Question

If $ \alpha ,\beta $ are the roots of the equation $ {{x}^{2}}+ax+b=0, $ then $ \frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}} $ is equal to

• 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}}} $

Answer 1

The Correct Option is A

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}}} $