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Source: — Data Sufficiency |

by tdadic84 » Wed May 20, 2009 7:59 am
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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by dmateer25 » Wed May 20, 2009 8:17 am
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
Last edited by dmateer25 on Wed May 20, 2009 8:17 am, edited 1 time in total.
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by DanaJ » Wed May 20, 2009 8:17 am
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.
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by iamcste » Wed May 20, 2009 10:54 am
k^2 + k - 2 > 0

Is K >-2 one of roots or K <-2 the root for this inequality
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by dmateer25 » Wed May 20, 2009 3:30 pm
iamcste wrote:k^2 + k - 2 > 0

Is K >-2 one of roots or K <-2 the root for this inequality
K < -2 is on of the roots
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by cramya » Wed May 20, 2009 3:47 pm
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
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