Mathematics - Quadratic Equation Question with Solution | TestHub

Let $f(x)=x^2-(2k+1)x+k^2+k-6$. For exactly one root in $(0,3)$, either $f(0)f(3)<0$ or one of $f(0)$ or $f(3)$ is zero and the other root lies in $(0,3)$. First, the discriminant $D=(2k+1{)}^2-4(k^2+k-6)=4k^2+4k+1-4k^2-4k+24=25>0$, so roots are always real and distinct. We have $f(0)=k^2+k-6=(k+3)(k-2)$ and $f(3)=3^2-3(2k+1)+k^2+k-6=9-6k-3+k^2+k-6=k^2-5k=k(k-5)$. Case 1: $f(0)f(3)<0$. This implies $(k+3)(k-2)k(k-5)<0$. The critical points are $-3,0,2,5$. Using the wavy curve method, $k\in (-3,0)\cup (2,5)$. The integer values for $k$ are $\{-2,-1\}$ and $\{3,4\}$. Case 2: $f(0)=0$. This means $k=-3$ or $k=2$. If $k=-3$, the equation is $x^2+5x=0$, roots are $0,-5$. Neither root is in $(0,3)$. If $k=2$, the equation is $x^2-5x=0$, roots are $0,5$. Neither root is in $(0,3)$. Case 3: $f(3)=0$. This means $k=0$ or $k=5$. If $k=0$, the equation is $x^2-x-6=0$, roots are $3,-2$. Neither root is in $(0,3)$. If $k=5$, the equation is $x^2-11x+24=0$, roots are $3,8$. Neither root is in $(0,3)$. Thus, only Case 1 provides valid integer values of $k$. The number of distinct integer values of $k$ is $4$.