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Is my approach right?

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VP
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Joined: 18 May 2008
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Is my approach right? [#permalink] New post 14 Jun 2008, 02:10
Is k^2 + k - 2 > 0

1. k < 1
2. k > -1
In solving above question, i followed the following approach:
k^2 + k - 2=(k-1)(k+2)
In order to above equation to be greater than 0, k should lie between -2 and 1
i.e. -2<k<1. Now
(1) says k<1 but nothing abt -2. SO Insufficient
(2) k>-1 so must be greater than -2 as well but says nothing abt 1. so Insufficient
Combining 1 and 2, we get the answer. So C
Is my approach correct? is there any other way of soving such type of problems?
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CEO
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Concentration: Entrepreneurship, Other
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Re: Is my approach right? [#permalink] New post 14 Jun 2008, 04:48
Expert's post
ritula wrote:
Is k^2 + k - 2 > 0

1. k < 1
2. k > -1
In solving above question, i followed the following approach:
k^2 + k - 2=(k-1)(k+2)
In order to above equation to be greater than 0, k should lie between -2 and 1
i.e. -2<k<1. Now
(1) says k<1 but nothing abt -2. SO Insufficient
(2) k>-1 so must be greater than -2 as well but says nothing abt 1. so Insufficient
Combining 1 and 2, we get the answer. So C
Is my approach correct? is there any other way of soving such type of problems?


k should be (-∞,-2)&(1,+∞) but your answer is correct :-D
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Director
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Re: Is my approach right? [#permalink] New post 14 Jun 2008, 04:52
Almost there
For (k-1)(k+2) to be +ve following things should be kept in mind

1) For All +ve K, (k+2) will be always +ve. So for (k-1) to be +ve we can write
k - 1 > 0 => k > 1
1) For All -ve K, (k-1) will be always -ve. So for (k+2) to be -ve we can write
k + 2 < 0 => k < -2
So Range of K = (-Infinity to -2) and (1 to Infinity)

Between -2 < k < 1 will be negative.
But after combining the reasoning given by you still holds.
Director
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Re: Is my approach right? [#permalink] New post 14 Jun 2008, 12:34
questions such as these, I find that sketching a quick graph after factorization will help you

attached is my sketch and its evident that C is the answer based on the sketch
Attachments

para.jpg
para.jpg [ 3.72 KiB | Viewed 528 times ]

Re: Is my approach right?   [#permalink] 14 Jun 2008, 12:34
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