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Is k^2 + k – 2 > 0?

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Bunuel
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Is k^2 + k – 2 > 0? [#permalink] New post   29 Oct 2017, 02:03
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A
B
C
D
E

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55% (hard)

Question Stats:

60% (01:53) correct 40% (01:36) wrong based on 110 sessions
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Is k^2 + k – 2 > 0?

(1) k < 1
(2) k > –1
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Official Answer
C
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saicharan1191
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Re: Is k^2 + k – 2 > 0? [#permalink] New post   29 Oct 2017, 03:30
For expression to be positive Either K > 1 or K < -2
Option C

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Re: Is k^2 + k – 2 > 0? [#permalink] New post   29 Oct 2017, 04:06
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Bunuel wrote:
Is k^2 + k – 2 > 0?

(1) k < 1
(2) k > –1



Hi...

Let's simplify the equation

\(k^2+k-2>0.........k^2+2k-k-2>0......k(k+2)-1(k+2)>0.....(k+2)(k-1)>0\)..
So both k+2 and k-1 should be of same SIGN..
So a)k between -2 and 1 will give us and NO
b) K<-2 and k>1 will give us and YES

Let's see the statements..
1) k<1...
It will give us falling in both above ranges, so ans can be yes or no
Insufficient
2) k>-1
Same as above
K=0 will give us no
K=5 will give YES
Insufficient

Combined
-1<k<1
This range falls within (a) above
ANS is always NO
Sufficient

C
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Re: Is k^2 + k – 2 > 0? [#permalink] New post   28 Jun 2018, 09:42
chetan2u wrote:
Bunuel wrote:
Is k^2 + k – 2 > 0?

(1) k < 1
(2) k > –1



Hi...

Let's simplify the equation

\(k^2+k-2>0.........k^2+2k-k-2>0......k(k+2)-1(k+2)>0.....(k+2)(k-1)>0\)..
So both k+2 and k-1 should be of same SIGN..
So a)k between -2 and 1 will give us and NO
b) K<-2 and k>1 will give us and YES

Let's see the statements..
1) k<1...
It will give us falling in both above ranges, so ans can be yes or no
Insufficient
2) k>-1
Same as above
K=0 will give us no
K=5 will give YES
Insufficient

Combined
-1<k<1
This range falls within (a) above
ANS is always NO
Sufficient

C


I don't understand this. Doesn't the formula simplify to the following
1) k^2 + k - 2 > 0
2) k^2 + k > 2
3) k(k+1) > 2
4) k > 2 and k > 1 => So K must be greater than 1 to be more than 0
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Re: Is k^2 + k – 2 > 0? [#permalink] New post   28 Jun 2018, 10:16
Expert Reply
surfingpirate wrote:
chetan2u wrote:
Bunuel wrote:
Is k^2 + k – 2 > 0?

(1) k < 1
(2) k > –1



Hi...

Let's simplify the equation

\(k^2+k-2>0.........k^2+2k-k-2>0......k(k+2)-1(k+2)>0.....(k+2)(k-1)>0\)..
So both k+2 and k-1 should be of same SIGN..
So a)k between -2 and 1 will give us and NO
b) K<-2 and k>1 will give us and YES

Let's see the statements..
1) k<1...
It will give us falling in both above ranges, so ans can be yes or no
Insufficient
2) k>-1
Same as above
K=0 will give us no
K=5 will give YES
Insufficient

Combined
-1<k<1
This range falls within (a) above
ANS is always NO
Sufficient

C


I don't understand this. Doesn't the formula simplify to the following
1) k^2 + k - 2 > 0
2) k^2 + k > 2
3) k(k+1) > 2
4) k > 2 and k > 1 => So K must be greater than 1 to be more than 0


By doing this you are missing out on the negative values...
It is not mentioned that k is positive..

For example if k =-3...
K(k+1)>2...
-3(-3+1)=-3*-2=6>2
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Re: Is k^2 + k – 2 > 0? [#permalink] New post   28 Jun 2018, 12:57
C[/quote]

I don't understand this. Doesn't the formula simplify to the following
1) k^2 + k - 2 > 0
2) k^2 + k > 2
3) k(k+1) > 2
4) k > 2 and k > 1 => So K must be greater than 1 to be more than 0[/quote]

By doing this you are missing out on the negative values...
It is not mentioned that k is positive..

For example if k =-3...
K(k+1)>2...
-3(-3+1)=-3*-2=6>2[/quote]
I see. I tested the negative case so ending up figuring that out. How would you represent the negative instance algebraically? Like k(k+1) > 2 (positive) and k(k+1) < -2 (negative)?
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Re: Is k^2 + k – 2 > 0? [#permalink] New post   28 Jun 2018, 22:39
Expert Reply
surfingpirate wrote:
C

I don't understand this. Doesn't the formula simplify to the following
1) k^2 + k - 2 > 0
2) k^2 + k > 2
3) k(k+1) > 2
4) k > 2 and k > 1 => So K must be greater than 1 to be more than 0
Quote:
By doing this you are missing out on the negative values...
It is not mentioned that k is positive..

For example if k =-3...
K(k+1)>2...
-3(-3+1)=-3*-2=6>2

I see. I tested the negative case so ending up figuring that out. How would you represent the negative instance algebraically? Like k(k+1) > 2 (positive) and k(k+1) < -2 (negative)?


hi..

you are not taking k(k+1)<-2..

you are working on the same equation..
k(k+1)>2..
but here k can take both negative values and positive values...

if k>0, k will have to be greater than 1... so k>1 fulfills the criteria
if k<0, k will have to be less than -2.... so k<-2 also fulfills the criteria
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Is k^2 + k – 2 > 0? [#permalink] New post   01 Sep 2019, 04:34
This is a yes/no based question. As it's an inequality, we can begin by simplifying the question.

Question: Is k^2 + k – 2 > 0?

This can be rewritten to say: Is k^2 + k > 2?

As we have no further information, let's get into the options.

(1) k < 1

Possible values for k here are 0, -1, -2, -3...

When k=0, the answer to our reframed question is NO
When k=-3, the answer to our reframed question is YES

Insufficient.

Scratch out A and D.

(2) k > -1

Possible values for k here are 0, 1, 2, 3...

When k=0, the answer to our reframed question is NO
When k=3, the answer to our reframed question is YES

Insufficient.

Scratch out B.

(1) & (2) combined tell us that -1 < k < 1, which means k=0. SUFFICIENT.

C is our answer.

Hope this helps!

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Is k^2 + k – 2 > 0? [#permalink]
01 Sep 2019, 04:34
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