Mathematics - Quadratic Equation Question with Solution | TestHub

For both roots of $x^2-(k-3)x+k=0$ to be greater than 2, three conditions must be met: (1) Discriminant $D\ge 0$, (2) $a\cdot f(2)>0$, and (3) the $x$ -coordinate of the vertex, $-\frac{b}{2a}$, must be greater than 2. From $D\ge 0$, we get $(k-3{)}^2-4k\ge 0⟹k^2-10k+9\ge 0⟹(k-1)(k-9)\ge 0$, so $k\in (-\infty ,1]\cup [9,\infty )$. From $a\cdot f(2)>0$, we have $1\cdot (2^2-(k-3)2+k)>0⟹4-2k+6+k>0⟹10-k>0⟹k<10$. From $-\frac{b}{2a}>2$, we have $\frac{k-3}{2}>2⟹k-3>4⟹k>7$. Combining these conditions, $k$ must be in $[9,10)$.