Question

If the two roots of the equation, $(a-1)(x^4+x^2+1)+(a+1)(x^2+x+1{)}^2=0$ are real and distinct, then the set of all values of $a$ is:

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

Answer 1

Answer: (B)

$(-\frac{1}{2},0)\cup (0,\frac{1}{2})$

Given equation is $(a-1)(x^4+x^2+1)+(a+1)(x^2+x+1{)}^2=0$

$\Rightarrow (a-1)(x^4+2x^2+1-x^2)+(a+1)(x^2+x+1{)}^2=0$

$\Rightarrow (a-1)(x^2+x+1)(x^2-x+1)+(a+1)(x^2+x+1{)}^2=0$

$\Rightarrow (x^2+x+1)\left[\right(a-1\left)\right(x^2-x+1)+(a+1\left)\right(x^2+x+1\left)\right]=0$

$\Rightarrow 2(x^2+x+1)(a{x}^2+x+a)=0$

Clearly $x^2+x+1=0$ won't have real solution.

So for given equation to posses two real an distinct solution discriminant of $a{x}^2+x+a=0$ has to be $>0$

$\Rightarrow (1{)}^2-4a^2>0\Rightarrow a^2<\frac{1}{4}$

$\Rightarrow (-\frac{1}{2},0)\cup (0,\frac{1}{2})$