Question

If the roots of the cubic equation $(m+1)x^3-\left(m^2+2\right)x^2-\left(m^2+2\right)x+m+1=0;(m\in R)$ form a non-constant arithmetic progression and one of the roots does not depend on $m$, then the sum of all possible values of $m$ is $S$. Find the value of $100S$.

Answer 1

Answer: 150

Now, $3a=\frac{m^2+2}{m+1}$....(1)
$a\left(a^2-d^2\right)=\frac{-(m+1)}{(m+1)}=-1$ ....(2)
and $a(a-d)+a(a+d)+(a-d)(a+d)=\frac{-\left(m^2+2\right)}{(m+1)}$....(3)
From (1) and (3), we get
$3a^2-d^2=-3a$ ....(4)
From (2) and (4), we get after eliminating 'd'
$3a^2-\left(a^2+\frac{1}{a}\right)=-3a\Rightarrow 2a^3+3a^2-1=0\Rightarrow (a+1)\left(2a^2+a-1\right)=0\Rightarrow (a+1{)}^2(2a-1)=0$
$\Rightarrow a=\frac{1}{2},-1.$
Case-I : If $a=-1$, then from equation (4), $d=0$, which is not possible.
Case-II: If $a=\frac{1}{2}$, we get
Put in equation (1)
$\frac{1}{2}=\frac{m^2+2}{3(m+1)}\Rightarrow 3m+3=2m^2+4$
$\Rightarrow 2m^2-3m+1=0\Rightarrow 2m^2-2m-m+1=0$
$2m(m-1)-1(m-1)=0\Rightarrow (m-1)(2m-1)=0$
$\text{ Hence, }m=1,\frac{1}{2}\Rightarrow S=1+\frac{1}{2}=\frac{3}{2}\Rightarrow 100S=150$