Mathematics - Quadratic Equation Question with Solution | TestHub

Let $f(x)=(m-2)x^2-(m-4)x-2$. Case 1: $m-2=0⟹m=2$. The equation becomes $0x^2-(2-4)x-2=0⟹2x-2=0⟹x=1$. Since the interval is $(0,1)$, $x=1$ is not in the interval. Thus, $m=2$ is not a solution. Case 2: $m\ne 2$. The equation is quadratic. For exactly one root in $(0,1)$, we examine the product $f(0)\cdot f(1)$.$f(0)=-2$.$f(1)=(m-2)(1{)}^2-(m-4)(1)-2=m-2-m+4-2=0$. Since $f(1)=0$, $x=1$ is a root of the equation. As $1\notin (0,1)$, we need the other root to be in $(0,1)$. Let the roots be $1$ and $\beta$. From Vieta's formulas, the product of roots is $1\cdot \beta =\frac{-2}{m-2}$. So $\beta =\frac{-2}{m-2}$. We need $0<\beta <1$, which means $0<\frac{-2}{m-2}<1$.

First inequality: $\frac{-2}{m-2}>0$. Since the numerator $-2$ is negative, the denominator $m-2$ must be negative. So $m-2<0⟹m<2$. Second inequality: $\frac{-2}{m-2}<1⟹\frac{-2}{m-2}-1<0⟹\frac{-2-(m-2)}{m-2}<0⟹\frac{-m}{m-2}<0⟹\frac{m}{m-2}>0$. This inequality holds when $m\in (-\infty ,0)\cup (2,\infty )$. Combining the conditions $m<2$ and $m\in (-\infty ,0)\cup (2,\infty )$, the intersection is $m\in (-\infty ,0)$.