If α, β are the zeros of the polynomial f(x) = ax2 + bx + c,

Solution:

We have to find the value of $\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$

Given $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $\mathrm{f}(\mathrm{x})=a x^{2}+b x+c$

$\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{3}}$

$=\frac{-b}{a}$

$\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$=\frac{c}{a}$

We have,

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)^{2}-\frac{2}{\alpha \beta}$

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{\beta}{\alpha \beta}+\frac{\alpha}{\beta \alpha}\right)^{2}-\frac{2}{\alpha \beta}$

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{\alpha+\beta}{\alpha \beta}\right)^{2}-\frac{2}{\alpha \beta}$

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{\frac{-b}{a}}{\frac{c}{a}}\right)^{2}-\frac{2}{\frac{c}{a}}$

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{-b}{a} \times \frac{a}{c}\right)^{2}-\frac{2}{c}$

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{-b}{c}\right)^{2}-\frac{2 a}{c}$

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{b^{2}}{c^{2}}\right)-\frac{2 a \times c}{c \times c}$

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{b^{2}}{c^{2}}\right)-\frac{2 a c}{c^{2}}$

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{b^{2}-2 a c}{c^{2}}\right)$

Hence, the correct choice is $(b)$