Question

If the roots of the equation $(a-1)(x^2+x+1{)}^2=(a+1\left)\right(x^4+x^2+1)$ are real and distinct, then $a$ belongs to

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

Answer 1

The correct option is D $\varphi$

$x^4+x^2+1=(x^2+1{)}^2-x^2$
$=(x^2+x+1)(x^2-x+1)$

$x^2+x+1\ne 0,\forall x\in R$
$(a-1)(x^2+x+1{)}^2=(a+1\left)\right(x^4+x^2+1)$
$\Rightarrow (a-1)(x^2+x+1)=(a+1)(x^2-x+1)$
$\Rightarrow x^2-ax+1=0...\left(1\right)$
The roots are real and distinct.
So, $D=a^2-4>0$
$\Rightarrow a\in (-\infty ,-2)\cup (2,\infty )...\left(2\right)$
Since, the given equation is bi-quadratic. So, it has $4$ roots.
And from $\left(1\right)$ and $\left(2\right),$ we get only two real and distinct roots.
Hence, no such value of $a$ exist for which roots of the given equation are real and distinct.