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 $100 S$.

Answer 1

Now, $3 a =\frac{ m ^2+2}{ m +1}$....(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
$3 a ^2- d ^2=-3 a$ ....(4)
From (2) and (4), we get after eliminating 'd'
$3 a^2-\left(a^2+\frac{1}{a}\right)=-3 a \Rightarrow 2 a^3+3 a^2-1=0 \Rightarrow(a+1)\left(2 a^2+a-1\right)=0 \Rightarrow(a+1)^2(2 a-1)=0 $
$\Rightarrow \quad a=\frac{1}{2},-1 .$
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 3 m +3=2 m ^2+4 $
$\Rightarrow 2 m ^2-3 m +1=0 \Rightarrow 2 m ^2-2 m - m +1=0 $
$2 m ( m -1)-1( m -1)=0 \Rightarrow( m -1)(2 m -1)=0 $
$\text { Hence, } m =1, \frac{1}{2} \Rightarrow S =1+\frac{1}{2}=\frac{3}{2} \Rightarrow 100 S =150$