Mathematics - Quadratic Equation Question with Solution | TestHub

Let $f(x)=x^2-(k-3)x+k$. For the roots to satisfy $0<\alpha <1<\beta <2$, the following conditions must hold: 1. The discriminant $D>0$. $D=(k-3{)}^2-4k=k^2-10k+9=(k-1)(k-9)>0$, which implies $k\in (-\infty ,1)\cup (9,\infty )$. 2. $f(0)>0⟹k>0$. 3. $f(1)<0$. $f(1)=1-(k-3)+k=1-k+3+k=4$. Since $f(1)=4$ is always positive, the condition $f(1)<0$ can never be satisfied. Therefore, there are no real values of $k$ for which the given conditions hold, and the set of such $k$ is $\varnothing$.