Question

If all the real numbers $x_1,x_2,x_3$, satisfying the equation $x^3-x^2+\beta x+\gamma =0$ are in A.P.

Then, all possible values of $\gamma$ belongs to

• A. $(-\frac{1}{9},\infty )$
• B. $(-\frac{1}{27},+\infty )$
• C. $(-\frac{2}{9},+\infty )$
• D. none of these

Answer 1

Answer: (B)

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

Comparing given equation with the general form of a cubic equation $(a{x}^3+b{x}^2+cx+d)=0$,

we have
$a=1,b=-1,c=\beta ,d=\gamma$

Let the roots of the equation be $(a_1-d),a_{1}and(a_1+d)$.

Sum of the roots$=\frac{-b}{a}=\frac{-(-1)}{1}=1=(a_1-d)+\left(a_1\right)+(a_1+d)=3a_1\Rightarrow a_1=\frac{1}{3}$

Product of roots$=\frac{-d}{a}=\frac{-\left(\gamma \right)}{1}=-\gamma =(a_1-d)\left(a_1\right)(a_1+d)=a_1\left({a_1}^2{-d}^2\right)=\left(\frac{1}{3}\right)({\left(\frac{1}{3}\right)}^2-b^2)=\left(\frac{1}{3}\right)(\frac{1}{9}-b^2)$

$\Rightarrow$ $-\gamma =\left(\frac{1}{3}\right)(\frac{1}{9}-b^2)$ $\Rightarrow$ $-3\gamma =(\frac{1}{9}-b^2)$ $\Rightarrow$ $b^2=\frac{1}{9}+3\gamma$

Since, square of a number is always non-negative,we have

$\frac{1}{9}+3\gamma \ge 0$ $\Rightarrow$ $\gamma \ge -\frac{1}{27}$ $\Rightarrow$ $\gamma \in (-\frac{1}{27},\infty )$