A) \[\frac{3}{2}\]
B) \[\frac{2}{3}\]
C) \[-\frac{3}{2}\]
D) \[-\frac{2}{3}\]
Correct Answer: C
Solution :
Let two numbers be x and y. \[x+y=3\] \[{{x}^{2}}+{{y}^{2}}=12\] \[\Rightarrow \]\[{{(x+y)}^{2}}={{(3)}^{2}}\]\[\Rightarrow \]\[{{x}^{2}}+{{y}^{2}}+2xy=9\] \[\Rightarrow \]\[12+2xy=9\]\[\Rightarrow \]\[2xy=-\,\,3\] \[\therefore \]\[xy=-\frac{3}{2}\]
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