Mathematics - Quadratic Equation Question with Solution | TestHub

Let $f(x)=x^2-(a^2-5a+6)x+(a^2-4a+3)$. For exactly one root to lie in $(1,2)$, we must have $f(1)f(2)<0$.$f(1)=1-(a^2-5a+6)+(a^2-4a+3)=a-2$.$f(2)=4-2(a^2-5a+6)+(a^2-4a+3)=-a^2+6a-5=-(a-1)(a-5)$. Thus, $(a-2)(-(a-1)(a-5))<0⟹(a-1)(a-2)(a-5)>0$. This inequality holds for $a\in (1,2)\cup (5,\infty )$. We must ensure that $f(1)\ne 0$ and $f(2)\ne 0$ to avoid roots being exactly 1 or 2. If $a=1,2,$ or $5$, one of $f(1)$ or $f(2)$ is zero, and in these cases, neither root lies strictly within $(1,2)$. Hence, the strict inequality is sufficient. The range of 'a' is $(1,2)\cup (5,\infty )$.