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Q.
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 $'a'$ is :
JEE MainJEE Main 2015Complex Numbers and Quadratic Equations
Solution:
$\left(a-1\right)\left(x^{4}+x^{2}+1\right)+\left(a+1\right)\left(x^{2}+x+1\right)^{2}=0$
$\Rightarrow \left(a-1\right)\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)\left(a+1\right)\left(x^{2}+x+1\right)^{2}=0$
$\Rightarrow \left(x^{2}+x+1\right)\left[\left(a-1\right)\left(x^{2}-x+1\right)+\left(a 1\right)\left(x^{2}+x+1\right)\right]=0$
$\Rightarrow \left(x^{2}+x+1\right)\left(ax^{2}+x +a\right)=0$
For roots to be distinct and real, $a\ne0$ and $1-4a^{2}>0$
$\Rightarrow a\ne0$ and $a^{2}< \frac{1}{4}$
$\Rightarrow a \in\,\left(-\frac{1}{2}, 0\right)\cup\left(0, \frac{1}{2}\right)$