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.