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# Is k^2 +k -2 > 0 ?

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Is k^2 +k -2 > 0 ? [#permalink]

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12 Feb 2013, 15:52
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Is k^2 + k - 2 > 0 ?

(1) k < 1
(2) k < -2
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Feb 2013, 15:57, edited 1 time in total.
Edited the question.

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Re: Is k^2 +k -2 > 0 ? [#permalink]

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12 Feb 2013, 16:11
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Is k^2 + k - 2 > 0 ?

First let's see for which ranges of k the given inequality holds true: $$k^2 + k - 2 > 0$$ --> $$(k+2)(k-1)>0$$ --> roots are -2 and 1 --> ">" sign indicates that the solution lies to the left of a smaller root and to the right of a larger root (check here for this technique: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476). Thus the given inequality holds true for: $$k<-2$$ and $$k>1$$.

So, the question asks whether $$k<-2$$ or $$k>1$$.

(1) k < 1. Not sufficient.
(2) k < -2. Sufficient.

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Re: Is k^2 +k -2 > 0 ? [#permalink]

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12 Feb 2013, 16:11
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rephrasing the statement we have

$$(x + 2) (x - 1) > 0$$ ====> $$x > -2$$ AND $$x > 1$$

on a number line we have

-------------------0 ----- $$1$$ +++++++

----- $$- 2$$ +++++0++++++++++++

So from 1) we have $$x > 1$$ so 2 intervals : one positive and one negative. two values. not suff

from2) $$x < - 2$$ the only values are positive. so is suff.

B is the answer

Bunuel ?? correct my approach, pretty straightforward
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Re: Is k^2 +k -2 > 0 ? [#permalink]

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14 Jan 2015, 09:58
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Re: Is k^2 +k -2 > 0 ? [#permalink]

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14 Jan 2015, 17:54
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Expert's post
Hi All,

If you spot the Quadratic expression in the prompt, then you can use an Algebra approach to get to the correct answer. This question can also be solved with a combination of TESTing VALUES and Number Properties.

We're asked if K^2 + K - 2 > 0. This is a YES/NO question

Fact 1: K < 1

IF....
K = 0
0^2 + 0 - 2 = -2 and the answer to the question is NO.

IF...
K = -3
(-3)^2 -3 - 2 = 4 and the answer to the question is YES.
Fact 1 is INSUFFICIENT

Fact 2: K < -2

With this Fact, we have an interesting "limit" issue.

Even though it's not permitted....IF K = -2, then
(-2)^2 - 2 + 2 = 0 which is NOT > 0

As K becomes "more negative" (re. -2.1, -3, -100, etc.)....
K^2 creates a "bigger positive" than (+K - 2) creates a "negative"

eg.
K = -3
(-3)^2 = +9 vs. (-3 - 2)

K = -2.1
(-2.1)^2 = +4.41 vs. (-2.1 - 2)
Etc.

Thus, the result of the calculation will ALWAYS be greater than 0 and the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT.

[Reveal] Spoiler:
B

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Re: Is k^2 +k -2 > 0 ? [#permalink]

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19 May 2016, 04:52
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Re: Is k^2 +k -2 > 0 ? [#permalink]

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19 May 2016, 05:00
anartey wrote:
Is k^2 + k - 2 > 0 ?

(1) k < 1
(2) k < -2

For the equation k^2+k-2 greater than 0, K should either be less than -2 or should be greater than 1 (explaination: k^2+K-2 can be written as (k+1)(k-2) and for this product to be positive either K+1 and K-2 both should be positive or negative)

Coming to options:

1. K<1 is not sufficient because it includes the range k<-2 and some numbers beyond it
2. k<-2 is sufficient (as explained above)

Hope this helps!

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Re: Is k^2 +k -2 > 0 ? [#permalink]

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19 May 2016, 09:51
Statement 1 gives solutions both > or < 0

Statement 2 is enough to reach the solution.

