A) \[\frac{1}{3}\]
B) 1
C) 3
D) \[\frac{2}{3}\]
Correct Answer: C
Solution :
\[\alpha +{{\alpha }^{2}}=-\frac{p}{3}\] and \[\alpha .{{\alpha }^{2}}=1.\] So\[\alpha =1,\]\[\omega ,\,{{\omega }^{2}}\]. If \[\alpha =1,p<0\]. If \[\alpha =\omega \,\,\text{or}\,\,{{\omega }^{2}},\] we have \[\omega +{{\omega }^{2}}=-\frac{p}{3}\,\] Þ \[-1=\frac{-p}{3}\,\,\,\,\,\Rightarrow \,\,p=3\].
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