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 $\beta$ belong to

• A. $(-\infty ,\frac{1}{3})$
• B. $(-\infty ,-\frac{1}{3})$
• C. $\left(\frac{1}{3},\infty \right)$
• D. $(-\frac{1}{3},\infty )$

Answer 1

The correct option is A $(-\infty ,\frac{1}{3})$

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$=\frac{-b}{a}=\frac{-(-1)}{1}=1=(a_1-d)+\left(a_1\right)+(a_1+d)=3a_1\Rightarrow a_1=\frac{1}{3}$..........$\left(i\right)$

Sum of bi-product of roots

$\frac{c}{a}=\frac{\beta }{1}=\beta =(a_1-d)\left(a_1\right)+(a_1-d)(a_1+d)+\left(a_1\right)(a_1+d)={a_1}^2-a_{1}d+{a_1}^2-d^2+{a_1}^2+a_{1}d=3{a_1}^2-d^2$

$=3{\left(\frac{1}{3}\right)}^2-d^2=\frac{1}{3}-d^2$..........from $\left(i\right)$

$\Rightarrow$ $\beta =\frac{1}{3}-d^2$ $\Rightarrow$ $d^2=\frac{1}{3}-\beta$

Since square of a number is always non-negative,we can imply $(\frac{1}{3}-\beta )\ge 0$ $\Rightarrow$ $\beta \le \frac{1}{3}\Rightarrow \beta \in (-\infty ,\frac{1}{3})$