Why in this inequality numerator is always negative?

Question

Starting from: $$\frac{x-1}{x-2}-\frac{2x-3}{x-3}>0$$ I get: $$\frac{-x^2+3x-3}{(x-2)(x-3)}>0$$ Numerator roots are not real: $$x_{1,2}=\frac{-3\pm \sqrt{3^2- 4 \cdot (-1) \cdot (-3)}}{-2}$$ Why can I conclude that $N(x)<0$, i.e. the numerator is always negative? So the solution is from the denominator: $D(x)<0 \Rightarrow x \in (2,3)$.

Answer 1

You have, for each $x\in\Bbb R$,$$-x^2+3x-3=-\left(x-\frac32\right)^2-\frac34<0.$$You can also say that, since $-x^2+3x-3=-3$ when $x=0$, if you had $-x^2+3x-3>0$ for some $x\in\Bbb R$, then $-y^2+3y-3=0$ for some $y$ between $0$ and $x$, by the Intermediate Value Theorem.

Answer 2

Every quadratic with real coefficients either has two real roots, or two non-real roots. Here, you have shown correctly that $-x^2+3x-3$ has two non-real roots, meaning that if you plot the graph $y=-x^2+3x-3$, it will not touch the $x-$axis. Furthermore, the coefficient of $x^2$ is negative, meaning that the parabola will be 'n-shaped':

You can also argue that if the parabola does not touch the $x$-axis, then the quadratic must be positive everywhere, or negative everywhere. If we plug in $x=0$, we find that $y=-3$, and so the quadratic must be negative everywhere.

Finally, the simplest member is probably to complete the square on $-x^2+3x-3$: \begin{align} -x^2+3x-3 &= -(x^2-3x+3) \\ &= -\left(\left(x-\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+3\right) \\ &= -\left(\left(x-\frac{3}{2}\right)^2+\frac{3}{4}\right) \\ &= -\left(x-\frac{3}{2}\right)^2 - \frac{3}{4} \end{align} The expression $$ \left(x-\frac{3}{2}\right)^2 $$ must always be positive for real $x$.* Hence, $$ -\left(x-\frac{3}{2}\right)^2 $$ is always negative. Therefore, $$ -\left(x-\frac{3}{2}\right)^2 - \frac{3}{4} $$ is always negative.

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*I should really say non-negative for real $x$, as if $x=\frac{3}{2}$ then the expression is zero.