A) - 1
B) 0
C) abc
D) \[a+2b+c\]
E) abc
Correct Answer: B
Solution :
Here \[\alpha +\beta =-\frac{2b}{a}\] \[\gamma +\alpha =-\frac{c}{2b}\], \[\alpha +\delta =-\frac{a}{c}\] and \[\alpha \beta =\frac{c}{a},\,\alpha \gamma =\frac{a}{2b},\,\,\alpha \delta =\frac{2b}{c}\] Þ \[\alpha +\delta =-\frac{1}{\alpha \beta },\,{{\alpha }^{2}}\beta +\alpha \beta \delta =-1\] .....(i) Þ \[\alpha +\beta =-\frac{1}{\alpha \gamma },\,{{\alpha }^{2}}\gamma +\alpha \beta \gamma =-1\] .....(ii) Þ \[\alpha +\gamma =-\frac{1}{\alpha \delta },\,{{\alpha }^{2}}\delta +\alpha \beta \gamma =-1\] .....(iii) Solve equations (i), (ii) and (iii), we get \[\alpha =-1\] \[\alpha +{{\alpha }^{2}}=(-1)+{{(-1)}^{2}}\] = \[-1+1=0\].You need to login to perform this action.
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