B is the answer
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Re: Is k^2 +k -2 > 0 ? [#permalink]

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29 Mar 2017, 12:44
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This is a DS question right? Not a PS question? I got confused because this is in the PS section

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Re: Is k^2 +k -2 > 0 ? [#permalink]

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29 Mar 2017, 12:50
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Top Contributor
joondez wrote:
This is a DS question right? Not a PS question? I got confused because this is in the PS section

Yes, it's a DS question.

Cheers,
Brent
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Re: Is k^2 +k -2 > 0 ? [#permalink]

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29 Mar 2017, 21:32
joondez wrote:
This is a DS question right? Not a PS question? I got confused because this is in the PS section

Moved to DS forum. Thank you for noticing.
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Re: Is k^2 +k -2 > 0 ? [#permalink]

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29 Mar 2017, 21:44
Sta: 1 gives solutions both > or < 0 when we substitute values. if we put -2 = k we will get 0
and if we put -2 as k we will get 0
Not sufficient

Sta: 2 is enough to reach the solution. Try with a value

B
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Re: Is k^2 +k -2 > 0 ? [#permalink]

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04 Apr 2017, 16:02
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anartey wrote:
Is k^2 + k - 2 > 0 ?

(1) k < 1
(2) k < -2

We need to determine whether k^2 + k - 2 > 0.

Factoring the expression, we have:

(k + 2)(k - 1) > 0

In order for (k + 2)(k - 1) to be greater than zero, either both (k + 2) and (k - 1) must be greater than zero or both (k + 2) and (k - 1) must be less than zero.

If both (k + 2) and (k - 1) are greater than zero, then k must be greater than 1. Similarly, if both (k + 2) and (k - 1) are less than zero, then k must be less than -2. That is:

k > 1 or k < -2

Statement One Alone:

k < 1

The information in statement one is not sufficient to answer the question. For example, if n = -1, then (k + 2)(k - 1) is less than zero; however, if k = -4, then (k + 2)(k - 1) is greater than zero.

Statement Two Alone:

k < -2

Since k is less than -2, regardless of which values we select for k, both (k + 2) and (k - 1) will always be negative. Thus, (k + 2)(k - 1) > 0. We have sufficient information to answer the question.

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Re: Is k^2 +k -2 > 0 ? [#permalink]

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06 Apr 2017, 09:40
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

For inequality questions we're not going to simply substitute all the values, but we'll be comparing the range of the original question and the range of conditions and investigate if the range of the question includes that of conditions. If it does, then the condition is sufficient.

anartey wrote:
Is $$k^2 + k - 2 > 0$$?

(1) $$k < 1$$
(2) $$k < -2$$

The first step is simplifying the original question.
$$k^2 + k - 2 > 0$$
$$(k+2)(k-1)>0$$
$$k < -2$$ or $$k > 1$$

Condition (1)
The range of the question, "$$k < -2$$ or $$k > 1$$" does not include the range of condition (1), "$$k < 1$$".

Thus this is NOT sufficient.

Condition (2)
The range of the question, "$$k < -2$$ or $$k > 1$$" includes the range of condition (2), "$$k < -2$$".

Normally for cases where we need 1 more equation, such as original conditions with "1 variable", or "2 variables and 1 equation", or "3 variables and 2 equations", we have 1 equation each in both 1) and 2). Therefore D has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) separately. Here, there is 59 % chance that D is the answer, while A or B has 38% chance. There is 3% chance that C or E is the answer for the case. Since D is most likely to be the answer according to DS definition, we solve the question assuming D would be our answer hence using 1) and 2) separately. Obviously there may be cases where the answer is A, B, C or E.
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Re: Is k^2 +k -2 > 0 ? [#permalink]

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06 Apr 2017, 22:48
Is k^2 + k - 2 > 0 ?

(1) k < 1
k= 1/2. NS

k = -4, Suff.
(2) k < -2

k< -2 . clearly suff.
k= -3. the exp = 4.

Ans. B
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Re: Is k^2 +k -2 > 0 ?   [#permalink] 06 Apr 2017, 22:48
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