分析求解三次方程的功能

干得好。包括处理退化案件。主要算法主要来自维基百科文章。

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;}