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 $\beta$ and $\gamma$ lie.

Answer 1

Since $x_1,x_2,x_3$ are in A.P. $2x_2=x_1+x_3$
$\therefore$ $3x_2=\sum x_1=1$ $\therefore$ $x_2=\frac{1}{3}$
But $x_2$ is a root of given equation
$\frac{1}{27}-\frac{1}{9}+\frac{1}{3}\beta +\gamma =0$
or $9\beta +27\gamma =2$ or $\beta +3\gamma =2/9$......(1)
Again $\sum x_1{x}_2=x_1{x}_2=x_2{x}_3=x_3{x}_1=\beta$
and $x_1{x}_2{x}_3=-\gamma$
putting $x_2=\frac{1}{3}$, we get
$\frac{1}{3}(x_1+x_3)x_1{x}_3=\beta$ and $\frac{1}{3}{x}_1{x}_3=-\gamma$
Eliminating $x_3$ from the above relation
$\frac{1}{3}(x_1-\frac{3\gamma }{x_1})-3\gamma =\beta$ or $x_1^2-3(\beta +3\gamma )x_1-3\gamma =0$......(2)
We have now to find the intervals for $\beta$ and $\gamma$ by the help of (1) and (2).
$\therefore$ $3{\left(\frac{2}{9}\right)}^2+4\gamma \ge 0$ by (1)
or $\gamma +\frac{1}{27}\ge 0$ $\therefore$ $\gamma \ge -\frac{1}{27}$.......(3)
and from (1)
$9\beta +27\gamma +1=3$
$3-9\beta =27\gamma +1\ge 0$, BY (3)
or $3\beta -1\le 0$ $\therefore$ $\beta \le \frac{1}{3}$
$\therefore$ $\beta ϵ]-\infty ,\frac{1}{3}],\gamma ϵ]-\frac{1}{27},\infty ]$