Determine where the range of a function is positive Consider the quadratic equation $y = -\dfrac{3}{4}x^2 + 4x - 4$, with roots at $x = \tfrac{4}{3}$ and $x = 4$. I'd like to find the values for $x$ where $y > 0$; that is where
$$
   -\dfrac{3}{4}x^2 + 4x - 4 > 0.
$$
Rewritting as
$$
(x - \tfrac{4}{3})(x - 4) > 0,
$$
the above inequality is true when 
A) both $(x - \tfrac{4}{3})$ and $(x - 4)$ are positive 
or 
B) both $(x - \tfrac{4}{3})$ and $(x - 4)$ are negative. 
Case A - both positive: 
$x - \tfrac{4}{3} > 0$ => $x > 4/3$ and $x - 4 > 0$ => $x > 4$. This is true when $x > 4$. 
Case B - both negative: 
$x - \tfrac{4}{3} < 0$ => $x < 4/3$ and $x - 4 < 0$ => $x < 4$. This is true when $x < \tfrac{4}{3}$. 
Therefore the solution is $(-\infty, \tfrac{4}{3}) \cup (4,\infty)$. Yet graphing this equation shows the positive range as $(4/3,4)$. The range $(-\infty, \tfrac{4}{3}) \cup (4,\infty)$ actually satisfies the equation $y = \dfrac{3}{4}x^2 - 4x + 4$.
Where have I run afoul?

$$-\dfrac{3}{4}x^2 + 4x - 4 > 0$$ can't be written as$$(x-4/3)(x-4) > 0$$ but as $$-\frac34(x-4/3)(x-4) > 0$$which gives$$(x-4/3)(x-4)\color{red}<0$$
