Solving A Special Quartic | Generalization

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[goldfing]

Since in the quartic the cubic power x^3 is anyway missing, I prefer Ferrari's method in this case.
x^4 + 12 x + 3 = 0
Subtract 12x and 3:
x^4 = -12x - 3

The binomic formula says
(x^2 + z)^2 = x^4 + 2zx^2 + z^2
So we can add 2zx^2 + z^2 in order to complete the square on the left side:
x^4 + 2zx^2 + z^2 = -12x - 3 + 2zx^2 + z^2
The left side is a complete square now, the right side can be rearranged:
(x^2 + z)^2 = (2z)x^2 - 12x + (z^2 - 3)

The right side is a quadratic polynomial of format ax^2 + bx + c with
a = 2z
b = -12
c = z^2 - 3

In order to make the right side a complete square, too, we must assure that its discriminant is zero:
b^2 - 4ac = 0
b^2 = 4ac

Plug in the coefficients from above:
(-12)^2 = 4*(2z)*(z^2 - 3)
144 = 8z*(z^2 - 3)
Divide by 8:
18 = z(z^2 - 3)
18 = z^3 - 3z
0 = z^3 - 3z - 18

This so-called "cubic resolvent" is the requirement for z. In order to find any integer solution, it is better to write
z^3 = 3z + 18
z^3 = 3*(z + 6)

The candidates for integers solutions must be divisors of the constant 18, namely plus or minus 1, 2, 3, 6, 9, 18.
It is easy to see that z = 3 is a solution, because 3^3 = 27 and 3*(3 + 6) = 3*9 = 27, too.
There is no need to calculate the other two (possibly complex) solutions for z.
It it is sufficient to plug z = 3 into the above quartic equation:

x^4 + 2zx^2 + z^2 = -12x - 3 + 2zx^2 + z^2
to get
x^4 + 6x^2 + 9 = 6x^2 - 12x + 6
x^4 + 6x^2 + 9 = 6*(x^2 - 2x + 1)

Now both sides can be written as perfect squares:
(by the way this is basically the same equation as in 2nd method at 5:40, which I saw later):
(x^2 + 3)^2 = 6*(x - 1)^2
(x^2 + 3)^2 = (sqrt(6)*(x - 1))^2
We take the square root on both sides, regarding that plus and minus is possible:
x^2 + 3 = +- sqrt(6)*(x - 1)
We now have two quadratic equations for x:

First case:
x^2 + 3 = -sqrt(6)*(x - 1)
x^2 + 3 = -sqrt(6)*x + sqrt(6)
x^2 + sqrt(6)*x + (3 - sqrt(6)) = 0
The discriminant is positive, namely
6 - 4*(3 - sqrt(6))
= 6 - 12 + 4*sqrt(6)
= -6 + 4*sqrt(6)
So we get two real solutions
x1, 2 = (-sqrt(6) +- sqrt(4*sqrt(6) - 6) / 2
Approximation values are:
x1 = -2.199162514
x2 = -0.2503272285

Second case:
x^2 + 3 = +sqrt(6)*(x - 1)
x^2 + 3 = +sqrt(6)*x - sqrt(6)
x^2 - sqrt(6)*x + (3 + sqrt(6)) = 0
The discriminant is negative, namely
6 - 4*(3 + sqrt(6))
= 6 - 12 - 4*sqrt(6)
= -6 - 4*sqrt(6)
So we get two complex solutions
x3, 4 = (sqrt(6) +- sqrt(4*sqrt(6) + 6) * i) / 2
Approximation values are:
x3 = 1.224744871 + 1, 987332318 i
x4 = 1.224744871 - 1, 987332318 i

[philipfoy]

Crazy thing is I factored this straight from scratch.

[gdtargetvn]

x⁴ + 12x = -3
Just complete the square on both sides
x⁴ + 6x² + 9 = 6x² - 12x + 6
(x² + 3)² = 6(x - 1)²
=> x² + 3 = ±(x-1)√6

[DonRedmond-jkhj]

Just for the fun of it there is a quintic formula, It involves ellipyic functions.

[HoSza]

A more general form x⁴+px+q=0 can still be solved with the same technique (difference of completed the squares) but it involves solving a cubic equation.

x⁴+px+q=0
x⁴+2ax²+a²-(2ax²-px+a²-q)=0

0=a³-aq-p²/8
(x²+a)²-2a(x-p/4a)²=0

In this case: p=12, q=3
a³-3a-18=0 => a=3
(x²+3)²-6(x²-2x+1)=0
...