Timeline for Is there an alternative formula for solving cubic equations?
Current License: CC BY-SA 3.0
7 events
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| Dec 29, 2015 at 15:56 | comment | added | YCor | To clarify: if we consider the smallest subfield of the real numbers containing the rationals and stable under taking positve $n$th roots of positive numbers, does it contain all real roots of degree 3 rational polynomials? (I expect a negative answer) | |
| Dec 29, 2015 at 7:34 | answer | added | Syafiq | timeline score: 1 | |
| Sep 9, 2012 at 8:58 | answer | added | Gene Ward Smith | timeline score: 16 | |
| Oct 7, 2011 at 17:48 | comment | added | Emil Jeřábek |
It is a simple exercise to reduce any cubic equation to $x^3+ax+b=0$ with $a\in\{-1,0,1\}$ by a linear substitution (first shift $x$ to get rid of the quadratic term, then scale $x$ to normalize the linear term).
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| Oct 7, 2011 at 14:45 | comment | added | user2035 | ... and $\cos\frac{\arccos B}3$ satisfies $4x^3-3x-B=0$. | |
| Oct 7, 2011 at 14:26 | comment | added | Gerald Edgar | The other classic solution for cubics uses trig functions. Then, at least, you can solve a cubic with real coefficients and three real roots without going to the complex numbers. planetmath.org/encyclopedia/ATrigonometricCubicFormula.html | |
| Oct 7, 2011 at 14:11 | history | asked | Ruslan_Sharipov | CC BY-SA 3.0 |