A) \[\frac{{{b}^{2}}-2ac}{{{a}^{2}}}\]
B) \[\frac{3abc-{{b}^{3}}}{{{c}^{3}}}\]
C) \[\frac{3abc-{{b}^{3}}}{{{a}^{2}}c}\]
D) \[\frac{{{b}^{3}}+3abc}{{{a}^{2}}c}\]
Correct Answer: D
Solution :
Since, \[\alpha \] and \[\beta \]are the zeroes of quadratic equation \[a{{x}^{2}}+bx-c=0\] \[\therefore \] \[\alpha +\beta =\frac{-b}{a}\] and \[\alpha \beta =\frac{-c}{a}\] Now, \[\frac{{{\alpha }^{2}}}{\beta }+\frac{{{\beta }^{2}}}{\alpha }=\frac{{{\alpha }^{2}}+{{\beta }^{3}}}{\alpha \beta }\] \[=\frac{(\alpha +\beta )\,({{\alpha }^{2}}+{{\beta }^{2}}-\alpha \beta )}{\alpha \beta }\] \[=\frac{-b}{a}\times \frac{-a}{c}[{{(\alpha +\beta )}^{2}}-3\alpha \beta ]=\frac{b}{c}\left[ \frac{{{b}^{2}}}{{{a}^{2}}}+\frac{3c}{a} \right]\] \[=\frac{b}{c}\left[ \frac{{{b}^{2}}+3ca}{{{a}^{2}}} \right]=\frac{{{b}^{3}}+3abc}{{{a}^{2}}c}\]
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