I need a well defined table / scheme for the nature (real / complex) roots of the generalized biquadratic equation.
$$ax^4+bx^2+c=0;a,b,c\in\mathbb{R},a\ne0$$ $$x^2=z$$ $$az^2+bz+c=0$$ $$x_{1,2}^2=z_1,x_{3,4}^2=z_2$$
Consider $S=z_1+z_2=-\frac{b}{a},P=z_1z_2=\frac{c}{a}$. Now suppose for example that $z_1,z_2<0$. The inequality is equivalent with the system $$ \begin{cases} z_1z_2>0 & (1)\\ z_1+z_2<0 & (2) \end{cases} $$ because (1) gives $z_1,z_2>0~\lor~z_1,z_2<0$ but (2) excludes the first possibility.
I used the same technique for other cases but I am unsure if my results are perfect or if I'm missing an edge case. Could anyone look over them please?
I have many biquads such as $x^4+2mx^2+m^2-1$ and I need to "discuss" over intervals of $m$ the nature of the roots, that's why I need it. Any suggestion is welcome.
Case I : $\Delta_z<0\Rightarrow x_{1,2,3,4}\in\mathbb{C-R}$
Case II : $\Delta=0\Rightarrow z_{1,2}=\frac{S}{2}$
$S<0\Rightarrow x_1=x_2\in\mathbb{C-R},x_3=x_4=\mathbb{C-R}$
$S>0\Rightarrow x_1=x_2\in\mathbb{R},x_3=x_4\in\mathbb{R}$
Case III : $\Delta_z>0$
a. $P<0\Rightarrow x_{1,2}\in\mathbb{R},x_{3,4}\in\mathbb{C-R}$
b. $P>0$
$S>0\Rightarrow x_{1,2,3,4}\in\mathbb{R}$.
$S<0\Rightarrow x_{1,2,3,4}\in\mathbb{C-R}$
c. $P=0$
$S=0\Rightarrow x_{1,2,3,4}=0$.
$S\ne0\Rightarrow x_{1,2}=0$
$S>0\Rightarrow x_{3,4}\in\mathbb{R}$
$S<0\Rightarrow x_{3,4}\in\mathbb{C-R}$