Question

If the roots of the equation $(a-1){(x^2+x+1)}^2=(a+1)(x^4+x^2+1)$ are real and distinct then the value of $a\in$

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

Answer 1

The correct option is B $(-\infty ,-2)\cup (2,\infty )$

We have $x^4+x^2+1={(x^2+1)}^2-x^2$
$=(x^2+x+1)(x^2-x+1)$
and $x^2+x+1={(x+\frac{1}{2})}^2+\frac{3}{4}\ne 0$ $\forall x$
Now, $(a-1){(x^2+x+1)}^2=(a+1)(x^4+x^2+1)$
$(a-1)(x^2+x+1)=(a+1)(x^2-x+1)$
$\Rightarrow x^2-ax+1=0$ which has real and distinct roots.
Therefore, $\Delta =a^2-4>0$
$\Rightarrow a^2>4$
$\Rightarrow a\in (-\infty ,-2)\cup (2,\infty )$