Question

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

Answer 1

Since, $x_1,x_2,x_3$ are in an AP. Let $x_1=a-d,x_2=a$ and
$x_3-a+d$ and $x_1,x_2,x_3$ be the roots of $x^3-x^2+\beta x+y=0$
$\therefore \sum \alpha =a-d+a+a+d=1$
$\Rightarrow a=1/3$ ...(i)
$\sum \alpha \beta =(a-d)a+a(a+d)+(a+d)(a+d)=\beta$...(ii)
and $\alpha \beta \gamma =(a-d)a(a+d)=-\gamma$...(iii)
From Eq.(i),
$3a=1$
$\Rightarrow a=1/3$
From Eq. (ii), $3a^2-d^2=\beta$
$\Rightarrow 3(1/3{)}^2-d^2=\beta$ [from Eq. (i)]
$\Rightarrow 1/3-\beta =d^2$
NOTE In this equation, we have two variables $\beta$ and y but we have
only one equation. So at first sight it looks that this equation cannot solve but we know that $d^2\ge 0,\forall d\in R,$ then $\beta$ can be solved .
$\Rightarrow \frac{1}{3}-\beta \ge 0[\because d^2\ge 0]$
$\Rightarrow \beta \le \frac{1}{3}\Rightarrow \beta \in [-\infty ,1/3]$
From Eq.(iii), $a(a^2-d^2)=-\gamma$
$\Rightarrow \frac{1}{3}(\frac{1}{9}-d^2)=-\gamma$
$\Rightarrow \frac{1}{27}-\frac{1}{3}-d^2=-\gamma \Rightarrow \gamma +\frac{1}{27}=\frac{1}{3}{d}^2$
$\Rightarrow \gamma +\frac{1}{27}\ge 0\Rightarrow \gamma \ge -1/27$
$\Rightarrow \gamma \in [-\frac{1}{27},\infty )$
Hence, $\beta \in (-\infty ,1/3]$ and $\gamma \in [-1/27,\infty )$