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# is K^2+K - 2 >0 1>K K>-1

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is K^2+K - 2 >0 1>K K>-1 [#permalink]

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24 Jul 2007, 20:40
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is K^2+K - 2 >0

1>K
K>-1
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25 Jul 2007, 00:05
ArvGMAT wrote:
is K^2+K - 2 >0

1>K
K>-1

I go for C.

stat1: k=0 no. K=-3 yes; insuff
stat2: k=0 no. k=4 yes; insuff.

stat1& 2 imply that k is either equal to 0 or a fraction. Either way, k^2+k-2 will always be less than 0. suff.
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25 Jul 2007, 01:59
ArvGMAT wrote:
is K^2+K - 2 >0

1>K
K>-1

k+2 * k-1 > 0

if k <1> -1, for most numbers it is positive but assume -1/2 to break it

For range -1 to + 1, it is always negative.

Ans : C
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25 Jul 2007, 04:00
(C) for me too

K^2+K - 2 > 0 ? (Not that 1 and -2 are the 2 roots of K^2+K - 2)
<=> (K+2)*(K-1) > 0 ?

For a*x^2 + b*x + c :
> the sign of the expression a*x^2 + b*x + c is the sign of a outside of roots (if they exist)
> the sign of the expression a*x^2 + b*x + c is the sign of -a between the roots (if they exist)

Here, we need to know whether:
o -2 < K < 1 and so (K+2)*(K-1) < 0
or
o K < -2 or K > 1 and so (K+2)*(K-1) > 0

From 1
1 > K

implies that K could be :
o K < -2 and so (K+2)*(K-1) > 0
o K = -2 and so (K+2)*(K-1) = 0
o 1 > K > -2 and so (K+2)*(K-1) < 0

INSUFF.

From 2
K > -1

implies that K could be :
o -1 < K < 1 and so (K+2)*(K-1) < 0
o K = 1 and so (K+2)*(K-1) = 0
o 1 > K and so (K+2)*(K-1) > 0

INSUFF.

Both (1) & (2)
-1 < K < 1
<=> -2 < -1 < K < 1.... Bingo, we are between the roots and (K+2)*(K-1) < 0

SUFF.
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25 Jul 2007, 09:03
ArvGMAT wrote:
is K^2+K - 2 >0

1. 1>K
2. K>-1

from i and ii: 1 > k > -1. so k is a both, -ve or +ve, fraction.

when k is a fraction, K^2 is a +ve and k could be -ve or +ve as above. so k^2 + k can not be > 2 and < 2. so its C.
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25 Jul 2007, 11:43
[quote="Fig"](C) for me too

K^2+K - 2 > 0 ? (Not that 1 and -2 are the 2 roots of K^2+K - 2)
<K> 0 ?

For a*x^2 + b*x + c :
> the sign of the expression a*x^2 + b*x + c is the sign of a outside of roots (if they exist)
> the sign of the expression a*x^2 + b*x + c is the sign of -a between the roots (if they exist)

Here, we need to know whether:
o -2 < K < 1 and so (K+2)*(K-1) < 0
or
o K <2> 1 and so (K+2)*(K-1) > 0

From 1
1 > K

implies that K could be :
o K <2> 0
o K = -2 and so (K+2)*(K-1) = 0
o 1 > K > -2 and so (K+2)*(K-1) <0> -1

implies that K could be :
o -1 < K < 1 and so (K+2)*(K-1) <0> K and so (K+2)*(K-1) > 0

INSUFF.

Both (1) & (2)
-1 < K < 1
<=> -2 < -1 < K < 1.... Bingo, we are between the roots and (K+2)*(K-1) <0> 0.[quote="Fig"]

Each stmnt is in-suff.
Combining both stmnts together, we have -1 < k < 1, which means K can be -0.5, 0 , 0.5

for -0.5, we get -1.5*1.5 which is < 0
for 0, we get -2 < 0
for 0.5, we get - .5 * 2.5 < 0

Hence C.
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25 Jul 2007, 23:38
St1:
If k = 1/2, eqn <0> 0. Insufficient.

St2:
If k = 0, eqn <0> 0. Insufficient.

Using both st1 and st2:

-1 < k < 1

K must be only a fraction, so eqn < 0. Sufficient.

Ans C
25 Jul 2007, 23:38
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