Find the solution set of $\displaystyle x + \frac{1}{4x} < 1$. The answer they gave was {x : x < 0}
$\displaystyle x+\frac{1}{4x}-1<0$
$\displaystyle \frac{4x^2}{4x}+\frac{1}{4x}-\frac{4x}{4x}<0$
$\displaystyle \frac{4x^2-4x+1}{4x}<0$
$\displaystyle
\frac{(2x-1)^2}{4x}<0
$
Note that this $\displaystyle (2x-1)^2$ is always positive so now lets look at the denominator $\displaystyle (4x)$ so the only possible solution here is $\displaystyle x<0$