Question

For what values of $m$ is the inequality
$\left|\frac{x^2+mx+1}{x^2+x+1}\right|<3$

Answer 1

since $x^2+x+1>0$ for all real x, the given inequality can be written as
$-3(x^2+x+1)<x^2+mx+1<3(x^2+x+1)$
$\therefore$ $4x^2+(m+3)x+4>0$........(1)
and $2x^2-(m-3)x+2>0$.........(2)
The inequality (1) will bold for all x if its
$\Delta ={(m+3)}^2-64<0$
and the sign will be same as the coefficient of Ist term i.e., $+ive$.
or $(m+11)(m-5)<0$ which implies that
$-11<m<5$........(3)
Similarly, (2) will hold for all x if its $\Delta <0$
$-1<m<7$.......(4)
Finally from (3) and (4), $-1<m<5$.