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# Is K^2+K - 2 > 0 A) K < 1 B) K > -2

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Director
Joined: 03 Jul 2003
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Is K^2+K - 2 > 0 A) K < 1 B) K > -2 [#permalink]

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29 Jun 2004, 19:36
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Is K^2+K - 2 > 0

A) K < 1
B) K > -2
Senior Manager
Joined: 21 Mar 2004
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29 Jun 2004, 20:07
Let E = k^2+k-2 = (k-1)(k+2)

for E>0 means k-1 and k+2 have to be of same sign...ie either both positive or both negative.

Case 1. Both k-1 and k+2 are positive

k-1>0 , or k>1
and
k+2>0 or K<-2

Case 2. Both k-1 and k+2 are negative

k-1<0 , or k<1
and
k+2<0 or K>-2

Now look at the question stem.
A. k<1
If E>0 , we need more info.......so A is insufficient.

B. k>-2
If E>0 , we need more info.......so B is insufficient.

taking both A and B , it boils down to Case 2.

Therefore answer is C.

- ash
_________________

ash
________________________
I'm crossing the bridge.........

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30 Jun 2004, 04:54
yeah (C) it is.
(k+2)(k-1)>0
implies
k+2 > , k-1>0 ---> k>-2 K>1 ---> k>-2 (extreme condition)
OR
k+2<0, k-1<0 ---> k<-2 k<1 ----> k<1 (extreme condition)

Hence -2<k<1

Hence (C)
30 Jun 2004, 04:54
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# Is K^2+K - 2 > 0 A) K < 1 B) K > -2

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