Question

The real numbers $x_1,x_2,x_3$ satisfying the equation
$x^3+x^2+\beta x+\gamma =0$ are in $A.P$. Find the intervals in which $\beta$ are $\gamma$ lie.

Answer 1

Since $x_1,x_2,x_3$ are in A.P.

Let $x_1=a-d,x_2=a$ and $x_3=a+d$ and $x_1,x_2,x_3$ are the roots of

$x^3-x^2+\beta x+\gamma =0$

$\therefore \sum \alpha =a-d+a+a+d=1$

$\Rightarrow a=\frac{1}{3}$ .......$\left(1\right)$

$\sum \alpha \beta =(a-d)a+a(a+d)+(a-d)(a+d)=\beta$ ........$\left(2\right)$

and $\alpha \beta \gamma =(a-d)a(a+d)=-\gamma$ ........$\left(3\right)$

From eqn$\left(1\right)$ we get

$3a^2-d^2=\beta$

$\Rightarrow 3{\left(\frac{1}{3}\right)}^2-d^2=\beta$

$\Rightarrow \frac{1}{3}-\beta =d^2$

Since $d^2\ge 0$ for all $d\in R$

$\Rightarrow \frac{1}{3}-\beta \ge 0$

$\Rightarrow \beta \le \frac{1}{3}$

$\Rightarrow \beta \in (-\infty ,\frac{1}{3}]$

From eqn$\left(3\right),a(a^2-d^2)=-\gamma$

$\Rightarrow \frac{1}{3}(\frac{1}{9}-d^2)=-\gamma$

$\Rightarrow \frac{1}{27}-\frac{1}{3}{d}^2=-\gamma$

$\Rightarrow \gamma +\frac{1}{27}=\frac{1}{3}{d}^2$

$\Rightarrow \gamma +\frac{1}{27}\ge 0$

$\Rightarrow \gamma \ge \frac{-1}{27}$

$\Rightarrow \gamma \in [\frac{-1}{27},\infty )$

Hence,$\beta \in (-\infty ,\frac{1}{3}]$ and $\gamma \in [\frac{-1}{27},\infty )$