Condition of complex roots on a quartic equation

Question

I have a quartic equation: $ax^4+bx^3+cx^2+dx+e=0$. How do I find its condition of the existence of complex root for $x$? (i.e. there is imaginary part of $x$)Is there any derivation available? Thank you.

Answer 1

This is borrowed from Wikipedia article Quartic function.

For a quartic equation $ax^4+bx^3+cx^2+dx+e=0$ with all coefficients real and $a\not=0$, we define the discriminant $\Delta$ as

$$\Delta:=256a^3e^3-192a^2bde^2-128a^2c^2e^2+144a^2cd^2e-27a^2d^4+144ab^2ce^2-6ab^2d^2e-80abc^2de+18abcd^3+16ac^4e-4ac^3d^2-27b^4e^2+18b^3cde-4b^3d^3-4b^2c^3e+b^2c^2d^2$$

and define the following polynomials

$$P:=8ac-3b^2$$

$$R:=b^3+8da^2-4abc$$

$$\Delta_0:=c^2-3bd+12ae$$

$$D:=64a^3e-16a^2c^2+16ab^2c-16a^2bd-3b^4$$

Now, we can determine if the roots are real or not:

• If $\Delta<0$ then the equation has two real roots, and two non-real roots which are provably complex conjugates.

• If $\Delta>0$ then the roots are distinct and are either all real or all non-real:

  • If $P<0$ and $D<0$ then all roots are real.
  • If $P>0$ or $D>0$ then the roots are two pairs of complex conjugate non-real numbers.
  • If $\Delta=0$ then there is at least one multiple root.
  • If $P<0$, $D<0$ and $\Delta_0\not=0$ then there is a real double root and two distinct real roots.
  • If $D>0$ or both of the following are true (1) $P>0$ and (2) either $D\not=0$ or $R\not=0$, then there is a real double root and two non-real complex conjugate roots.
  • If $\Delta_0=0$ and $D\not=0$ there is a real triple root and a real simple root.
  • Finally, if $D=0$, then
    • If $P<0$ there are two real double roots.
    • If $P>0$ and $R=0$ there are two complex conjugate double roots.
    • If $\Delta_0=0$ then there is a real quadruple root, namely $x=-b/4a$.
  • The cases left out here are not possible.