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), $a (a^2-d^2) = - \gamma$
$\Rightarrow \frac{1}{3} \bigg(\frac{1}{9} -d^2 \bigg) = - \gamma$
$\Rightarrow \frac{1}{27}- \frac{1}{3} -d^2 = - \gamma \Rightarrow \gamma + \frac{1}{27} = \frac{1}{3} d^2$
$\Rightarrow \gamma + \frac{1}{27} \ge 0 \Rightarrow \gamma \ge - 1/27$
$\Rightarrow \gamma \in \bigg[- \frac{1}{27}, \infty \bigg)$
Hence, $\beta \in (- \infty, 1/3]$ and $ \gamma \in [- 1/27, \infty)$