A) \[\frac{1}{{{\alpha }^{3}}+\alpha \beta },\frac{1}{{{\beta }^{3}}+\alpha \beta }\]
B) \[\frac{1}{{{\alpha }^{2}}+\alpha \beta },\frac{1}{{{\beta }^{2}}+\alpha \beta }\]
C) \[\frac{1}{{{\alpha }^{4}}+\alpha \beta },\frac{1}{{{\beta }^{4}}+\alpha \beta }\]
D) None of these
Correct Answer: B
Solution :
Multiplying the second equation by \[\frac{c}{{{a}^{3}}}\], we get \[\frac{{{b}^{2}}{{c}^{2}}}{{{a}^{3}}}{{x}^{2}}-\frac{{{b}^{2}}c}{{{a}^{2}}}x+c=0\] \[\Rightarrow \,\,\,a{{\left( \frac{bc}{{{a}^{2}}}x \right)}^{2}}-b\left( \frac{bc}{{{a}^{2}}} \right)x+c=0\] \[\Rightarrow \,\,\,\frac{bc}{{{a}^{2}}}\,\,x=\alpha ,\,\,\beta \] \[\Rightarrow \,\,\,\,(\alpha +\beta )\alpha \beta x=\alpha ,\,\,\beta \] \[\Rightarrow \,\,x=\frac{1}{(\alpha +\beta )\alpha },\,\,\frac{1}{(\alpha +\beta )\beta }\]
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