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

Answer: (A)

$\frac{{{a}^{2}}-2b}{{{b}^{2}}}$

Answer 2

Since, $\alpha$ and $\beta$ are the roots of
${{x}^{2}}+ax+b=0,$
then $\alpha +\beta =-a,\,\,\alpha \beta =b$
$\therefore$ $\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{{{(\alpha \beta )}^{2}}}$
$=\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{{{(\alpha \beta )}^{2}}}=\frac{{{a}^{2}}-2b}{{{b}^{2}}}$