Solving Cubic Equations

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If you thought the Quadratic Formula was complicated, the method for solving Cubic Equations is even more complex. We will use the example from the Cubic Equation Calculator:

2x3   - 4x2   - 22x + 24 = 0

Cubic equations have to be solved in several steps. First we define a variable 'f':

f = (3c/a) - (b²/a²)
3
"Plugging in" the numbers from the above equation, we get:
f = ((3 * -22/2) - (16/4)) / 3       =     - 12.333333...

Next we define 'g':

g = (2b³/a³) - (9bc/a²) + (27d/a)
27
From this point on, you are expected to "plug in" the numbers:
g = 4.07407407407407....

Then we define 'h':

h = (g²/4) + (f³/27)

h = -65.333333...

If h > 0, there is only 1 real root and is solved by another method.
(SCROLL down for this method)

For the special case where f=0, g=0 and h = 0, all 3 roots are real and equal.
(SCROLL to the bottom for this method)

When h <= 0, as is the case here, all 3 roots are real and we proceed as follows:

ALL 3 Roots Are Real

i = ((g²/4) - h) ½

i = 8.33563754151978...


j = (i) 1/3
j = 2.0275875100994063...

NOTE: The following trigonometric calculations are in radians
k = arc cosine (- (g / 2i))
k = 1.817673356517739...

L = j * -1
L = -2.0275875100994...

M = cosine (K/3)
M = 0.8219949365268...

N = (Square Root of 3) * sine (K/3)
N = 0.9863939238321...

P = (b/3a) * -1
P = 0.6666666666666...

x1 = 2j * cosine(k/3)   -(b/3a)
x1 = 4

x2 = L * (M + N) + P
x2 = -3

x3 = L * (M - N) + P
x3 = 1

When Only 1 Root Is Real
3x3   - 10x2   + 14x + 27 = 0
f = (3c/a) - (b²/a²)
3
f =   .962962962962962...

g = (2b³/a³) - (9bc/a²) + (27d/a)
27
g = 11.441700960219478...

h = (g²/4) + (f ³/27)
h = 32.761202560585275...

R = -(g/2) + (h)½
R = .002889779596782...

S = (R)1/3
S = .142436591824886...

T = -(g/2) - (h)½
T = -11.4445907398163...

U = (T)1/3
U = -2.25354770293599...

X1 = (S + U) - (b/3a)
X1 = -1

X2 = -(S + U)/2 - (b/3a) + i*(S-U)*(3)½/2
X2 = 2.16666666666... + i*2.07498326633146

X3 = -(S + U)/2 - (b/3a) - i*(S-U)*(3)½/2
X3 = 2.16666666666... - i*2.07498326633146

When All 3 Roots Are Real and Equal

x3   + 6x2   + 12x + 8 = 0
f = (3c/a) - (b²/a²)
3
f =   ((3*12/1)-(36/1)) / 3
f =   0

g = (2b³/a³) - (9bc/a²) + (27d/a)
27
g = ((2*216/1) - (9*6*12/1) + (27*8/1)) / 27
g = (432 - 648 + 216) / 27
g = 0

h = (g²/4) + (f³/27)
h=0

x1 = x2 = x3= (d/a)1/3 * -1
x1 = x2 = x3= (8/1)1/3 * -1
x1 = x2 = x3= -2



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