用c++解一元三次方程的根.求出x3-3x-1=0在(-8,8)的所有实根x1=[-q/2+(q^2/4+p^3/27)^(1/2)]^(1/3)++[-q/2-(q^2/4+p^3/27)^(1/2)]^(1/3)x2=w[-q/2+(q^2/4+p^3/27)^(1/2)]^(1/3)++w^2[-q/2-(q^2/4+p^3/27)^(1/2)]^(1/3)x2=w^2[-q/2+(q^2/4+p^3/27)^
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![用c++解一元三次方程的根.求出x3-3x-1=0在(-8,8)的所有实根x1=[-q/2+(q^2/4+p^3/27)^(1/2)]^(1/3)++[-q/2-(q^2/4+p^3/27)^(1/2)]^(1/3)x2=w[-q/2+(q^2/4+p^3/27)^(1/2)]^(1/3)++w^2[-q/2-(q^2/4+p^3/27)^(1/2)]^(1/3)x2=w^2[-q/2+(q^2/4+p^3/27)^](/uploads/image/z/13887128-56-8.jpg?t=%E7%94%A8c%2B%2B%E8%A7%A3%E4%B8%80%E5%85%83%E4%B8%89%E6%AC%A1%E6%96%B9%E7%A8%8B%E7%9A%84%E6%A0%B9.%E6%B1%82%E5%87%BAx3-3x-1%3D0%E5%9C%A8%EF%BC%88-8%2C8%EF%BC%89%E7%9A%84%E6%89%80%E6%9C%89%E5%AE%9E%E6%A0%B9x1%3D%5B-q%2F2%2B%28q%5E2%2F4%2Bp%5E3%2F27%29%5E%281%2F2%29%5D%5E%281%2F3%29%2B%2B%5B-q%2F2-%28q%5E2%2F4%2Bp%5E3%2F27%29%5E%281%2F2%29%5D%5E%281%2F3%29x2%3Dw%5B-q%2F2%2B%28q%5E2%2F4%2Bp%5E3%2F27%29%5E%281%2F2%29%5D%5E%281%2F3%29%2B%2Bw%5E2%5B-q%2F2-%28q%5E2%2F4%2Bp%5E3%2F27%29%5E%281%2F2%29%5D%5E%281%2F3%29x2%3Dw%5E2%5B-q%2F2%2B%28q%5E2%2F4%2Bp%5E3%2F27%29%5E)
用c++解一元三次方程的根.求出x3-3x-1=0在(-8,8)的所有实根x1=[-q/2+(q^2/4+p^3/27)^(1/2)]^(1/3)++[-q/2-(q^2/4+p^3/27)^(1/2)]^(1/3)x2=w[-q/2+(q^2/4+p^3/27)^(1/2)]^(1/3)++w^2[-q/2-(q^2/4+p^3/27)^(1/2)]^(1/3)x2=w^2[-q/2+(q^2/4+p^3/27)^
用c++解一元三次方程的根.
求出x3-3x-1=0在(-8,8)的所有实根
x1=[-q/2+(q^2/4+p^3/27)^(1/2)]^(1/3)+
+[-q/2-(q^2/4+p^3/27)^(1/2)]^(1/3)
x2=w[-q/2+(q^2/4+p^3/27)^(1/2)]^(1/3)+
+w^2[-q/2-(q^2/4+p^3/27)^(1/2)]^(1/3)
x2=w^2[-q/2+(q^2/4+p^3/27)^(1/2)]^(1/3)+
+w[-q/2-(q^2/4+p^3/27)^(1/2)]^(1/3)
其中w=(-1+√3i)/2.
但是要求实根啊
我编的程序只能求出一个根啊 他要求出所有的根.
用c++解一元三次方程的根.求出x3-3x-1=0在(-8,8)的所有实根x1=[-q/2+(q^2/4+p^3/27)^(1/2)]^(1/3)++[-q/2-(q^2/4+p^3/27)^(1/2)]^(1/3)x2=w[-q/2+(q^2/4+p^3/27)^(1/2)]^(1/3)++w^2[-q/2-(q^2/4+p^3/27)^(1/2)]^(1/3)x2=w^2[-q/2+(q^2/4+p^3/27)^
看看这个帖子,说不定你会有所启发,共同学习啊.