Solve Quartic with Casio FX-115MS

[Albert Chan]

y^4 + A y^2 + B y + C = 0
y^4 + 2D y^2 + D^2 = (2D - A) y^2 - B y + (D^2 - C)

For some value of D, RHS can be made a perfect square:

(y^2 + D)^2 = (2D - A) * (y - (B/2)/(2D-A))^2

Let E = √(2D-A), F = -B/(2E), Quartic => (y^2 + E y + D+F)(y^2 - E y + D-F) = 0

Example: f(x) = x^4 + 2 x^3 + 3 x^2 + 4 x + 5 = 0

Depress the Quartic (remove cubic term) by substitute x = y + k, where k = -b/(4a)
We get depressed quartic: y^4 + A y^2 + B y + C

k = -b/(4a) = -2/4 = -0.5
C = f(k) = 3.5625
B = f(k−1) = 4.0625
A = f(k+1) = 8.0625
B = (A - B) / 2 = 2
A = A - 1 - B - C = 1.5

Find D such that RHS is perfect square, i.e. discriminant = B^2 - 4(2D - A)(D^2 - C) = 0
We do not want complex values for E, so 2D ≥ A, D^2 ≥ C:

D ≥ max(A/2, √C) ≈ max(0.75, 1.904) = 1.904

A trick to force Casio FX-115MS to solve for D: 0ABC + B^2 - 4(2D - A)(D^2 - C)

D = 1.99130646 (Casio SOLVE, with guessed D = 2)
E = √(2D-A) = 1.575630959
F = -B/(2E) = -0.6346663818

Quadratic Solver, enter a=1, b=+E, c=D+F, roots = −0.78781548 ± 0.857896758 i
Quadratic Solver, enter a=1, b=−E, c=D-F , roots = +0.78781548 ± 1.416093080 i

x roots = y roots + k = -1.28781548 ± 0.857896758 i, 0.28781548 ± 1.416093080 i

For calculator without Quad Solver, we can use this:

y^2 + E y + (D+F) = 0 ==> y = −E/2 ± √(−F - D/2 - A/4)
y^2 − E y + (D−F) = 0 ==> y = +E/2 ± √(+F - D/2 - A/4)

Since C=(D+F)(D-F), we might derive 1 from the other, hopefully better accuracy too.