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When a cubic equation has a zero discriminant, and thus has real roots with multiplicity, is there any closed-form solution available for such roots? All sources I found just mention either the Cardano-method roots (which involve complex numbers even if the result is real), or the Viète/Descartes trigonometric form for the real roots, but they require a negative non-zero discriminant (if the discriminant is zero, you get an indetermination at the arccos of infinity).

Is there any closed-form (without complex numbers) for the roots when the discriminant is zero?

cesss
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    Any double root of $P(x)$ will be a root of the derivative $P'(x)$ as well, which is a quadratic and can be solved directly, or find $\gcd(P,P')$ by long division. – dxiv Nov 19 '16 at 00:37
  • I'm trying to understand this, but the derivative roots are different to the double root that you can see in Claude's answer. The simple root is different too, – cesss Nov 19 '16 at 17:07
  • The expression posted by Claude may look different, but gives in fact the same double root. Note that the expression is only valid if $\Delta = 0$ so at that point there is an additional known relationship between $a,b,c,d$. Because of that, such expressions don't necessarily have a unique representation in terms of the coefficients. For an unrelated trivial example, if you have two variables $u,v$ and you know that $u+v=1$ then the expressions $u-v$ and $u^2 - v^2$ are equivalent. – dxiv Nov 19 '16 at 17:55
  • For a concrete example, say $P(x)=(x-1)^2(x+2)= x^3+x^2-5 x+3$. Then $P'(x)=3 x^2+2 x-5$ and calculating the $\gcd$ using the Euclidian algorithm gives in the end $\gcd(P(x),P'(x))=x-1$. Therefore $x=1$ is a root of $P'(x)$ and a double root of $P(x)$. Alternatively, if you first verify that the discriminant $\Delta=0$ then the same $x=1$ is given by Claude's formula $\big(9 \cdot 1 \cdot 3 - 1 \cdot (-5)\big) / \big(2 \cdot (1 \cdot 1 - 3 \cdot 1 \cdot (-5))\big) = 32 / (2 \cdot 16) = 1$. – dxiv Nov 19 '16 at 18:01
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    Thanks a lot, I understand it now. – cesss Nov 19 '16 at 23:20

1 Answers1

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For the cubic equation $$ax^3+bx^2+cx+d=0$$ if $\Delta=0$ and $\Delta_0=b^2-3ac=0$ are equal to $0$, the equation has a triple root $$x=-\frac{b}{3a}$$ If $\Delta=0$ and $\Delta_0\neq 0$ , there are a double root
$$x_{1,2}=\frac{9ad-bc}{2(b^2-3ac)}$$ and a simple root, $$x_3=\frac{4abc-9a^2d-b^3}{a(b^2-3ac)}$$

Have look here.

Claude Leibovici
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