Question

If the polynomial equation $(x^2+x+1{)}^2-(m-3\left)\right(x^2+x+1)+m=0,m\in R$ has two distinct real roots, then $m$ lies in the interval

• A. $(\frac{45}{4},\infty )$
• B. $(-\infty ,\frac{-45}{4})$
• C. $(\frac{9}{2},\infty )$
• D. $(9,\infty )$

Answer 1

The correct option is B $(-\infty ,\frac{-45}{4})$

$(x^2+x+1{)}^2-(m-3\left)\right(x^2+x+1)+m=0$
Let $t=x^2+x+1={(x+\frac{1}{2})}^2+\frac{3}{4}$
$\Rightarrow t\in [\frac{3}{4},\infty )$
Given equation becomes $t^2-(m-3)t+m=0\cdots \left(1\right)$
Let its roots be $t_1$ and $t_2$
(i) For every $t>\frac{3}{4}$, there exist two 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 will have two distinct roots iff for equation $\left(1\right)$, roots are of form
$t_1<\frac{3}{4}$ and $t_2>\frac{3}{4}$
i.e., $\frac{3}{4}$ lies in between the roots of equation $\left(1\right)$.
$\Rightarrow f\left(\frac{3}{4}\right)<0$
$\Rightarrow \frac{9}{16}-(m-3)\frac{3}{4}+m<0$
$\Rightarrow m<-\frac{45}{4}$