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y = x^2 + 6x 7, x= 1). x is positive integer 2). y^2=y

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Manager
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y = x^2 + 6x 7, x= 1). x is positive integer 2). y^2=y [#permalink] New post 12 Sep 2006, 18:50
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C
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E

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y = x^2 + 6x – 7, x=

1). x is positive integer

2). y^2=y
Manager
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Joined: 03 Jul 2006
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Kudos [?]: 9 [0], given: 0

 [#permalink] New post 12 Sep 2006, 19:11
why not 'A'

from the stem, x = -7 or 1

and A says x has to be +ve then x must be 1 ?
VP
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 [#permalink] New post 12 Sep 2006, 19:14
rkatl wrote:
why not 'A'

from the stem, x = -7 or 1

and A says x has to be +ve then x must be 1 ?


here the eq is not =0 but equal to y.

How can you deduce anything from 1 only?

either it has to be C or E
CEO
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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 [#permalink] New post 12 Sep 2006, 20:25
C

St1: clearly INSUFF

St2: y = 0 or y = 1.
then there are two equations
x^2 + 6x-7 = 0 gives us x = 1 or -7 and
x^2+6x-8 = 0 gives us two non-integer roots.: INSUFF

Together:
x = 1.: SUFF
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

SVP
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 [#permalink] New post 12 Sep 2006, 23:27
the key here is statment b

y^2 =y thus y is either = {-1,1,0}....insuff

both together

x is positive and y = x^2 + 6x – 7

if y = -1 thus x^2 + 6x-6 = 0....this equation has two non intiger answers one +ve and one -ve ( test with (b^2 - 4ac)....insuff

y = 1 thus x^2 + 6x-8 = 0 this equation has non intiger answers one +ve and one -ve....insuff

if y = 0 the equation has two intiger roots and as it is mentioned in one that x is positive intiger

thus C is my answer
  [#permalink] 12 Sep 2006, 23:27
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y = x^2 + 6x 7, x= 1). x is positive integer 2). y^2=y

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