Q66 / 200 · Manipal (MET) 2013 · Mathematics

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

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

Solution

$x^4+x^2+1=\left(x^2+1\right)^2-x^2$ $=\left(x^2+x+1\right)\left(x^2-x+1\right)$ $x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4} \neq 0$ $\begin{aligned} \text { Now, }(a-1)\left(x^2+x+1\right)^2 & =(a+1)\left(x^4+x^2+1\right) \\ (a-1)\left(x^2+x+1\right) & =(a+1)\left(x^2-x+1\right)\end{aligned}$ $\Rightarrow x^2-a x+1=0$ which has real and distinct roots $\begin{array}{ll}\therefore & \Delta=a^2-4\gt0 \\ \Rightarrow & a^2\gt4 \\ \Rightarrow & a \in(-\infty,-2) \cup(2, \infty)\end{array}$

Hence, the correct answer is option: 'B'.

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