Please note that there may be other methods apart from this large formulas to solve cubics and quartics. But I wanted to show here that the formulas do exist.
The general form of the 3rd degree equation (or Cubic) is: ax3 + bx2 + cx + d = 0
Cubics have 3 roots.
The 3 roots can be represented this way:
First root (of three):
Second root (of three):
Third root (of three):
The second and third formula are equal except for a "+ or -" sign at
the beginning, and another "+ or -" sign in the middle. Note that
the second and third formula contain the imaginary unit "i".
Now, the same three formulas in ASCII. The differences
between the second and third formula are highlighted here in yellow:
| x = -b/(3*a) -
(2^(1/3)*(-b^2 + 3*a*c))/(3*a*(-2*b^3 + 9*a*b*c - 27*a^2*d +
Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)) +
(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 +
9*a*b*c - 27*a^2*d)^2])^(1/3)/(3*2^(1/3)*a)
x = -b/(3*a) + ((1 + i*Sqrt[3])*(-b^2 + 3*a*c))/(3*2^(2/3)*a*(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)) - (1 - i*Sqrt[3])*(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)/(6*2^(1/3)*a) x = -b/(3*a) + ((1 - i*Sqrt[3])*(-b^2 + 3*a*c))/(3*2^(2/3)*a*(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)) - (1 + i*Sqrt[3])*(-2*b^3 + 9*a*b*c - 27*a^2*d + Sqrt[4*(-b^2 + 3*a*c)^3 + (-2*b^3 + 9*a*b*c - 27*a^2*d)^2])^(1/3)/(6*2^(1/3)*a) |
The formulas on top were obtained with
Wolfram's program Mathematica. There seemed not to be an
option, so I copied them by hand into ASCII.
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