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

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Schools: Dartmouth College
Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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New post 28 Jun 2018, 09:21
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dave13 wrote:
hey GMATGuruNY great explanation ! :-) i have just one question

Algebra:
\(|x| < x^2\) (as you see \(x^2\) is without bars / brackets )

\(|x| < |x| * |x|\) Here you rewrite \(x^2\)as |x| * |x| and not as x*x , can you please explain why ? :-)

many thanks ! :)


\(x^2 = x * x = |x| * |x|\)

For example:
\(3^2 = 3 * 3 = |3| * |3| = 9\)
\((-3)^2 = -3 * -3 = |-3| * |-3| = 9\)

Given inequality:
\(|x| < x^2\)

Here, I chose to replace x^2 with |x| * |x| so that |x| would appear on both sides, enabling me to divide both sides by |x|:
\(|x| < x^2\)
\(|x| < |x| * |x|\)
\(\frac{|x|}{|x|} < |x| * \frac{|x|}{|x|}\)
\(1 < |x|\)
\(|x| > 1\)
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Re: If |x| < x^2, which of the following must be true ?  [#permalink]

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New post 02 Sep 2018, 17:57
Bunuel wrote:
Skywalker18 wrote:
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 \(x>1\).


Bunuel can u show how you reached 1<|x|1<|x|
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Re: If |x| < x^2, which of the following must be true ? &nbs [#permalink] 02 Sep 2018, 17:57

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