Question

If the two roots of the equation, $(a+1)(x^2+x+1{)}^2+(a+1)(x^2+x+1{)}^2=0$ are real and distinct, then the set of all values of ${}^{\prime }{a}^{\prime }$ is :

• A. $\left(-\frac{1}{2},0\right)$
• B. $\left(-\infty ,-2\right)\cup \left(2,\infty \right)$
• C. $\left(-\frac{1}{2},0\right)\cup \left(0,\frac{1}{2}\right)$
• D. $\left(0,\frac{1}{2}\right)$

Answer 1

Answer: (C)

$\left(a-1\right)\left(x^4+x^2+1\right)+\left(a+1\right){\left(x^2+x+1\right)}^2=0$
$\Rightarrow \left(a-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\left(a+1\right){\left(x^2+x+1\right)}^2=0$
$\Rightarrow \left(x^2+x+1\right)\left[\left(a-1\right)\left(x^2-x+1\right)+\left(a1\right)\left(x^2+x+1\right)\right]=0$
$\Rightarrow \left(x^2+x+1\right)\left(a{x}^2+x+a\right)=0$
For roots to be distinct and real, $a\ne 0$ and $1-4a^2>0$
$\Rightarrow a\ne 0$ and $a^2<\frac{1}{4}$
$\Rightarrow a\in \left(-\frac{1}{2},0\right)\cup \left(0,\frac{1}{2}\right)$