Question

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

• A. 2
• B. 6
• C. 8
• D. 64

Answer 1

Answer: (B)

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