K^2 + K -2 >0

[ManSab]

Is K^2 + K-2 >0 ?
1) k<1
2) k>-1

[welcome]

K^2+k-2>0

(K-1)(k+2)>0
so eithere K>1 & K>-2
Or k<1 & k<-2

Plot the options on Number line

I) no suff
II) no suff

I+II = Suff Ans C.

[x2suresh]

agree with C

[sestri]

IMO A
'K' is not an integer as per the Q stem
so assuming the max val of 'K' to be 0.999 and substituting it we get the value of K^2+K- 2 as as a negative decimal less than 0.
so 'A' is sufficient.

[luckylucky]

Is K^2 + K-2 >0 ?
1) k<1
2) k>-1

The above equation can be deduced to (k + 2) ( k -1) > 0

This is satisfied by k > -2 and K > 1 i.e k > 1

1) k < 1

plug in 0

-2 > 0 not satisfied

plug in -10

100 - 10 - 2 > 0

88 > 0 satisfies

Hence A alone is not sufficient

2) k > -1

plug in 0

-2 > 0 doesnt satisfy

plug in 10

100 + 10 - 2 > 0

108 > 0 satisfies

Hence B alone is not sufficient


Combinig both A and B

-1 < k < 1

plug in 0

- 2 > 0 doesnt satisfy

plug in 0.9

0.81 + 0.9 - 2 > 0

-1.1 > 0 doesnt satisfy

We can be able to answer the question only when both the statements are combined

Hence C

[mjjking]

agrre with C. when -1<k<1 we can answer to the answer with a definite "no", hence suff.