Question

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

विकल्प

• $$\frac{b^2-2ac}{a^2}$$
• $$\frac{b^2-2ac}{c^2}$$
• $$\frac{b^2+2ac}{a^2}$$
• $$\frac{b^2+2ac}{c^2}$$

Answer 1

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

Given ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ are the zeros of the quadratic polynomial f(x) = ax² + bx + c, 

${\displaystyle \alpha +ß=-\frac{-\text{coefficient of x}}{\text{coefficient of}{x}^3}}$

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

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

${\displaystyle =\frac{c}{a}}$

We have,

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

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

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

${\displaystyle \frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}={\left(\frac{\frac{-6}{a}}{\frac{c}{a}}\right)}^2-\frac{2}{\frac{c}{a}}}$

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

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

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

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

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

${\displaystyle \frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{b^2-2ac}{c^2}}$

Hence, the correct choice is ${\displaystyle \left(b\right).}$