Question

If $\alpha, \beta$ and $\gamma$ are the negative roots of the equation $27 x^3+216 x^2+36 p x+32 p=0, p \in R-\{0\}$, then $\left|\frac{ p }{\gamma}\right|$ equals

• A. 2
• B. 6
• C. 8
• D. 64

Answer 1

Answer: (B)

$\text { A.M. }=\frac{(-\alpha)+(-\beta)+(-\gamma)}{3}=\frac{216}{27 \cdot 3}=\frac{8}{3} $
$\text { H.M. }=\frac{3}{\left(\frac{-1}{\alpha}\right)+\left(\frac{-1}{\beta}\right)+\left(\frac{-1}{\gamma}\right)}=\frac{-3 \alpha \beta \gamma}{\Sigma \alpha \beta}=\frac{3 \cdot 32 p }{36 p }=\frac{8}{3}$
$\Theta \text { A.M. }=\text { H.M. }$
$\therefore -\alpha=-\beta=-\gamma=\frac{8}{3} \Rightarrow \alpha=\beta=\gamma=\frac{-8}{3} $
$ \alpha \beta \gamma=\frac{-32 p}{27} \Rightarrow\left(\frac{-8}{3}\right)^3=\frac{-32 p}{27} \Rightarrow p=16 $
$\therefore \left|\frac{p}{\gamma}\right|=\left|\frac{16}{8 / 3}\right|=6 $