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SVP
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Let us refresh modulus that you all hate so much:  [#permalink] New post 04 Jul 2003, 07:27
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A
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C
D
E

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||C–1|+|C||=|C+3|–C
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 [#permalink] New post 04 Jul 2003, 12:45
Hi, Stolyar

I think the only answer is 2, but you are right, I don┬┤t like very much this kind of questions, so I may overlook something.
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 [#permalink] New post 05 Jul 2003, 06:12
There is a need to square both parts, open up all brakets and moduls, and then check each root. I also have 2 only.
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 [#permalink] New post 07 Jul 2003, 05:46
The key to solving this type of problem is to solve a different equation in every important interval. As a hint, I will tell you that the important intervals are -inf to -3, -3 to 0, 0 to 1, 1 to inf.

BTW, have you tried C = -1?
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AkamaiBrah
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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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Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE
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 [#permalink] New post 07 Jul 2003, 05:53
IMO, you don't need to square anything

The key to solving this type of problem is to try and solve a different equation (without any absvalue signs) in every important interval. As a hint, I will tell you that the important intervals are -inf to -3, -3 to 0, 0 to 1, 1 to inf. . (Note that in each interval, the sign of the expression in each absvalue sign will not change).

BTW, have you tried C = -1?

BTW again, there is no reason for the pair of absvalue signs surrounding the LHS of the equation because the addition of two absolute values will always be non-negative.
_________________

Best,

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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 [#permalink] New post 07 Jul 2003, 06:03
AkamaiBrah wrote:
IMO, you don't need to square anything

The key to solving this type of problem is to try and solve a different equation (without any absvalue signs) in every important interval. As a hint, I will tell you that the important intervals are -inf to -3, -3 to 0, 0 to 1, 1 to inf. . (Note that in each interval, the sign of the expression in each absvalue sign will not change).

BTW, have you tried C = -1?

BTW again, there is no reason for the pair of absvalue signs surrounding the LHS of the equation because the addition of two absolute values will always be non-negative.



sure Akam we know this method, but are too lazy to do all the stuff.
  [#permalink] 07 Jul 2003, 06:03
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