absolute value inequalities When answer this kind of inequality
$|2x^2-5x+2| < |x+1|$
I am testing the four combinations when both side are +, one is + and the other is - and the opposite and when they are both -.
When I check the negative options, I need to flip the inequality sign?
Thanks   

Another way - no tricks, just systematically looking at all cases, where we can write the inequality without the absolute value sign, to convince you that all possibilities are covered..
Following the definition of the absolute value function, RHS is easy to rewrite as,
$$|x+1| = \begin{cases} x+1 & x \ge -1\\ -x-1 &x < -1 \end{cases}$$
Noting $2x^2-5x+2 = (2x-1)(x-2)$, we have the LHS as
$$|2x^2-5x+2| = \begin{cases} 2x^2-5x+2 & x \le \frac12 \text{ or } x \ge 2\\ 
-2x^2+5x-2 &\frac12 < x < 2 \end{cases}$$
Thus we can break the number line into for regions (using break points at $x = -1, \frac12, 2$)  and write the inequality in each region without any absolute value symbols.  Then we solve the inequality in each region, and combine results in the end.  Details follow. 
Region 1: $x < -1$
Here the inequality is $2x^2-5x+2 < -x - 1$ which is equivalent to
$2x^2 - 4x + 3 < 0$ or $2(x-1)^2+1 < 0$, which is not possible.
Region 2: $-1 \le x \le \frac12$
Here we have the inequality as  $2x^2-5x+2 < x + 1 \iff 2x^2 - 6x + 1 < 0$ which is true when $\frac12(3-\sqrt7) < x <  \frac12(3+\sqrt7)$.  Taking the part of this solution which is in the region being considered, we have solutions for $x \in (\frac12(3-\sqrt7), \frac12]$   
Region 3: $\frac12 < x < 2$
Here the inequality is $-2x^2+5x-2 < x + 1 \iff 2(x- 1)^2 + 2 > 0$ which holds true for all $x$.  So the entire region is a solution.
Region 4: $2 \le x$
Here the inequality is $2x^2-5x+2 < x + 1$, which is exactly the same form as Region 2, and has the same solutions.  Hence $2 \le x <  \frac12(3+\sqrt7)$ is the solution from this Region.
Combining solutions from all regions, we have $\frac12(3-\sqrt7) < x <  \frac12(3+\sqrt7)$ or $|x - \frac32| < \frac{\sqrt7}{2}$ as the complete solution set (which others have already pointed out). 
