Solve for $(5x-1)(x-3)<0$ The inequality $(5x-1)(x-3)<0$ is true when 
$(5x-1)<0$ and $(x-3)>0$ 
or 
$(5x-1)>0$ and $(x-3)<0$.
If I solve for $x $ in the first scenario, $x < \frac{1}{5}$ and $x > 3$ which is wrong. But if I solve for $x$ in the second scenario, $x > \frac{1}{5}$ 
and $x < 3$ which is correct. 
Why is such kind of contradiction occurring in the first scenario?

$f(x)=(5x-1)(x-3)$ polynomial of degree 2 has 2 zeroes: $5x-1=0$, and $x=3$.
1) $\lim_{x \rightarrow \pm \infty}f(x)= \infty$.
Hence for 
$5x-1<0$, or $x <1/5$: $f(x)>0$, and
$x>3$: $f(x) >0$.
2) For $x \in (1/5,3)$:
$f(x)= 5(x-1/5)(x-3) <0$ , since $(x-1/5)>0$ and $(x-3) <0$.
