Mathematics - Quadratic Equation Question with Solution | TestHub

The given inequality is $\frac{(x-3)(x+1)}{(x-1)(x-3)}\le 0$. For the expression to be defined, $x^2-4x+3\ne 0$, which means $(x-1)(x-3)\ne 0$, so $x\ne 1$ and $x\ne 3$. If $x\ne 3$, we can simplify the inequality to $\frac{x+1}{x-1}\le 0$. This inequality holds when $x+1$ and $x-1$ have opposite signs, or $x+1=0$. This occurs for $x\in [-1,1)$. Considering the restrictions $x\ne 1$ and $x\ne 3$, the solution set is $[-1,1)$. The integers in this interval are $-1$ and $0$. Thus, there are 2 such integers.