Mathematics - Quadratic Equation Question with Solution | TestHub

Factorizing the numerator and denominator, we get $\frac{(x-1)(x-2)}{(x-2)(x-3)}\ge 0$. For this expression to be defined, $x\ne 2$ and $x\ne 3$. For $x\ne 2$, we can simplify the expression to $\frac{x-1}{x-3}\ge 0$. The critical points are $x=1$ and $x=3$. Using a sign chart: for $x<1$, both $(x-1)$ and $(x-3)$ are negative, so the fraction is positive. For $1\le x<3$, $(x-1)$ is non-negative and $(x-3)$ is negative, so the fraction is non-positive (excluding $x=2$). For $x>3$, both $(x-1)$ and $(x-3)$ are positive, so the fraction is positive. Combining these, the solution is $x\in (-\infty ,1]\cup (3,\infty )$. This interval naturally excludes $x=2$ and $x=3$ (as $3$ is an open interval boundary).