ax^3+bX^2+cx+d=0的计算通式是什么
来源:百度知道 编辑:UC知道 时间:2024/05/02 19:30:19
x1= 1/6/a*(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)-2/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)-1/3*b/a
x2= -1/12/a*(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)+1/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)-1/3*b/a+1/2*i*3^(1/2)*(1/6/a*(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)+2/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3))
x3= -1/12/a*(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)+1/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)-1/3*b/a-1/2*i