Question

If range of $a$ for which equation $(x^2+x+2{)}^2-(a-3\left)\right(x^2+x+2\left)\right(x^2+x+1)+(a-4\left)\right(x^2+x+1{)}^2=0$ has atleast one real root is $(p,\frac{q}{r}]$, then $p+q+r$
is

Answer 1

$(x^2+x+2{)}^2-(a-3\left)\right(x^2+x+2\left)\right(x^2+x+1)+(a-4\left)\right(x^2+x+1{)}^2=0$
$\Rightarrow {\left(\frac{x^2+x+2}{x^2+x+1}\right)}^2-(a-3)\left(\frac{x^2+x+2}{x^2+x+1}\right)+(a-4)=0$
$\text{Let}\frac{x^2+x+2}{x^2+x+1}=t>1\Rightarrow t^2-(a-3)t+(a-4)=0$
$\Rightarrow (t-1)(t-a+4)=0$
$t\ne 1,t=a-4$
So $\frac{x^2+x+2}{x^2+x+1}=a-4$
$\Rightarrow (a-5)x^2+(a-5)x+(a-6)=0$
Since $x\in R,D\ge 0$
$\Rightarrow (a-5{)}^2-4(a-5\left)\right(a-6)\ge 0\Rightarrow 5<a\le \frac{19}{3}$
$p=5,q=19,r=3$
$p+q+r=5+19+3=27$