Question

If the roots of the equation $(\lambda +1)x^3-({\lambda }^2+2)x^2-({\lambda }^2+2)x+(\lambda +1)=0;\lambda ϵR$ are in a $A.P$. with non-zero common difference and one of the roots is independent of $\lambda$, then the sum of all possible values of $\lambda$ is $m$. Find the value of $4m$.

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is C $6$

$(\lambda +1)(x^3+1)=({\lambda }^2+2)x(x+1)$
$(x+1)\left[\right(\lambda +1\left)\right(x^2-x+1)-x({\lambda }^2+2\left)\right]=0$
$(x+1)\left[\right(\lambda +1)x^2-x({\lambda }^2+\lambda +3)+\lambda +1]=\mathrm{0.....}\left(1\right)$
From the above equation, $x=-1$ is a root of the equation.
Since the three roots are in A.P., $-1$ is either the middle term or any one of the end terms of the A.P.
Case-1: $-1$ is the middle term.
Let the other two roots be $a$ and $c$.
$a+c=2(-1)$
$\frac{{\lambda }^2+\lambda +3}{\lambda +1}=-2$
${\lambda }^2+3\lambda +5=0\Rightarrow \lambda$ is imaginary, which contradicts the statement in question. Hence, case 1 is not possible
Case 2: Let the first term of A.P. be $-1$, and the other two roots be $b$ and $c$.
$-1+c=2b$
$c-2b=\mathrm{1....}\left(2\right)$
$bc=1$ (from the quadratic equation in $\left(1\right)$)
$(c-2b{)}^2+8bc=1+8=(c+2b{)}^2$
$c+2b=\pm \mathrm{3....}\left(3\right)$
Solving $\left(2\right)$ and $\left(3\right)$, we get $c=2\Rightarrow b=\frac{1}{2}$....(c= -1 is not possible as the A.P. as non zero common difference)
$\frac{{\lambda }^2+\lambda +3}{\lambda +1}=b+c$
$\Rightarrow \lambda +\frac{3}{\lambda +1}=\frac{5}{2}\Rightarrow 2{\lambda }^2+2\lambda +6=5\lambda +5$
$\Rightarrow 2{\lambda }^2-3\lambda +1=0$
$\Rightarrow \lambda =1,\frac{1}{2}$
$\Rightarrow 4m=4(1+\frac{1}{2})=6$