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14 Jun 2008, 03:10
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Is k^2 + k - 2 > 0
1. 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. -2-1 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?
1. 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. -2-1 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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14 Jun 2008, 05:48
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ritula
k should be (-∞,-2)&(1,+∞) but your answer is correct
14 Jun 2008, 05:52
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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.
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.
