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# Math Basic Question

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Joined: 07 Jul 2003
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Schools: Haas, MFE; Anderson, MBA; USC, MSEE

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07 Jun 2009, 21:11
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mxb908 wrote:
Now when i open the equation i get (x-1)*(x+1)<0

thus implying that... x-1<0 or x+1<0
this implies the two equations being: x<1 or x<-1

Your inference in not correct. (x-1)*(x+1) < 0 iff (x-1) and (x+1) are of opposite signs.

x < 1 doesn't always work: counterexample x = -3
x < -1 never works.

for the inequality to work, both x<1 AND x>-1; or |x| < 1
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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Joined: 03 Jun 2009
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08 Jun 2009, 09:56
(x^2-1) < 0

(x+1)(x-1) < 0;

Product of 2 numbers is < 0, only when one number is +ve and the other -ve.

Therefore from the inequality, the Solution set shall be

(x+1) < 0 & (x-1) > 0; or ; (x+1) > 0 & (x-1) < 0;

=> x < -1 & x > 1 ; or ; x > -1 & x < 1
=> X takes all values except those between (-1 & 1) ; or X takes all values between (-1 & 1)
If we check with any value in First the Solution set ie., 2 or -2 and substitute in x^2-1 .,ie 4-1 which is not less than zero. Therefore this is not the Solution Set.

If we check with any value in Second the Solution set ie., 0 or 0.5 and substitute in x^2-1 .,ie 0-1=-1 or 0.25-1=-0.75 which is less than zero. Therefore this is the Solution Set.
Re: Math Basic Question   [#permalink] 08 Jun 2009, 09:56
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