I need to solve a cubic equation (ax^3 + bx^2 + c*x + d = 0) analytically and in real numbers, preferably in pure javascript (no libs). As there could be 1 to 3 roots, I think an array of numbers is a reasonable result type.
P.S. Provided my own solution below, hope it'll be useful.
Here you go. Includes handling degenerate cases. Main algorithm is mostly from wikipedia article.
function cuberoot(x) {
var y = Math.pow(Math.abs(x), 1/3);
return x < 0 ? -y : y;
}
function solveCubic(a, b, c, d) {
if (Math.abs(a) < 1e-8) { // Quadratic case, ax^2+bx+c=0
a = b; b = c; c = d;
if (Math.abs(a) < 1e-8) { // Linear case, ax+b=0
a = b; b = c;
if (Math.abs(a) < 1e-8) // Degenerate case
return [];
return [-b/a];
}
var D = b*b - 4*a*c;
if (Math.abs(D) < 1e-8)
return [-b/(2*a)];
else if (D > 0)
return [(-b+Math.sqrt(D))/(2*a), (-b-Math.sqrt(D))/(2*a)];
return [];
}
// Convert to depressed cubic t^3+pt+q = 0 (subst x = t - b/3a)
var p = (3*a*c - b*b)/(3*a*a);
var q = (2*b*b*b - 9*a*b*c + 27*a*a*d)/(27*a*a*a);
var roots;
if (Math.abs(p) < 1e-8) { // p = 0 -> t^3 = -q -> t = -q^1/3
roots = [cuberoot(-q)];
} else if (Math.abs(q) < 1e-8) { // q = 0 -> t^3 + pt = 0 -> t(t^2+p)=0
roots = [0].concat(p < 0 ? [Math.sqrt(-p), -Math.sqrt(-p)] : []);
} else {
var D = q*q/4 + p*p*p/27;
if (Math.abs(D) < 1e-8) { // D = 0 -> two roots
roots = [-1.5*q/p, 3*q/p];
} else if (D > 0) { // Only one real root
var u = cuberoot(-q/2 - Math.sqrt(D));
roots = [u - p/(3*u)];
} else { // D < 0, three roots, but needs to use complex numbers/trigonometric solution
var u = 2*Math.sqrt(-p/3);
var t = Math.acos(3*q/p/u)/3; // D < 0 implies p < 0 and acos argument in [-1..1]
var k = 2*Math.PI/3;
roots = [u*Math.cos(t), u*Math.cos(t-k), u*Math.cos(t-2*k)];
}
}
// Convert back from depressed cubic
for (var i = 0; i < roots.length; i++)
roots[i] -= b/(3*a);
return roots;
}
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