2010年7月28日 · The Quartic Formula is just the end result of this methodology, written in terms of the original coefficients. Because of that, the method is far easier to remember than the formula, which is why I find it annoying when people cite just the formula and tell you, "don't bother, use a computer instead."

While they do start getting awkward quickly, the next few ordinals are fairly well-defined, largely because of their occasional usage in solving cubic and quartic equations and in defining algebraic curves and surfaces: the Sextic, the Septic, and the Octic. Beyond that, they just don't show up often enough to be worth explicitly naming.

What about a set of expressions in the quartic's coefficients that discriminate between all cases? There are 9 cases: 4 distinct real roots. 3 distinct real roots with one of them being a double root. 2 distinct double roots both real. Triple root and and a distinct fourth root. Quadruple root. 2 distinct real roots and two complex roots.

2015年9月15日 · You have to test them to see if they work. If you test all the possibilities, you'll find that there are four which work ($-1, 2, -2$, and $3$). Since any quartic polynomial has at most four real roots, you can be sure that you've found all of them. Unfortunately, many quartic polynomials have roots which are not rational.

2020年3月14日 · Well, instead of looking to the quartic formula directly, why not follow the math of its derivation. Its infinitely easier to "complete the square", is it were, than to plug and play with the quadratic formula. Okay, thats not true for quadratics, but it …

2019年12月9日 · Finding the distance of closest approach of two ellipses involves solving a quartic equation. The eigenvalues of a 4×4 matrix are the roots of a quartic polynomial which is the characteristic polynomial of the matrix. The characteristic equation of a fourth-order linear difference equation or differential equation is a quartic equation.

$\begingroup$ In this particular problem, the suppressed quartic equation reduces to a bi-quadratic equation having no "u" term. It has only u^4 , u^2 and constant term. It has only u^4 , u^2 and constant term.

2018年8月24日 · If you didn't have normalization, this would be exactly solvable in the n=3 case by Sums of Squares methods.. In general this is hard, but there exist approximation algorithms with $1 - Cn^{-2}$ approximation ratios for quartic optimization with quadratic constraints (exactly what you have here).

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2013年10月13日 · If a quartic equation has a rational solution, then it must be factor of the last coefficient. So you can check all its factor. If you find one then you can factorize it and reduce it to third degree polynomial, which are much easier to solve. Also you can repeat the process again and hopefully bring it to a second degree polynomial.