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Given the cubic equation: For the general cubic equation (1) with real coefficients, the general formula for the roots, in terms of the coefficients, is as follows if $(2 b^3-9 a b c+27 a^2 d)^2-4 (b^2-3 a c)^3=-27 a^2 \Delta>0$, i.e. if there are two non real roots:
Wikipedia's article on quartic functions has a lengthy process by which to get the solutions, but does not give an explicit formula. Beware that in the cubic and quartic formulas, depending on how the formula is expressed, the correctness of the answers likely depends on a particular choice of definition of principal roots for nonreal complex ...
Using Vieta's formula to find the sum of the roots for a given cubic equation. 4 Using Vieta's formula to evaluate $\frac{x_1}{x_2} + \frac{x_2}{x_3} + \frac{x_3}{x_1}$
As with quadratics, first we divide through to make the polynomial monic and then shift the unknown by a constant so the second highest power has zero coefficient, giving x3 + px + q = 0. We now use Cardano's method. There exist complex numbers u, v with u + v = x, uv = − p 3 (since if only you knew x we'd just have to solve t2 − xt − p 3 ...
This answer has been awarded bounties worth 50 reputation by Teoc. A general form for the cubic equation is, ax3 + bx2 + cx + d = 0 (1) (1) a x 3 + b x 2 + c x + d = 0. To find the roots of this equation we first try to get rid of the quadratic term x2 x 2. The substitution x = y − b 3a x = y − b 3 a helps in achieving our goal.
a) Show that the “depressed” cubic equation $𝑎𝑥^3 + 𝑏𝑥 + 𝑐 = 0$ can be solved geometrically for its real roots on a rectangular Cartesian coordinate system on which the cubic curve $𝑦 = 𝑥^3$ has been carefully drawn, by merely drawing the line $𝑎𝑦 + 𝑏𝑥 + 𝑐 = 0$. Explain.
$\begingroup$ Finding eigenvalues of a $3\times 3$ matrix in general requires solving a cubic equation. This kind of problem is very common in teaching, but mysteriously one seems to only encounter examples where the eigenvalues can (also) be found without solving a cubic equation, or at least without using the general formula for doing so ...
Now let X = x + 1 X = x + 1, and divising by 3, your equation is now something like X3 + pX + t = 0 X 3 + p X + t = 0, which can be solved by Cardano method. To depress a cubic means to write it in the form y3 + py + q = 0 y 3 + p y + q = 0 by performing a convenient substitution.
3. You can solve cubics using a similar idea to 'completing the square'. The first step is to note that (x + y)3 = x3 + 3x2y + 3xy2 + y3 and use this to remove the quadratic factor. For your example, lets just factorize x3 + 4x2 + 5x − 3. Once we do this we can multiply by 2 to get the original equation.
Our vertex point is −b 3a − b 3 a if there are 3 3 real roots, and −b a − b a if there is 1 1 real root, as the sum of real roots of a cubic equation is equal to −b a − b a. The discriminant of a quadratic equation is obtained by replacing x with the vertex point, −b 2a − b 2 a. In this case I replaced x with −b 3a − b 3 a ...