A) Zero
B) Positive
C) Negative
D) None of these
Correct Answer: B
Solution :
\[\alpha +\beta =-b/a,\alpha \beta =c/a\] \[(1+\alpha +{{\alpha }^{2}})(1+\beta +{{\beta }^{2}})\] \[=1+(\alpha +\beta )+({{\alpha }^{2}}+{{\beta }^{2}})+\alpha \beta +\alpha \beta (\alpha +\beta )+{{\alpha }^{2}}{{\beta }^{2}}\] \[=1+(\alpha +\beta )+{{(\alpha +\beta )}^{2}}-\alpha \beta +\alpha \beta (\alpha +\beta )+{{(\alpha \beta )}^{2}}\] \[=1-\frac{b}{a}+\frac{{{b}^{2}}}{{{a}^{2}}}-\frac{c}{a}+\left( \frac{c}{a} \right)\,\,\left( -\frac{b}{a} \right)+\frac{{{c}^{2}}}{{{a}^{2}}}\] \[=({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)/{{a}^{2}}\] \[=[(a-{{b}^{2}})+{{(b-c)}^{2}}+{{(c-a)}^{2}}]/2{{a}^{2}}\] which is positive. Trick: It is almost clear that for every different values of a,b,c the function is zero, positive or negative. Therefore let \[a=1,b=0,c=-4\] so that \[\alpha =2,\ \beta =-2\]. Obviously the expression has positive value.
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