MHB Roots of p ( z ) in IR [ z ] (Lava's question at Yahoo Answers)

[Fernando Revilla]

Here is the question:

Solve the equation.

Here is a link to the question:

The equation z3 + az2 + bz + c = 0, where a, b, c are real, have a purely imaginary root (i.e. the real part o? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Lava,

Denote $z_1,z_2,z_3$ the roots of $p(z)=z^3+az^2+bz+c\in\mathbb{R}[z]$. If $\beta i$ is a root of $p(z)$, then $-\beta i$ is also a root of $p(z)$ ($p(z)$ is a real polynomial). By hypothesis $\beta\ne 0$ and $p(z)$ has a real root $\alpha$ (odd degree). If $z_1=\alpha$, $z_2=\beta i$, $z_3=-\beta i$, and using the Cardano-Vieta relations: $$\left \{ \begin{matrix} z_1+z_2+z_3=-a\\z_1z_2+z_1z_3+z_2z_3=b\\z_1z_2z_3=-c\end{matrix}\right.\Leftrightarrow \left \{ \begin{matrix} \alpha=-a\\\beta^2=b\\\alpha\beta^2=-c\end{matrix}\right.$$ As $\beta^2=b=c/a$, necessarily $c=ba.$ So the roots of $p(z)$ are $z_1=-a$, $z_2=+\sqrt{|b|}i$, $z_3=-\sqrt{|b|}i$.