Question

Let equation $x^3+p{x}^2+qx-q=0$ where $p,q\in R-\left\{0\right\}$ has 3 real roots $\alpha ,\beta ,\lambda$ in $H.P.$, then
$\frac{p}{q}$ has the minimum value equal to

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

Answer 1

The correct option is B $-\frac{1}{3}$

Given $x^3+p{x}^2+qx-q=0$

Roots of equation: $\alpha ,\beta ,\lambda$ for HP

General cubic equation $a{x}^3+b{x}^2+cx+d$

$\alpha +\beta +\lambda =(-1{)}^1\frac{b}{a}=(-1)\frac{p}{1}=-p\alpha \beta +\beta \lambda +\alpha \lambda ={(-1)}^2\frac{c}{a}=q\alpha \beta \lambda ={(-1)}^3\frac{-q}{1}=q$

AM $\ge$ GM

$\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }\ge 3{\left[\frac{1}{\alpha \beta \gamma }\right]}^{1/3}\frac{\gamma \beta +\alpha \gamma +\alpha \beta }{\alpha \beta \gamma }\ge 3{\left[\frac{1}{\alpha \beta \gamma }\right]}^{1/3}\frac{q}{q}\ge 3{\left[\frac{1}{\alpha \beta \gamma }\right]}^{1/3}1\ge 3{\left[\frac{1}{\alpha \beta \gamma }\right]}^{1/3}\left(1\right)$

$\alpha +\beta +\gamma \ge 3{\left(\alpha \beta \gamma \right)}^{1/3}\left(2\right)\alpha \beta +\beta \gamma +\gamma \alpha \ge 3{\left(\alpha \beta \gamma \right)}^{2/3}\left(3\right)$

Dividing $\left(2\right)$ by $\left(3\right)$

$\frac{\alpha +\beta +\gamma }{\alpha \beta +\beta \gamma +\gamma \alpha }\ge \frac{1}{{\left(\alpha \beta \gamma \right)}^{1/3}}\frac{-p}{q}\ge \frac{1}{{\left(\alpha \beta \gamma \right)}^{1/3}}\left(4\right)$

Dividing $\left(4\right)$ by $\left(1\right)$

$\frac{-p}{q}\ge \frac{1}{3}\frac{p}{q}\ge -\frac{1}{3}$