Question

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

• A. $\frac{b^2-2ac}{c^2}$
• B. $\frac{b^2-2ac}{a^2}$
• C. $\frac{b^2-4ac}{{2a}^2}$
• D. $\frac{b^2+2ac}{a^2}$

Answer 1

The correct option is A $\frac{b^2-2ac}{c^2}$

Given, $\alpha ,\beta$ are the zeros of the polynomial $f\left(x\right)=a{x}^2+bx+c$
$\Rightarrow \alpha +\beta =\frac{-b}{a}$
$\alpha \beta =\frac{c}{a}$
Now,$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}$
$=\frac{(\alpha +\beta {)}^2-2(\alpha \beta )}{(\alpha \beta {)}^2}$
$=\frac{(\frac{-b}{a}{)}^2-2(\frac{c}{a})}{(\frac{c}{a}{)}^2}$
$=\frac{\left(\frac{b^2}{a^2}\right)-2\left(\frac{c}{a}\right)}{\left(\frac{c^2}{a^2}\right)}$
$=\frac{b^2-2ac}{c^2}$
Option A is correct.