Question

Find the minimum positive integral value of $a$ for which roots of $(a-1)(x^2+x+1{)}^2=(a+1\left)\right(x^4+x^2+1)$ are real and distinct.

(Also find the entire interval of $a$)

Answer 1

Given $(a-1)(x^2+x+1{)}^2=(a+1\left)\right(x^4+x^2+1)$

Dividing by $x^2\Rightarrow (a-1)(x+\frac{1}{x}+1{)}^2=(a+1\left)\right(x^2+\frac{1}{x^2+1})$--(1)

Let $P=x+\frac{1}{x}\Rightarrow x^2+\frac{1}{x^2}=(x+\frac{1}{x}{)}^2-2=p^2-2$

(1) because $(a-1)(p+1{)}^2-(a+1\left)\right(p^2-1)=0$

$(a-1)(p^2+1+2p)-(a+1)(p^2-1)=0$

$-p^2+(a-1)p-1=0$

$D=(a-1{)}^2-4>0$ (for real & distance roots )

$(a-3)(a+1)>0$

$aϵ(-\infty ,-1)\cup (3,\infty )$