# How to count the real roots of a quartic equation?

Suppose I have a quartic equation with real coefficients, such as:

$$a x^4 +b x^3+c x^2+d x +e=0$$

I want to know the number of its real roots. Search engines lead me to symbolic expressions for all the roots, and these can be produced by CAS packages like Mathematica, but these results are too long and complex (in both senses) to be of use to me.

I would hope there is a compact/efficient method to count the real roots of a real quartic equation, similar to the way the discriminant of a quadratic polynomial tells us the number of real roots of a real quadratic equation.

• since you wrote $a x^2$ instead of $a x^4,$ you did not investigate a quartic. – Will Jagy Jul 13 '15 at 3:20
• @WillJagy, OK, thanks:) I have edited it. – xyz Jul 13 '15 at 3:22
• Is $e=2.718...?$ – user253055 Jul 13 '15 at 4:09
• @ShutaoTang: I believe you are asking about methods that provide a count of real roots of a given real quartic polynomial, as your mention of the discriminant of a quadratic real polynomial illustrates. There is a definition of discriminant for any real polynomial that gives some information about the number of real roots, though not necessarily a full answer, and there are other well-known methods for finding how many real roots there are. With your encouragement I will edit your Question to clarify that this is what you are asking. – hardmath Jul 13 '15 at 12:15
• You will find a complete discussion here: en.wikipedia.org/wiki/… – Yves Daoust Jul 30 '15 at 6:48

General (and simple) method for polynoms is Sturm sequence, if you know coefficients. In general case there is no methods. But if you know smth about coefficients, you can use discriminant.

Basically, the answer is no. There is not a simple and quick way to do it. There are ways, but they require a bit of work or are not guaranteed.

• Wikipedia gives the complete discussion of the roots of the quartic. en.wikipedia.org/wiki/…. It is based on five polynomial expressions and is exact. it can be simplified to the discussion of real roots alone. – Yves Daoust Jul 30 '15 at 6:54
• It is exact yes, but it is also not simple unless you have a CAS to sort through the mess. Then again, the asker apparently does have such a thing available. – Terra Hyde Jul 30 '15 at 14:02
• Not at all, the case analysis is very short, less than a handful of sign tests. – Yves Daoust Jul 30 '15 at 14:06

I believe that Vieta's Formulae can help here. First, you divide the whole equation by $$a$$. We can redefine variables $$b, c, d,$$ and $$e$$ as $$\frac{b}{a}, \frac{c}{a}, \frac{d}{a},$$ and $$\frac{e}{a}$$ respectively. Basically, you expand the following, where $$p, q, r,$$ and $$s$$ are roots:

$$(x-p)(x-q)(x-r)(x-s)$$

And get:

\begin{align}b &= -(p+q+r+s) \\ c &= pq + pr + ps + qr + qs + rs \\ d &= -(pqr + pqs + prs + qrs)\\ e &= pqrs\end{align}

You now have $$4$$ variables and $$4$$ equations, so it is now a matter of solving this system. Admittedly, it is rather tedious, and requires quite a lot of work.