Question

For the zeroes $\alpha$ & $\beta$ of polynomial $P\left(x\right)=a{x}^2+bx+c,\frac{1}{\alpha }+\frac{1}{\beta }=$ ....................

• A. $-\frac{b}{c}$
• B. $-\frac{b}{a}$
• C. $\frac{c}{a}$
• D. none

Answer 1

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

$P\left(x\right)=a{x}^2+bx+c$

$\alpha$ and $\beta$ are the roots of $P\left(x\right)$

$\therefore \alpha +\beta =\frac{-b}{a}$ and $\alpha \beta =\frac{c}{a}$

Now, $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-\frac{b}{a}}{\frac{c}{a}}=-\frac{b}{c}$

Thus, $\frac{1}{\alpha }+\frac{1}{\beta }=-\frac{b}{c}$