Pretty straightforward question, I just had a question for the conclusion.
I rearranged, and factored and have the quadratic:
$$2x + 1/x < 3$$
(multiply both sides by x and rearrange)
$$2x^x - 3x + 1 < 0$$
$$(2x-1)(x-1) < 0$$
so $x<1/2$ and $x<1$?
Something tells me I've made an error at the end.
On
In terms of avoiding mistakes it's often easier to factor than to cancel: rather than multiplying through by $x$, you factor
$$ \frac{1}{x} (2x^2 - 3x + 1) < 0 $$ $$ \frac{(2x-1)(x-1)}{x} < 0 $$
and then use the usual methods to finish the problem. It sounds like you need a reminder on those too: the method I learned is to find the roots and poles:
and then make a sign table
$$ \begin{array}{c|c|c|c} & x < 0 & 0 < x < 1/2 & 1/2 < x < 1 & 1 < x \\ \hline \\ 2x-1 & - & - & + & + \\\hline \\ x-1 & - & - & - & + \\\hline \\ x & - & + & + & + \end{array} $$
and then you can work out which regions have a product with the correct sign.
On
When working with inequalities that have unknown stuff in a denominator, it can be a good idea to multiply by the square of the denominator:
$$2x+{1\over x}\lt3$$ becomes $$2x^3+x\lt3x^2$$
A key point is the inequality stays pointing in the same direction, because you've multiplied both sides by something that's guaranteed to be positive (if $x\not=0$). Moving the $3x^2$ to the other side and factoring as usual gives
$$x(2x-1)(x-1)\lt0$$
Since the roots of the cubic are $0$, $1/2$, and $1$, by inspection, it's easy to see that the inequality is satisfied when $x\lt0$ and when $1/2\lt x\lt 1$.
Your main problem is when you multiply by $x$. If $x$ is negative, this will switch the inequality from $< $ to $>$.
If $x$ is positive, your analysis shows that $x$ must be between $1/2$ and $1$ (for the quadratic to be negative).
If $x$ is negative, the same analysis shows that $x$ must be outside the interval between $1/2$ and $1$, because after switching the inequality you will get $2x^2 - 3x + 1 > 0$.
Thus the complete solution to your question is