Difference between revisions of "User:Temperal/The Problem Solver's Resource4"
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The three roots <math>r, s, t</math> of a cubic polynomial equation <math>x^3 + ax^2 + bx + c = 0</math> are given implicitly by | The three roots <math>r, s, t</math> of a cubic polynomial equation <math>x^3 + ax^2 + bx + c = 0</math> are given implicitly by | ||
− | <math>r | + | <math>r=- \frac {a}{3} + \left(\frac { - 2a^3 + 9ab - 27c + \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}+\left(\frac { - 2a^3 + 9ab - 27c - \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}</math> |
− | + | <math>s=- \frac {a}{3} - \frac {1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c + \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}</math>+ \frac { - 1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c - \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}<math> | |
− | <math>s | + | </math>t=- \frac {a}{3} + \frac { - 1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c + \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3} - \frac {1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c - \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}$ |
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− | <math>t | ||
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Quartic formulas are listed [http://www.josechu.com/ecuaciones_polinomicas/cuartica_solucion.htm here]. | Quartic formulas are listed [http://www.josechu.com/ecuaciones_polinomicas/cuartica_solucion.htm here]. |
Revision as of 16:25, 8 December 2007
AlgebraThis is a collection of algebra laws and definitions. Obviously, there is WAY too much to cover here, but we'll try to give a good overview. Elementary AlgebraDefinitions
Factor TheoremIff a polynomial has roots , then , and are all factors of . Quadratic FormulaFor a quadratic of form , where are constants, the equation has roots Fundamental Theorems of Algebra
Third-degree and Quartic FormulasIf third-degree polynomial has roots , then: The three roots of a cubic polynomial equation are given implicitly by + \frac { - 1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c - \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}$$ (Error compiling LaTeX. Unknown error_msg)t=- \frac {a}{3} + \frac { - 1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c + \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3} - \frac {1 + i\sqrt {3}}{2} \left(\frac { - 2a^3 + 9ab - 27c - \sqrt {(2a^3 - 9ab + 27c)^2 + 4( - a^2 + 3b)^3}}{54}\right)^{1/3}$ Quartic formulas are listed here. The general quintic equation (or an equation of even higher degree) does not have a formula. DeterminantsThe determinant of a by (said to have order ) matrix is . General Formula for the DeterminantLet be a square matrix of order . Write , where is the entry on the row and the column , for and . For any and , set (called the cofactors) to be the determinant of the square matrix of order obtained from A by removing the row number i and the column number j multiplied by (-1)^{i+j}. Thus:
Cramer's LawConsider a set of three linear equations (i.e. polynomials of degree one) Let , , , , , and . This can be generalized to any number of linear equations. Abstract AlgebraIncomplete. Diophantine EquationsIncomplete. |