my approach...lemme kno if it is correct
i changed the stem to be:
K (k-1) > 2
Statement 1:
K < 1
is k = 0 then 0>2...not tru
if k = -4 then this is > 2..so tru..
there INSUFF
Statement 2:
K > -1
Same thing...INSUFF
Together...
0 is not true...there..suff together
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K
Source: Beat The GMAT — Data Sufficiency |
Is k^2 + k - 2 > 0 ?
To get the boundary points set it equal to 0.
(k+2)(k-1)=0
k=-2
k=1
<-----\-2\------------------\1\------>
when k is less than -2 the answer will be positive.
when k is between -2 and 1 the answer will be negative.
when k is greater than 1, the answer will be positive.
So are asking is k>1 or is k<-2
1. k < 1
k can be -3 and the answer will be yes
k can be 0 and the answer will be no
Insuff
2. k > -1
k can be 0 and the answer will be no
k can be 10 and the answer will be yes
Insuff
1 and 2)
-1<k<1
So now we know that it falls between the interval where k is always going to be negative (between -2 and 1). Therefore, we have a definitive no to the question.
Suff
C
To get the boundary points set it equal to 0.
(k+2)(k-1)=0
k=-2
k=1
<-----\-2\------------------\1\------>
when k is less than -2 the answer will be positive.
when k is between -2 and 1 the answer will be negative.
when k is greater than 1, the answer will be positive.
So are asking is k>1 or is k<-2
1. k < 1
k can be -3 and the answer will be yes
k can be 0 and the answer will be no
Insuff
2. k > -1
k can be 0 and the answer will be no
k can be 10 and the answer will be yes
Insuff
1 and 2)
-1<k<1
So now we know that it falls between the interval where k is always going to be negative (between -2 and 1). Therefore, we have a definitive no to the question.
Suff
C
Last edited by dmateer25 on Wed May 20, 2009 8:17 am, edited 1 time in total.
I think it's better to notice that you're dealing with a quadratic equation here.
k^2 + k - 2 = (k + 2)(k - 1).
Then you just apply that age-old rule:
When we have a quadratic equation:
ax^2 + bx + c
then the sign of it will be a's sign sign when not between the roots and the opposite of a's sign between the roots
In this particular case, a = 1, so a is positive. The roots will be -2 and 1, so if k is in (-2, 1), the sign of the quadratic will be minus. When k is (-infinite, -2) U (1, infinite), then the sign will be plus.
This is why you need both stmts to solve this one: if k is between -1 and 1, then it will also be between -2 and 1, so you can safely say that the quadratic is negative.
k^2 + k - 2 = (k + 2)(k - 1).
Then you just apply that age-old rule:
When we have a quadratic equation:
ax^2 + bx + c
then the sign of it will be a's sign sign when not between the roots and the opposite of a's sign between the roots
In this particular case, a = 1, so a is positive. The roots will be -2 and 1, so if k is in (-2, 1), the sign of the quadratic will be minus. When k is (-infinite, -2) U (1, infinite), then the sign will be plus.
This is why you need both stmts to solve this one: if k is between -1 and 1, then it will also be between -2 and 1, so you can safely say that the quadratic is negative.
Is k (k+1) > 2
Together:
-1<k<1
k+1 will never be bigger than 2 and if u r multiplying it by a fraction then it will greater than 2 if k is between -1 and 1 (since it's like dividing k+1 by something.
C
Together:
-1<k<1
k+1 will never be bigger than 2 and if u r multiplying it by a fraction then it will greater than 2 if k is between -1 and 1 (since it's like dividing k+1 by something.
C
