Question

If two roots of the equation $(a-1)\left(x^2+x+1\right)^2-(a+1)\left(x^4+x^2+1\right)=0$ are real and distinct, then 'a' lies in the interval

• A. $(-2,2)$
• B. $(-\infty,-2) \cup(2, \infty)$
• C. $(2, \infty)$
• D. $(-3,3)$

Answer 1

Answer: (B)

$(a-1)\left(x^2+x+1\right)^2-(a+1)\left(x^4+x^2+1\right)=0$..........(1)
$\because x^4+x^2+1=\left(x^2+x+1\right)\left(x^2-x+1\right)$
$\therefore (1) \text { becomes }$
$\Rightarrow \left(x^2+x+1\right)\left[\left(x^2+x+1\right)(a-1)-(a+1)\left(x^2-x+1\right)\right]=0$
$\Rightarrow \left(x^2+x+1\right)\left(x^2-a x+1\right)=0$
Here two roots are imaginary and for other two roots to be real $D>0$
$\Rightarrow a^2-4>0 \Rightarrow a \in(-\infty,-2) \cup(2, \infty)$