A) 2
B) -2
C) 1
D) -1
Correct Answer: A
Solution :
Let \[\alpha ,\beta \] be the roots of the equation \[a{{x}^{2}}+bx+c=0\]then\[\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}\]. Given \[\alpha +\beta =\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}\,\,\Rightarrow -\frac{b}{a}=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{{{\alpha }^{2}}{{\beta }^{2}}}\] Þ \[-\frac{b}{a}\frac{{{c}^{2}}}{{{a}^{2}}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \]Þ \[-\frac{b{{c}^{2}}}{{{a}^{3}}}=\frac{{{b}^{2}}}{{{a}^{2}}}-\frac{2c}{a}\] Þ \[\frac{2}{a}=\frac{{{b}^{2}}}{{{a}^{2}}c}+\frac{bc}{{{a}^{3}}}\Rightarrow 2=\frac{{{b}^{2}}}{ac}+\frac{bc}{{{a}^{2}}}\].
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