Mathematics - Quadratic Equation Question with Solution | TestHub

Let $f(x)=x^2-(m+1)x+m^2+m-8$. For both roots to be less than 2, three conditions must be met: 1) Discriminant $D\ge 0$: $(m+1{)}^2-4(m^2+m-8)\ge 0⟹-3m^2-2m+33\ge 0⟹3m^2+2m-33\le 0$. The roots of $3m^2+2m-33=0$ are $m=3$ and $m=-11/3$. So, $m\in [-11/3,3]$. 2) $f(2)>0$: $2^2-(m+1)(2)+m^2+m-8>0⟹m^2-m-6>0⟹(m-3)(m+2)>0$. So, $m\in (-\infty ,-2)\cup (3,\infty )$. 3) Axis of symmetry $-b/(2a)<2$: $\frac{m+1}{2}<2⟹m+1<4⟹m<3$. Combining these conditions: $m\in [-11/3,3]$ AND $m\in (-\infty ,-2)\cup (3,\infty )$ AND $m\in (-\infty ,3)$. The intersection of these intervals is $m\in [-11/3,-2)$.