A) \[-\frac{b}{a}\]
B) \[\frac{c}{d}\]
C) \[-\frac{c}{d}\]
D) \[-\frac{c}{a}\]
Correct Answer: C
Solution :
| [c] Since \[\alpha ,\beta ,\gamma \] are the zeroes of the polynomial |
| \[f(x)=a{{x}^{3}}+b{{x}^{2}}+cx+d.\] |
| Then, \[\alpha +\beta +\gamma =-\frac{b}{a},\] \[\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a},\] \[\alpha \beta \gamma =-\frac{d}{a}\] |
| Now, \[\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{\beta \gamma +\gamma \alpha +\alpha \beta }{\alpha \beta \gamma }=\frac{\frac{c}{a}}{\frac{-d}{a}}=-\frac{c}{d}\] |
You need to login to perform this action.
You will be redirected in
3 sec
