Mathematics - Quadratic Equation Question with Solution | TestHub

Let $f(x)=x^2-(m+3)x+3m$. For both roots to be greater than 1, we need: 1) Discriminant $D\ge 0$: $(m+3{)}^2-4(3m)=m^2-6m+9=(m-3{)}^2\ge 0$, which is true for all real $m$. 2) Vertex $x$ -coordinate $x_v>1$: $x_v=\frac{m+3}{2}>1⟹m+3>2⟹m>-1$. 3) $f(1)>0$: $f(1)=1-(m+3)+3m=2m-2>0⟹m>1$. Combining $m>-1$ and $m>1$, the intersection is $m>1$. So, $m\in (1,\infty )$.