Question

Two of the roots of the equation $(\lambda -2)(x^2+x+1{)}^2-(\lambda +2\left)\right(x^4+x^2+1)=0$ are equal, then the value(s) of ${}^{\prime }{\lambda }^{\prime }$ is/are

• A. $\{-4,4\}$
• B. $\{-2,2\}$
• C. $\left\{2\right\}$
• D. $\{4,5\}$

Answer 1

The correct option is A $\{-4,4\}$

Given, $(\lambda -2)(x^2+x+1{)}^2-(\lambda +2\left)\right(x^4+x^2+1)=0$

$\Rightarrow (\lambda -2)(x^2+x+1{)}^2-(\lambda +2\left)\right((x^2+1{)}^2-x^2)=0$

$\Rightarrow (x^2+x+1)\left[\right(\lambda -2\left)\right(x^2+x+1)-(\lambda +2\left)\right(x^2-x+1\left)\right]=0$

But, $(x^2+x+1)=0$ can't have any real roots.

$\therefore (\lambda -2)(x^2+x+1)-(\lambda +2)(x^2-x+1)=0$

This will have equal roots.

Solving this, we get

$-4x^2+2\lambda x-4=0$

$\Rightarrow 2x^2-\lambda x+2=0$

It has equal roots.

So, discriminant, $D=0$

$\Rightarrow D={\lambda }^2-16=0$

$\Rightarrow \lambda =\pm 4$