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Re: GRE Math Essentials - INEQUALITIES / Absolute Value / Modulus [#permalink]
rx10 wrote:
Hey, COolguy101

Take \(x = -2\), so \(x^2>x\)

COolguy101 wrote:
24. If x^2>x, then either x > 1 or x is negative (x < 0), how x to be negative is possible. I am getting x>0.

But sir, I take x to the left and solve then I got x>1 and x>0, where do I make mistake. I can not even figure it out.
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Re: GRE Math Essentials - INEQUALITIES / Absolute Value / Modulus [#permalink]
\(x^2 > x\)

\(x^2 - x > 0\)

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

What you did was just divided with \(x\) but without knowing the sign we can't do that. inequality may change by doing so.


COolguy101 wrote:
But sir, I take x to the left and solve then I got x>1 and x>0, where do I make mistake. I can not even figure it out.
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Re: GRE Math Essentials - INEQUALITIES / Absolute Value / Modulus [#permalink]
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According to also GreenlightTestPrep statement, 26 is absolutely correct

Quote:
If \(x^2>x\), then either \(x > 1\) or x is negative \((x < 0)\)

Rearrange to get: \(x^2 - x > 0\)

Factor: \(x(x - 1) > 0\)

So, either \(x < 0\) or \(x > 1\)
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Re: GRE Math Essentials - INEQUALITIES / Absolute Value / Modulus [#permalink]
Carcass wrote:
According to also GreenlightTestPrep statement, 26 is absolutely correct

Quote:
If \(x^2>x\), then either \(x > 1\) or x is negative \((x < 0)\)

Rearrange to get: \(x^2 - x > 0\)

Factor: \(x(x - 1) > 0\)

So, either \(x < 0\) or \(x > 1\)

Got it sir!
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Re: GRE Math Essentials - INEQUALITIES / Absolute Value / Modulus [#permalink]
Here's a video on Inequalities:

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Re: GRE Math Essentials - INEQUALITIES / Absolute Value / Modulus [#permalink]
Watch the following video to learn the Basics of Inequalities



Watch the following video to learn How to Solve Inequality Problems

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Re: GRE Math Essentials - INEQUALITIES / Absolute Value / Modulus [#permalink]
Carcass wrote:
According to also statement, 26 is absolutely correct

Quote:
If \(x^2>x\), then either \(x > 1\) or x is negative \((x < 0)\)

Rearrange to get: \(x^2 - x > 0\)

Factor: \(x(x - 1) > 0\)

So, either \(x < 0\) or \(x > 1\)


Hello Sir, could you please help explain why it's x<0 and not x>0

when I got to x(x - 1) > 0
I break it down to
x>0 and x-1> 0 resulting in x>0 and x>1 as an answer
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Re: GRE Math Essentials - INEQUALITIES / Absolute Value / Modulus [#permalink]
Expert Reply
Which number of the above you refer to ? 24-25-26..........................which one ?
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Re: GRE Math Essentials - INEQUALITIES / Absolute Value / Modulus [#permalink]
Carcass

number 24. I got the same answer as coolguy. X>0, X>1
I mean it makes sense that If x^2>x then x>1 or x<0 but I just can't algebraically solve it

thank you,sir
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GRE Math Essentials - INEQUALITIES / Absolute Value / Modulus [#permalink]
Expert Reply
\(x^2 > x \)

It looks as though we might be able to divide both sides by x to give x >1

But, in fact, we cannot do this. The two inequalities \(x^2 > x\) and x > 1 are not the same.

This is because in the inequality x > 1, x is clearly greater than 1. But in the inequality \(x^2 > x\) we have to take into account the possibility that x is negative, since if x is negative, \(x^2\) (which must be positive or zero) is always greater than x.

In fact the complete solution of this inequality is x >1 or x<0. The second part of the solution must be true since if x is negative, \(x^2\) is always greater than x.

According to the explanation above, in fact x must be

>1 for example 2 and 4>2 this is true

<0 for example -1 and 1>-1 this is true

x=0 and 0> 0 impossible this is not true

x is >0 AND <1 for example 1/2 . 1/4>1/2 impossible. This is NOT true

resource https://www.mathcentre.ac.uk/
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GRE Math Essentials - INEQUALITIES / Absolute Value / Modulus [#permalink]
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