作业帮1元3次方程根的判别式
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/29 17:57:25
1元3次方程根的判别式
0=ax^3+bx^2+cx+d
的解的个数用a,b,c,d的式子表示
并作说明~
0=ax^3+bx^2+cx+d
的解的个数用a,b,c,d的式子表示
并作说明~
x1=1/6/a*(36*c*b*a-108*d*a^2-8*b^3+12*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(1/3)-2/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(1/3)-1/3*b/a
x2=-1/12/a*(36*c*b*a-108*d*a^2-8*b^3+12*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(1/3)+1/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(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*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(1/3)+2/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(1/3))
x3=-1/12/a*(36*c*b*a-108*d*a^2-8*b^3+12*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(1/3)+1/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(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*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(1/3)+2/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(1/3))
确实很复杂,但其中包括复根,因而只要复根系数为零,则为实根,由此可得实根条件和解个数的判断法
x2=-1/12/a*(36*c*b*a-108*d*a^2-8*b^3+12*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(1/3)+1/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(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*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(1/3)+2/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(1/3))
x3=-1/12/a*(36*c*b*a-108*d*a^2-8*b^3+12*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(1/3)+1/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(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*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(1/3)+2/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*a*(12*c^3*a-3*c^2*b^2-54*c*b*a*d+81*d^2*a^2+12*d*b^3)^(1/2))^(1/3))
确实很复杂,但其中包括复根,因而只要复根系数为零,则为实根,由此可得实根条件和解个数的判断法