A) < 0
B) > 0
C) = 0
D) None of these
Correct Answer: B
Solution :
| \[\alpha +\beta =-\,b/a\] & \[\alpha \beta =c/a\] Then \[(1+\alpha +{{\alpha }^{2}})\,\,(1+\beta +{{\beta }^{2}})\] |
| \[=1+(\alpha +\beta )+(\alpha \beta )+{{(\alpha \beta )}^{2}}+({{\alpha }^{2}}+{{\beta }^{2}})+\alpha \beta (\alpha +\beta )\]\[=1-\frac{b}{a}+\frac{c}{a}+\frac{{{c}^{2}}}{{{a}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}}-\frac{2c}{a}-\frac{bc}{{{a}^{2}}}\] |
| \[=\frac{{{a}^{2}}+ab+ac+{{c}^{2}}+{{b}^{2}}-2ac-bc}{{{a}^{2}}}\] |
| \[=\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac}{{{a}^{2}}}\] |
| \[=\frac{(2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab-2bc-2ac)}{2{{a}^{2}}}\] |
| \[=\frac{{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(a-c)}^{2}}}{2{{a}^{2}}}>0\] |
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