Question

If $\alpha ,\beta ,\gamma$ are the roots of $x^3+a{x}^2+bx+c=0$, then $\Sigma (\frac{\alpha }{\beta }+\frac{\beta }{\alpha })=$

• A. $\frac{ab}{c}$
• B. $\frac{ab}{c}-1$
• C. $\frac{ab}{c}-2$
• D. $\frac{ab}{c}-3$

Answer 1

The correct option is D $\frac{ab}{c}-3$

As $\alpha ,\beta ,\gamma$ are roots of $x^3+a{x}^2+bx+c=0$, gives
$S_1=\alpha +\beta +\gamma =-a{S}_2=\alpha \beta +\beta \gamma +\gamma \alpha =b{S}_3=\alpha \beta \gamma =-c$
Now,
$\sum (\frac{\alpha }{\beta }+\frac{\beta }{\alpha })=\sum \left(\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }\right)=\frac{({\alpha }^2+{\beta }^2)}{\alpha \beta }+\frac{({\beta }^2+{\gamma }^2)}{\beta \gamma }+\frac{({\alpha }^2+{\gamma }^2)}{\alpha \gamma }=\frac{{\alpha }^2\gamma +{\beta }^2\gamma +\alpha {\beta }^2+\alpha {\gamma }^2+{\alpha }^2\beta +\beta {\gamma }^2}{\alpha \beta \gamma }=\frac{{S_{1}S}_2-3S_3}{S_3}=\frac{ab-3c}{c}=\frac{ab}{c}-3$