Prove that, if $|\alpha| < 2\sqrt{2}$, then there is no value of $x$ for which $$x^2-\alpha|x| + 2 < 0\qquad\qquad(*)$$

Find the solution set of (*)  for $\alpha=3$.

For $\alpha > 2\sqrt{2}$, then the sum of the lengths of the intervals in which $x$ satisfies (*) is denoted by $S$. Find $S$ in terns of $\alpha$ and deduce that $S < 2\alpha$.