Is x^2 + y^2 > 6?
(1) (x + y)^2 > 6
(2) xy = 2
OA : E
I can anyone please explain the answer for this ?
(1) (x + y)^2 > 6
(2) xy = 2
OA : E
I can anyone please explain the answer for this ?
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Thank you.. thats a really nice explanationayushiiitm wrote:stem 1 says (x+y)^2>6
i.e x^2+y^2+2xy>6
you may be tempted here to conclude that then (x+y)^2<6
think of the case now like this
x^2+y^2+2xy>6
then may be x^2+y^2 was equal to 4 and 2xy was equal to 3 so x^2+y^2 <6, even though x^2+y^2+2xy>6
or may be x^2+y^2 was equal to 8 and 2xy was equal to 1 so x^2+y^2 >6 even though x^2+y^2+2xy>6
insufficient
stem 2 is not sufficient as we see
using both stem 1 and 2
x^2+y^2+2xy>6....putting value of xy
x^2+y^2>4
so x^2+y^2 could be equal to 5
or x^2+y^2 could be equal to 10
so insufficient
(1) (x + y)² > 6 implies x² + y² + 2xy > 6naki009 wrote:Is x^2 + y^2 > 6?
(1) (x + y)^2 > 6
(2) xy = 2
OA : E
I can anyone please explain the answer for this ?
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