问题三十七：C++怎么解一元四次方程？（3）——怎么解一元四次方程

37.3 怎么解一元四次方程？

用“费拉里”方法求解：将四次方程化为两个二次方程，然后求解二次方程。

----------------------------------------------main.cpp ------------------------------------------

main.cpp

float* roots_quartic_equation(float a, float b, float c, float d, float e) {
    //the first element is the number of the real roots, and other elements are the real roots.
    //Ferrari's solution.
    float *roots = new float[5];
    if (a == 0) {
        roots = roots_cubic_equation(b, c, d, e);
    }
    else {
        float b1 = b/a;
        float c1 = c/a;
        float d1 = d/a;
        float e1 = e/a;
        if ((b1 == 0) && (c1 == 0) && (d1 == 0)) {
        //in this special case, such as a=1, b=c=d=0, e=-1, the roots should be +1 and -1
            if (e1 > 0) {
                roots[0] = 0.0;
            }
            else {
                roots[1] = sqrt(sqrt(-e1));
                roots[2] = -sqrt(sqrt(-e1));
                roots[0] = 2.0;
            }
        }
        else {
            float *roots_y = new float[4];
            roots_y = roots_cubic_equation(-1.0, c1, 4*e1-b1*d1, d1*d1+e1*b1*b1-4*e1*c1);
            float y = roots_y[1];
            float B1, B2, C1, C2;
            if (b1*b1-4*c1+4*y == 0) {
                B1 = b/2;
                B2 = b/2;
                C1 = y/2;
                C2 = y/2;
            }
            else {
                B1 = b/2 - sqrt(b1*b1-4*c1+4*y)/2;
                B2 = b/2 + sqrt(b1*b1-4*c1+4*y)/2;
                C1 = y/2 - (b1*y -2*d1)/(2*sqrt(b1*b1-4*c1+4*y));
                C2 = y/2 + (b1*y -2*d1)/(2*sqrt(b1*b1-4*c1+4*y));
            }
            float *roots_x1 = new float[3];
            float *roots_x2 = new float[3];
            roots_x1 = roots_quadratic_equation(1.0, B1, C1);
            roots_x2 = roots_quadratic_equation(1.0, B2, C2);
            if (roots_x1[0] != 0) {
                for (int i=1; i<roots_x1[0]+1; i++) {
                    roots[i] = roots_x1[i];
                }
            }
            if (roots_x2[0] != 0) {
                int roots_x1_number = int(roots_x1[0]);
                for (int j=1; j<roots_x2[0]+1; j++) {
                    roots[roots_x1_number+j] =roots_x2[j];
                }
            }
            roots[0] = roots_x1[0] + roots_x2[0];
        }
    }
    return roots;
}

    int main(){
        float *r;
//        r = roots_quartic_equation(1.0, -4.0, 6.0, -4.0, 1.0);//1,1,1,1
//        r = roots_quartic_equation(1.0, -10.0, 35.0, -50.0, 24.0);//1,2,3,4
//        r = roots_quartic_equation(1.0, 0.0, -5.0, 0.0, 4.0);//1,-1,2,-2
//        r = roots_quartic_equation(1.0, 0.0, 0.0, 0.0, 0.0);//0,0,0,0
        r = roots_quartic_equation(1.0, 0.0, 0.0, 0.0, -16.0);//-2,2

        for (int i=0; i<(r[0]+1); i++) {
            std::cout << "r[" << i << "]=" << r[i] << endl;
        }
}

另一种解一元四次方程的方法：“笛卡尔”方法（即“待定系数法”）参考“问题四十”中的“第六步”