I do a procedure for solving algebraic inequalities of the second ($x^2+bx+c>0$) degree for my student. I know it is possible to solve the inequality by factorisation, Solving a quadratic inequality, see the first response by Casebash. I try another method.
($\Delta=\left(\frac{b}{2}\right)^2 +c$)

$\left(x+\frac{b}{2}\right)^2-\Delta>0$

$\left(x+\frac{b}{2}\right)^2>\Delta$

$\sqrt{\left(x+\frac{b}{2}\right)^2}\gtrless\pm\sqrt{\Delta}$

$x+\frac{b}{2}\gtrless\pm\sqrt{\Delta}$ and then together,

$x+\frac{b}{2}>+\sqrt{\Delta}$ and $x+\frac{b}{2}<-\sqrt{\Delta}$

$x>-\frac{b}{2}-\sqrt{\Delta}$ and $x<-\frac{b}{2}-\sqrt{\Delta}$

How to explain the $\pm$ on the right but not left of the inequality? I'm confused because $|a|=\sqrt{a^2}$.

It is possible to clarify the explanation? or it's a dead end.

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You can. Then the combinations where both sides are $+$ or both are $-$ give you one thing, the combinations where the signs are different give you the other. –  André Nicolas Jun 5 '12 at 16:08
The notation $A > \pm B$ is quite meaningless and should be avoided. –  Hans Lundmark Jun 5 '12 at 20:06
@hans-lundmark How can i write this? –  La Raison Jun 5 '12 at 20:29
Write what? It's completely unclear what is meant. Write what you mean instead. You can for example use two separate inequalities: "$A > B$ or $A < -B$" (if that's what you mean). Or use absolute values: "$|A|>B$". –  Hans Lundmark Jun 5 '12 at 20:44

It is highly undesirable to include $\pm$ in an inequality. You are correct that $\sqrt{a^2}=|a|$ for real $a$, but it seems what you're confused about is what to do with an inequality containing an absolute value.

In the case that $\Delta<0$, then the inequality $\left(x+\frac{b}{2}\right)^2>\Delta$ holds for all real $x$, and we're done.

In the case that $\Delta\geq 0$, then we proceed as follows: $$\left(x+\frac{b}{2}\right)^2>\Delta$$ $$\left|x+\frac{b}{2}\right|>\sqrt{\Delta}$$ $$x+\frac{b}{2}>\sqrt{\Delta}\quad \mathrm{or}\quad x+\frac{b}{2}<-\sqrt{\Delta}$$ $$x>-\frac{b}{2}+\sqrt{\Delta}\quad \mathrm{or}\quad x<-\frac{b}{2}-\sqrt{\Delta}$$

Does that help to clarify things?

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At first, I did not want to use the absolute value, but I think we have to, directly or in a disguised manner. Students must master the absolute value function in this way. –  La Raison Jun 6 '12 at 12:34