Question

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

Answer 1

From the question, the real roots of $x^3-x^2+\beta x+\gamma =0$ are $x_1,x_2$ $x_3$ and they are in A.P.

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

let $x_1=a-d,x_2=a,x_3=a+d.$

Now,$x_1+x_2+x_3=-\frac{-1}{1}=1$

$\Rightarrow a-d+a+a+d=1$

$\Rightarrow a=\frac{1}{3}$

$x_1{x}_2+x_2{x}_3+x_3{x}_1=\frac{\beta }{1}=\beta$

$\Rightarrow (a-d)a+a(a+d)+(a+d)(a-d)=\beta$

$x_1{x}_2{x}_3=-\frac{\gamma }{1}=-\gamma$

$\Rightarrow (a-d)a(a+d)=-\gamma$

From (1) and (2), we get

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

$\Rightarrow \Rightarrow 3\frac{1}{9}-d^2=\beta ,\text{so}\beta =\frac{1}{3}-d^2<\frac{1}{3}$

From (1) and (3), we get

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

$\Rightarrow \gamma =\frac{1}{3}(d^2-\frac{1}{9})>\frac{1}{3}(-\frac{1}{9})=-\frac{1}{27}$

$\gamma \in (-\frac{1}{27},+\infty )$