Is there any way to tell whether a quartic equation has double or triple root(s)?
$$x^4 + a x^3 + b x^2 + c x + d = 0$$

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    $\begingroup$ Do you mean construct a quartic with double or triple roots, or do you mean determine whether a quartic has a double or triple root? $\endgroup$ – Brian M. Scott May 4 '12 at 9:32
  • $\begingroup$ @Brian M. Scott : I mean determine whether a quartic has a double or triple root $\endgroup$ – mrbm May 4 '12 at 9:34
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    $\begingroup$ In the meantime: you'll want to see this paper. $\endgroup$ – J. M. is a poor mathematician May 4 '12 at 9:37
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    $\begingroup$ @Brian M. Scott: Thank you, Can you now give a appropriate answer too? :) $\endgroup$ – mrbm May 4 '12 at 9:39
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    $\begingroup$ Ahem, a quartic equation can have as few as 0 roots or as many as 4, gentlemen. $\endgroup$ – Neil May 4 '12 at 9:48

To answer your question: yes.

As Neil mentioned in the comments, a quartic can have between 0 and 4 different roots. Other than the obvious way of checking the number of roots by actually solving the equation, the following also applies, as listed in J.M.'s link to this monthly article:

For an equation of the type $x^4+qx^2+rx+s=0$

enter image description here

If you were looking for a quick way, easy to memorize, I'm afraid this won't be of much help after all. Also, notice the absence of a $x^3$-term.

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    $\begingroup$ The removal of the cubic term can be easily arranged, of course; by Vieta, the sum of the roots of $ax^4+bx^3+cx^2+dx+e$ is $-\dfrac{b}{a}$; thus, letting $x=u-\dfrac{b}{4a}$ results in a quartic in $u$ without a $u^3$ term, or, in more common parlance, a depressed quartic. $\endgroup$ – J. M. is a poor mathematician May 4 '12 at 11:27

The quartic $x^4 + a x^3 + b x^2 + c x + d = 0$ has a multiple root if it has a common root with $4x^3 + 3a x^2 + 2b x +c = 0$.

Any common factor can be determined using Euclid's algorithm. The "worst case" is that there are two double roots and the common factor is a quadratic - but that's easy to factor. A triple root will give a quadratic common factor too, but recognisable as having two equal roots.

This is not a simple criterion in terms of the coefficients, but it is a practical way of computing multiple roots when they exist and if the factorisation is not obvious.

The question is tagged pre-calculus - the easiest way to prove that this works is with some simple calculus (the second equation is obtained by differentiating the first). But the criterion can be applied without calculus, and is valid in many situations where limits do not make sense - and this motivates generalisations and extensions of the notion of derivative in algebra.


Maybe I misunderstood the question, but I think this might be useful for you.

Let $f(x)\in F[x]$ be a polynomial and $f'(x)$ be the formal derivative of $f(x)$ where $F$ is any field.

Then $f(x)$ has multiple roots if and only if $\gcd(f(x),f'(x))\ne 1$.

If $F$ is a field of characteristic zero then you know more: If you denote $g(x)=\gcd(f(x),f'(x))$ and divide $f(x)$ by $g(x)$, i.e. you find the polynomial $h(x)$ such that $f(x)=g(x)\cdot h(x)$, then the polynomials $f(x)$ and $h(x)$ have the same roots, but the multiplicity of each root of $h(x)$ is one.

You are asking about polynomial over $\mathbb C$; for this field the above results are true. Moreover, $\mathbb C$ is algebraically closed, hence every quartic polynomial has 4 roots in $\mathbb C$, if we count them with multiplicities.

So in your case, you can compute $g(x)=\gcd(f(x),f'(x))$, e.g. using Euclidean algorithm. If $g(x)=1$, then all (complex) roots of $f(x)$ have multiplicity one. If $g(x)=f'(x)$ then $f(x)$ has only one root of multiplicity 4. In the remaining two cases there is a root of multiplicity 2 or 3.

(However, if you're only interested in multiplicity of real roots, the situation is slightly more complicated.)

  • $\begingroup$ That criterion for multiple roots should be $\neq 1$? $\endgroup$ – Mark Bennet May 4 '12 at 10:51
  • $\begingroup$ Thanks Mark, I've corrected that. (And sorry for the typo in your name in the comment before.) $\endgroup$ – Martin Sleziak May 4 '12 at 11:00
  • $\begingroup$ "e.g. using Euclidean algorithm" is not as helpful as the rest of the explanation. It is easy to find the information linked to, which describes the Euclidean algorithm for known integer expressions. How does one proceed with algebraic expressions? I am experimenting, in a similar vain to the technique for algebraic long division, but as yet am having no luck. Is there a more helpful link that someone could provide? My searches haven't been that successful, yet. $\endgroup$ – user37225 Aug 4 '12 at 17:53

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