# Radical solution to a polynomial quartic equation

Consider the following quartic equation:

$$x^4 + rx^3 + r^2x^2 + r^3x + r^4 - 1 = 0$$

By Lodovico Ferrari solution, this equation must possess four radical solution provided that $r$ is a rational number, my question is simply if we assume $r$ is an algebraic number (which is a more general set of numbers that contains all rational numbers), can this equation still possess four radical roots

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• The equation is not biquadratic. A biquadratic equation is of the form $ax^4+bx^2+c=0$. The equation $ax^4+bx^3+cx^2+dx+e=0$ is called a quartic. – Peter May 31 '15 at 7:14
• Do you mean that the roots are radical for every algebraic $r$ ? Then the answer is no. Do you mean that for some algebraic irrational $r$ the roots are radical ? Then the answer is yes. – Peter May 31 '15 at 7:18
• Actually as you mentioned I used to get radical roots for some algebraic (r), but suspected that can be generalised for any algebraic (r), also I re edited the question based on your valuable remarks – bassam karzeddin May 31 '15 at 7:33
• The question as it stands has a problem: it doesn't state which is the field over which it is considered. If the field is $\Bbb C$, then its roots will always be expressed as radicals. Further, Cardano's formulae are for equations of degree $3$, not $4$. – Alex M. Jun 1 '15 at 17:03

The roots of the equation are radical if and only if $r$ itself can be expressed by radicals.
• @bassamkarzeddin If the roots are expressible by radicals, then so are the symmetric functions of the roots. In particular $-r$ is the sum of the roots, and so is expressible by radicals. – Mark Bennet May 31 '15 at 11:16
• I do not follow: the $4$th degree equation with complex coefficients is always solvable by radicals, irrespective of other properties of its coefficients. What am I missing? – Alex M. Jun 1 '15 at 17:06
• @Peter, the OP's expression is quartic in both $x$ and $r$, so the "other direction" is more or less automatic: If, for a given number $r$, there is a root $x$ that can be expressed in radicals (of rational numbers), then the quartic in $r$, using those radical expressions as coefficients, is solvable in radicals (because it's a quartic), and one of its four roots must be the original number $r$. – Barry Cipra Nov 12 '15 at 16:19