A) 0
B) positive
C) negative
D) none of these
Correct Answer: B
Solution :
Since, \[\alpha ,\beta \]are the roots of the equation\[a{{x}^{2}}+bx+c=0,\]then \[\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}\] So, \[(1+\alpha +{{\alpha }^{2}})(1+\beta +{{\beta }^{2}})\] \[=1+\beta +{{\beta }^{2}}+\alpha +\alpha \beta +\alpha {{\beta }^{2}}+{{\alpha }^{2}}+{{\alpha }^{2}}\beta +{{\alpha }^{2}}{{\beta }^{2}}\] \[=1+(\alpha +\beta )+\alpha \beta +{{\alpha }^{2}}{{\beta }^{2}}+\alpha {{\beta }^{2}}\] \[+\,{{\alpha }^{2}}\beta +{{\alpha }^{2}}+{{\beta }^{2}}\] \[=1+(\alpha +\beta )+\alpha \beta +{{\alpha }^{2}}{{\beta }^{2}}+\alpha \beta (\alpha +\beta )\] \[+\,{{(\alpha +\beta )}^{2}}-2\alpha \beta \] \[=1-\frac{b}{a}+\frac{c}{a}+\frac{{{c}^{2}}}{{{a}^{2}}}+\frac{c}{a}\left( -\frac{b}{a} \right)+{{\left( -\frac{b}{a} \right)}^{2}}-2\left( \frac{c}{2} \right)\] \[=1-\frac{b}{a}+\frac{{{c}^{2}}}{{{a}^{2}}}-\frac{bc}{{{a}^{2}}}-\frac{c}{a}\] \[=\frac{{{a}^{2}}-ab+{{c}^{2}}-bc+{{b}^{2}}-ac}{{{a}^{2}}}\] Which is positive for all values of a, b and c.
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