Mathematics - Quadratic Equation Question with Solution | TestHub

For exactly one root to lie in $(0,2)$, we consider two cases. Case 1: $f(0)f(2)<0$. Here $f(0)=k^2-3k+2=(k-1)(k-2)$ and $f(2)=4-2(k-2)+(k^2-3k+2)=k^2-5k+10$. The discriminant of $k^2-5k+10$ is $25-40=-15<0$, so $k^2-5k+10>0$ for all real $k$. Thus, we need $(k-1)(k-2)<0$, which gives $1<k<2$. No integer values of $k$ satisfy this. Case 2: One of the boundary points is a root. If $f(0)=0$, then $(k-1)(k-2)=0⟹k=1$ or $k=2$. If $k=1$, $f(x)=x^2+x=0$, roots are $0,-1$. Neither root lies in $(0,2)$. If $k=2$, $f(x)=x^2=0$, roots are $0,0$. Neither root lies in $(0,2)$. If $f(2)=0$, then $k^2-5k+10=0$, which has no real solutions for $k$. For real roots of $f(x)=0$, the discriminant $D=(k-2{)}^2-4(k^2-3k+2)=-3k^2+8k-4\ge 0$, which implies $k\in [2/3,2]$. All integer $k$ values considered (1, 2) are within this range. Since no integer $k$ satisfies either case, the number of such integer values is 0.