Solve Quartic with Casio FX-115MS

[Albert Chan]

An old algorithm for solving Quartics: Brown's method
(source: W.S. Brown, "Solution of Biquadratic Equations", Aircraft Engineering, 16,14 (1944)

Quartics: x^4 + A x^3 + B x^2 + C x + D = 0

Step 1: get Y, the largest real root of resolvent cubics

y^3 - B y^2 + (AC-4D) y + D(4B-A²)-C² = 0

it will work for any root of cubics, but later quadratics might involve complex numbers.

Step 2: solve 2 quadratics

c^2 - A c + (B-Y) = 0
d^2 - Y d + D = 0

Step 3: make sure roots, c1 d2 + c2 d1 = C. If not, swap d1, d2.
This may required calculation of c1 d1 + c2 d2 as well, due to numerical imprecision.

I have a slight improvement in this step, see example below

Step 4: solve 2 more quadratics, the 4 roots is the roots of Quartics:

x^2 + c1 x + d1 = 0
x^2 + c2 x + d2 = 0

Example: solve x^4 + 2 x^3 + 3 x^2 + 4 x + 5 = 0

A = 2
B = 3
C = 4
D = 5
E = AC-4D = -12
F = D(4B-A²)-C² = 24

y^3 - B y^2 + E y + F = 0 → Y = 4.482612919

c^2 - A c + (B-Y) = 0        → [c1,c2] = [-0.575630959, 2.575630959]
d^2 - Y d + D = 0             → [d1,d2] = [2.088157364, 2.394455555]

if B and Y nearly equal, we may use another form for B-Y

Y^2 (Y-B) + E Y + F = 0 → B-Y = (E Y + F) / Y²

Calculate AY/2
If AY/2 > C, make sure roots of c,d sorted the same way
If AY/2 < C, make sure roots of c,d sorted opposite way

For this case, AY/2 = Y > C, thus c,d roots sorted the same way.

x^2 + c1 x + d1 = 0 → x = 0.2878154795 ± 1.416093081 i
x^2 + c2 x + d2 = 0 → x = -1.287815480 ± 0.8578967571 i