Q90 / 160 · AP EAMCET 2012 · Mathematics
If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+p x^2+q x+r=0$, then the coefficient of $x$ in the cubic equation whose roots are $\alpha(\beta+\gamma), \beta(\gamma+\alpha)$ and $\gamma(\alpha+\beta)$ is
- A$2 q$
- Correct answerB$q^2+p r$
- C$p^2-q r$
- D$r(p q-r)$
Solution
Since, $\alpha, \beta$ and $\gamma$ are the roots of $f(x)=x^3+p x^2+q x+r=0$.
$\therefore \quad \alpha+\beta+\gamma=-p, \quad \alpha \beta+\beta \gamma+\gamma \alpha=q$
and
$\alpha \beta \gamma=-r$
Let $x=\alpha(\beta+\gamma)=\alpha \beta+\alpha \gamma+\beta \gamma-\frac{\alpha \beta \gamma}{\alpha}$
$\therefore \alpha$ satisfies the given equation.
$\begin{aligned} \therefore \quad f(\alpha)=\alpha^3+p \alpha^2+q \alpha+r & =0 \\ \frac{r^3}{(x-q)^3}+\frac{p r^2}{(x-q)^2}+\frac{q r}{(x-q)}+r & =0 \\ \Rightarrow(x-q)^3+q(x-q)^2+p r(x-q)+r^2 & =0\end{aligned}$
$\therefore$ Coefficient of $x=3 q^2-2 q^2+r p$
$=q^2+p r$
Hence, the correct answer is option: 'B'.