Question

If $\alpha ,\beta ,\gamma$ are the roots of $x^3+p{x}^2+qx+r=0$, then $\sum \frac{{\beta }^2+{\gamma }^2}{\beta \gamma }=$

• A. $\frac{pq}{r}-1$
• B. $\frac{pq}{r}-2$
• C. $\frac{pq}{r}-3$
• D. $\frac{pq}{r}-4$

Answer 1

The correct option is C $\frac{pq}{r}-3$

As $\alpha ,\beta ,\gamma$ are the roots of $x^3+p{x}^2+qx+r=0$
Gives
$\alpha +\beta +\gamma =-p\alpha \beta +\beta \gamma +\gamma \alpha =q\alpha \beta \gamma =-r$

Now
$\sum \frac{{\beta }^2+{\gamma }^2}{\beta \gamma }=\frac{{\beta }^2+{\gamma }^2}{\beta \gamma }+\frac{{\alpha }^2+{\gamma }^2}{\gamma \alpha }+\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{\alpha {\beta }^2+{\gamma }^2\alpha +{\alpha }^2\beta +\beta {\gamma }^2+\gamma {\alpha }^2+{\beta }^2\gamma }{\alpha \beta \gamma }=\frac{(\alpha +\beta +\gamma )(\alpha \beta +\beta \gamma +\gamma \alpha )-3\alpha \beta \gamma }{\alpha \beta \gamma }=\frac{-pq-3(-r)}{-r}=\frac{pq}{r}-3$