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 the value of $a$ $\in$

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

Answer 1

Answer: (B)

$(-\infty ,-2)\cup (2,\infty )$

$x^4+x^2+1=(x^2+1{)}^2-x^2$
$x^4+x^2+1=(x^2+x+1)(x^2-x+1)$
$x^2+x+1={(x+\frac{1}{2})}^2+\frac{3}{4}\ne 0$ $\forall x$
Therefore, we can cancel this factor and we get
$(a-1)(x^2+x+1)=(a+1)(x^2-x+1)$
or $x^2-ax+1=0$
It has real and distinct roots if $D=a^2-4>0$.