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

Answer: (A)

$-\frac{b}{c}$

Given: $P\left(x\right)=a{x}^2+bx+c$
Let $\alpha$ and $\beta$ are the zeros of $P\left(x\right)$
$\therefore \alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$
Now, $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }$
$=\frac{-\frac{b}{a}}{\frac{c}{a}}$

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