Mathematics - Quadratic Equation Question with Solution | TestHub

Let $f(x)=x^2-(m+1)x+(m+2)$. For both roots to be greater than 2, we need three conditions: 1) Discriminant $D\ge 0$: $D=(-(m+1){)}^2-4(1)(m+2)=m^2+2m+1-4m-8=m^2-2m-7\ge 0$. The roots of $m^2-2m-7=0$ are $m=\frac{2\pm \sqrt{4-4(1)(-7)}}{2}=1\pm 2\sqrt{2}$. So $m\in (-\infty ,1-2\sqrt{2}]\cup [1+2\sqrt{2},\infty )$. 2) $a\cdot f(2)>0$: Since $a=1$, $f(2)=2^2-(m+1)(2)+(m+2)=4-2m-2+m+2=4-m>0\Rightarrow m<4$. 3) Vertex $x_v=-\frac{b}{2a}>2$: $x_v=\frac{m+1}{2}>2\Rightarrow m+1>4\Rightarrow m>3$. Combining these conditions: $m\in (-\infty ,1-2\sqrt{2}]\cup [1+2\sqrt{2},\infty )$, $m<4$, and $m>3$. The intersection of $m<4$ and $m>3$ is $(3,4)$. The intersection of $(3,4)$ with $(-\infty ,1-2\sqrt{2}]\cup [1+2\sqrt{2},\infty )$ (where $1+2\sqrt{2}\approx 3.828$) is $[1+2\sqrt{2},4)$.