- General cubic equation.

- The quadratic term disappears

- Vieta's substitution

- Summing up

- Iteration method

A xThe coefficients A, B, C, D are real or complex numbers with A not 0. Dividing through by A, the equation comes to the form^{3}+ B x^{2}+ C x + D = 0

x^{3}+ b x^{2}+ c x + d = 0

x = y + rThe cubic equation becomes:

(y + r)Now we choose y such that the quadratic term disappears^{3}+ b (y + r)^{2}+ c (y + r) + d = 0 <=> y^{3}+ (3 r + b) y^{2}+ (3 r^{2}+ 2 r b + c) y + r^{3}+ r^{2}b + r c + d = 0

choose r = -b/3 So, with the substitution b x = y - - 3 the equation x^{3}+ b x^{2}+ c x + d = 0 comes in the form y^{3}+ e y + f = 0

1 y = z + s - zThe constant s is an undefined constant for the moment.

The equation

yNow we choose s = -e/3.^{3}+ e y + f = 0 becomes s (z + -)^{3}+ e (z + (s/z)) + f = 0 z expanding and multiplying through by z^{3}, we have z^{6}+ (3 s + e) z^{4}+ f z^{3}+ s (3 s + e) z^{2}+ s^{3}= 0

The equation becomes

zThis is an easy to solve quadratic equation.^{6}+ f z^{3}- e^{3}/27 = 0 With z^{3}= u u^{2}+ f u -e^{3}/27 = 0

A xThe coefficients A, B, C, D are real or complex numbers with A not 0. Dividing through by A, the equation comes to the form^{3}+ B x^{2}+ C x + D = 0

xThis is an easy to solve quadratic equation.^{3}+ b x^{2}+ c x + d = 0 With the substitution b x = y - - 3 comes y^{3}+ e y + f = 0 To reduce the last equation we use the Vieta substitution e y = z - --- 3 z The equation becomes z^{6}+ f z^{3}- e^{3}/27 = 0 With z^{3}= u u^{2}+ f u - e^{3}/27 = 0

45 xEach solution yields three values of z. To calculate these values, we bring the u-values in polar form.^{3}+ 24 x^{2}- 7 x - 2 = 0 <=> 3 8 2 7 2 x + -- x - -- x - -- = 0 15 45 45 With the substitution 8 x = y - --- 45 comes 3 169 506 y - --- y - ----- = 0 675 91125 Now we leave the fractional notation <=> y^{3}- 0.25037037037 y - 5.55281207133e-3 = 0 To reduce the last equation we use the Vieta substitution 0.0834567901235 y = z + --------------- z Then we have z^{6}- 5.55281207133e-3 z^{3}+ 5.81279532442e-4 = 0 With z^{3}= u u^{2}- 5.55281207133e-3 u + 5.81279532442e-4 = 0 The solutions for u are u_{1}= 2.77640603567e-3 + 0.0239493444997 i and u_{2}= 2.77640603567e-3 - 0.0239493444997 i

u_{1}= 0.024109739369 (cos(1.45538324457) + i sin(1.45538324457)) u_{2}= 0.024109739369 (cos(1.45538324457) - i sin(1.45538324457)) The six values of z are in polar form z_{1}= 0.288888888889 (cos(0.48512774819) + i sin (0.48512774819) ) z_{2}= 0.288888888889 (cos(2.57952285058) + i sin (2.57952285058) ) z_{3}= 0.288888888889 (cos(-1.6092673542) + i sin (-1.6092673542) ) z_{4}= 0.288888888889 (cos(0.48512774819) - i sin (0.48512774819) ) z_{5}= 0.288888888889 (cos(2.57952285058) - i sin (2.57952285058) ) z_{6}= 0.288888888889 (cos(-1.6092673542) - i sin (-1.6092673542) ) With 0.0834567901235 y = z + --------------- z we find three real y-values y_{1}= 0.511111111112 y_{2}= - 0.488888888888 y_{3}= - 0.022222222221 Finally, with the substitution 8 x = y - --- 45 we find the three roots of the given equation x_{1}= 0.333333333334 x_{2}= -0.666666666666 x_{3}= -0.199999999999 The exact roots are x_{1}= 1/3 x_{2}= -2/3 x_{3}= 1/5

Use this link
for the** theory, optimalisation procedure and examples** of this iteration method.

Relying on the optimized procedure of the iteration method, we solve now the equation

x^{3} + 2 x^{2} + 3 x - 4 = 0. We follow the procedure literally.

- By plotting x
^{3}+ 2 x^{2}+ 3 x - 4 we see that 0.77 is an approximation of the only real root. -
We write x = x + r( x
^{3}+ 2 x^{2}+ 3 x - 4 ) -
We choose the r-value such that 1 + r( 3.(0.77)
^{2}+ 4.(0.77) +3) = 0. r_{o}= -0.13 is a good approximation. -
We apply iteration on x = x - 0.13 ( x
^{3}+ 2 x^{2}+ 3 x - 4 ) starting with x= 0.77

We find a very good result using only 5 steps.0.77619671 0.776041122953 0.776045557563 0.776045431542 0.776045435125

- By plotting x
^{3}- 2.7 x^{2}+ 4.5 x - 6 we see that 2 is a rough approximation of the only real root. -
We write x = x + r( x
^{3}- 2.7 x^{2}+ 4.5 x - 6 )

Relying on the optimized procedure of the iteration method, we choose the r-value such that

1 + r(3*4 - 5.4*2 + 4.5) = 0. r_{o}= -0.175 is a good approximation. -
We apply iteration on x = x - 0.175 ( x
^{3}- 2.7 x^{2}+ 4.5 x - 6 )

We start with x = 2. We find a very good result using only 5 steps.1.96500000 1.96421257 1.96417892 1.96417747 1.96417741

- Cubic Equations (pdf)

solving cubic equations with one simple root; Using graphs to solve cubic equations

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