x can’t be 0.

Since |x| = x when x is positive and -x when x is negative, let’s rewrite the inequality without the absolute value sign. That is, if x > 0, then we have x < x^2, and if x < 0, then we have -x < x^2.

Case 1: If x > 0,

x < x^2

Dividing both sides by x, we have:

1 < x

Case 2: If x < 0,

-x < x^2

Dividing both sides by x (switching the inequality sign since x is negative), we have:

-1 > x

Thus, we have x > 1 or x < -1; in that case, x^2 must be greater than 1.

Answer: E
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If |x| < x^2, which of the following must be true ? 28 Sep 2025, 09:05
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iurequi
Hey scott, I want to understand one thing, In case 2, you divided both sides by x, but then changed the inequality sign, Can you explain me that part, as of what I know is we change the inequalities when we multiply or divide by any -ve number. But here you divided by x.

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The inequality sign flipped in Case 2 because x is negative there.

When you divide both sides of an inequality by a negative number, the inequality reverses direction. Since Case 2 assumes x < 0, dividing by x means dividing by a negative, so the sign must switch.
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