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# Is k^2 + k - 2 > 0 1, k<1 2, k>-1

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Is k^2 + k - 2 > 0 1, k<1 2, k>-1 [#permalink]

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22 Sep 2005, 03:54
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Is k^2 + k - 2 > 0

1, k<1
2, k>-1
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22 Sep 2005, 04:22
macca wrote:
Is k^2 + k - 2 > 0

1, k<1
2, k>-1

STATEMENT #1:
k=0, false
k=-5, true
Insufficient

STATEMENT #2:
k=0, false
k=+5, true
Insufficient

COMBINED -1<k<1 :
k=-0.5, false
k=0, false
k=+0.5, false

C
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22 Sep 2005, 04:42

Statement 1 is sufficient any number we pick for k<1 gives us k^2>0 and k-2 is always <0 for any k<1 so the answer to the question is k^2+k-2>0 is NO. therefore statement 1 is sufficient.

Statement 2 is insufficient , given k>-1 if we pick 0 or 1 or a 0<k<1 we get k^2+k-2 always negative, however if we pick a number k>2 we get k^2+k-2 positive => statement 2 is insufficient to answer the quesiton. Answer A
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22 Sep 2005, 05:54
It is ok to use examples to show insufficient.
But you cannot deduce sufficiency from just picking 3 examples.
You have to use analysis to be sure you are right.

k^2 + k - 2 = (k-1)(k+2)

That tells you the function crosses the line at k=1, k=-2
and negative between those values.

So if (1) and (2) are true we are within -2<k<1

C
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22 Sep 2005, 06:57
Just rearrange the question stem to get k^2 + k > 2

Statement 1 says that k<1.

If k is between 0 and 1, then there's no way that k^2 (which is between 0 and 1) plus k (also between 0 and 1) can be greater than 2. So it woudl be less than 2.

If k is less than 0, then k^2 can be greater than 1, but k cannot be. However, take the example of k being equal to -5. If k=-5, then k^2 is 25, and 25+(-5) = 20, which is greater than 2.

Therefore, statement 1 is insufficient, so the answer cannot be A.

Similar reasoning with statement 2. If k is greater than -1, clearly it can be a number between 0 and 1, which would make the question stem false, or it can be something like 5, which would make the question stem true.

However, taken together, we get that k is between -1 and 1. No value between -1 and 1 would make the question stem true. We already saw that it's false for numbers between 0 and 1. For numbers between -1 and 0, it's also false for similar reasons; the square of k will be a positive number between 0 and 1, and k will be a negative number with a greater absolute value than k^2. So adding them together yields a negative, which is less than 2.

Therefore, taken together, we can answer that no, the expression is not greater than 0.

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22 Sep 2005, 07:06
I would go with c on this ....

k^2+k-2 > 0?

(1) K <1

well lets say K=-3

9-3-2=4 which is greater than 0

say K=1/2

1/4+1/2-2 is <0

(2) K>-1

say K=3

9+3-2 is >0

say K=-1/2

1/4-1/2-2 is <0

combining them is sufficient....

C it is

Last edited by FN on 22 Sep 2005, 07:11, edited 1 time in total.
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22 Sep 2005, 07:48
If a question with an inequality in like this looks too hard, try considering it with an = instead and see what you can deduce.

It is all about finding the roots / factorising expressions like :

x^2 + ax + b = c
d x^2 + ex + f = 0

then you can draw the curve -
do it on paper, if you can't visualise it in your head.

Then to work out the area with "<" or ">" instead of "="
it is either above or below the curve you have drawn.
22 Sep 2005, 07:48
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