A) \[\frac{{{a}^{2}}-2b}{{{b}^{2}}}\]
B) \[\frac{{{b}^{2}}-2a}{{{b}^{2}}}\]
C) \[\frac{{{a}^{2}}+2b}{{{b}^{2}}}\]
D) \[\frac{{{b}^{2}}+2a}{{{b}^{2}}}\]
Correct Answer: A
Solution :
Since, \[\alpha \] and \[\beta \] are the roots of \[{{x}^{2}}+ax+b=0,\] then \[\alpha +\beta =-a,\,\,\alpha \beta =b\] \[\therefore \] \[\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{{{(\alpha \beta )}^{2}}}\] \[=\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{{{(\alpha \beta )}^{2}}}=\frac{{{a}^{2}}-2b}{{{b}^{2}}}\]
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