Polynomials Question - None

Question

If $\alpha$, $\beta$ are the zeros of the polynomial f(x) = ax² + bx + c, then $\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=$

Accepted Answer

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 f(x) = ax² + bx + 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,
$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)^{2}-\frac{2}{\alpha \beta}$
$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{\beta}{\alpha \beta}+\frac{\alpha}{\beta \alpha}\right)^{2}-\frac{2}{\alpha \beta}$
$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{\alpha+\beta}{\alpha \beta}\right)^{2}-\frac{2}{\alpha \beta}$
$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{\frac{-b}{a}}{\frac{c}{a}}\right)^{2}-\frac{2}{\frac{c}{a}}$
$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{-b}{a} \times \frac{a}{c}\right)^{2}-\frac{2}{\frac{c}{a}}$
$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{-b}{c}\right)^{2}-\frac{2 a}{c}$
$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{b^{2}}{c^{2}}\right)-\frac{2 a \times c}{c \times c}$
$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{b^{2}}{c^{2}}\right)-\frac{2 a c}{c^{2}}$
$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\left(\frac{b^{2}-2 a c}{c^{2}}\right)$

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