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This means that only one cube root needs to be computed, and leads to the second formula given in § Cardano's formula. The other roots of the equation can be obtained by changing of cube root, or, equivalently, by multiplying the cube root by each of the two primitive cube roots of unity , which are − 1 ± − 3 2 . {\displaystyle {\frac {-1 ...
In 1535, Niccolò Fontana Tartaglia became famous for having solved cubics of the form x 3 + ax = b (with a,b > 0). However, he chose to keep his method secret. In 1539, Cardano, then a lecturer in mathematics at the Piatti Foundation in Milan, published his first mathematical book, Pratica Arithmeticæ et mensurandi singularis (The Practice of Arithmetic and Simple Mensuration).
The solutions of this equation are the x-values of the critical points and are given, using the quadratic formula, by =. The sign of the expression Δ 0 = b 2 – 3ac inside the square root determines the number of critical points. If it is positive, then there are two critical points, one is a local maximum, and the other is a local minimum.
Cardano's formula for solution in radicals of a cubic equation was discovered at this time. It applies in the casus irreducibilis, but, in this case, requires the computation of the square root of a negative number, which involves knowledge of complex numbers, unknown at the time.
Proposed in 1873, the van der Waals equation of state was one of the first to perform markedly better than the ideal gas law. In this equation, usually is called the attraction parameter and the repulsion parameter (or the effective molecular volume). While the equation is definitely superior to the ideal gas law and does predict the formation ...
and thus is a biquadratic equation, ... (e.g. conversion to a reduced cubic and application of Cardano's formula). Any of the three possible roots will do ...
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This equation may be solved using the methods of Cardano or Lagrange, but an affine change to A will simplify the expression considerably, and lead directly to a trigonometric solution. If A = pB + qI, then A and B have the same eigenvectors, and β is an eigenvalue of B if and only if α = pβ + q is an eigenvalue of A.