Question

The number of positive integral values of $m$ less than $17$ for which the equation $(x^2+x+1{)}^2-(m-3\left)\right(x^2+x+1)+m=0,m\in R$ has $4$ distinct real roots is

Answer 1

$(x^2+x+1{)}^2-(m-3\left)\right(x^2+x+1)+m=0\cdots (i)$
Assuming
$t=x^2+x+1$
$={(x+\frac{1}{2})}^2+\frac{3}{4}$
$\Rightarrow t\in [\frac{3}{4},\infty )$
$f\left(t\right)=t^2-(m-3)t+m=0\cdots \left(ii\right)$
Let the roots of the equation $\left(ii\right)$ be $t_1,t_2$

(i) For every $t>\frac{3}{4}$, there exists 2 distinct real roots for $x^2+x+1=t$
(ii) For every $t<\frac{3}{4}$, there exists no real roots For $x^2+x+1=t$
Given equation $\left(i\right)$ will have$4$ distinct real roots iff both roots of equation $\left(ii\right)$
$t_1,t_2>\frac{3}{4}$

So, the requird condition are,
$\left(i\right)D>0$
$\Rightarrow m^2-10m+9>0\Rightarrow (m-1)(m-9)>0\Rightarrow m\in (-\infty ,1)\cup (9,\infty )$
$\left(ii\right)f\left(\frac{3}{4}\right)>0\Rightarrow \frac{9}{16}-\frac{3(m-3)}{4}+m>0$
$\Rightarrow m>-\frac{45}{4}$
$\left(iii\right)-\frac{b}{2a}>\frac{3}{4}\Rightarrow \frac{(m-3)}{2}>\frac{3}{4}\Rightarrow m>\frac{9}{2}$
$\therefore m\in (9,\infty )$

Hence the number of inetgral values of $m$ less than $17$ is $7.$