搜搜考试网t^4=6t^2 -t -12 那么 t等于多少?
来源：学生作业帮 编辑：搜搜考试网作业帮 分类：数学作业 时间：2024/04/29 16:37:50
t^4=6t^2 -t -12 那么 t等于多少?
和 这个问题的手工计算 计算量可不小呀 我也不想花费那个冤枉时间
下面我使用Matlab帮你计算下吧 很简单的
程序如下：
%by dynamic
%2009.3.5
>> solve('t^4=6*t^2 -t -12')%下面是解析解 够复杂的吧,恩接着后面我将它转化为数值解了 你可以看看 但是没有实根哦
ans =
1/4*2^(1/2)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+1/4*(-(-32*(708+4*i*671^(1/2))^(1/3)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+2*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)*(708+4*i*671^(1/2))^(2/3)+160*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+8*2^(1/2)*(708+4*i*671^(1/2))^(1/3))/(708+4*i*671^(1/2))^(1/3)/((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2))^(1/2)
1/4*2^(1/2)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)-1/4*(-(-32*(708+4*i*671^(1/2))^(1/3)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+2*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)*(708+4*i*671^(1/2))^(2/3)+160*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+8*2^(1/2)*(708+4*i*671^(1/2))^(1/3))/(708+4*i*671^(1/2))^(1/3)/((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2))^(1/2)
-1/4*2^(1/2)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+1/4*(-(-32*(708+4*i*671^(1/2))^(1/3)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+2*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)*(708+4*i*671^(1/2))^(2/3)+160*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)-8*2^(1/2)*(708+4*i*671^(1/2))^(1/3))/(708+4*i*671^(1/2))^(1/3)/((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2))^(1/2)
-1/4*2^(1/2)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)-1/4*(-(-32*(708+4*i*671^(1/2))^(1/3)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+2*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)*(708+4*i*671^(1/2))^(2/3)+160*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)-8*2^(1/2)*(708+4*i*671^(1/2))^(1/3))/(708+4*i*671^(1/2))^(1/3)/((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2))^(1/2)
>> vpa(ans,5)
ans =
1.7982-.61035*i
1.7982+.61035*i
-1.7982-.30740*i
-1.7982+.30740*i
>> roots([1 0 -6 1 12])%当然熟悉Matlab的人,可以使用这个命令,求得的结果是一样的
ans =
1.7982 + 0.6103i
1.7982 - 0.6103i
-1.7982 + 0.3073i
-1.7982 - 0.3073i
可以看出该方程没有实根 希望能帮上你
下面我使用Matlab帮你计算下吧 很简单的
程序如下：
%by dynamic
%2009.3.5
>> solve('t^4=6*t^2 -t -12')%下面是解析解 够复杂的吧,恩接着后面我将它转化为数值解了 你可以看看 但是没有实根哦
ans =
1/4*2^(1/2)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+1/4*(-(-32*(708+4*i*671^(1/2))^(1/3)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+2*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)*(708+4*i*671^(1/2))^(2/3)+160*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+8*2^(1/2)*(708+4*i*671^(1/2))^(1/3))/(708+4*i*671^(1/2))^(1/3)/((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2))^(1/2)
1/4*2^(1/2)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)-1/4*(-(-32*(708+4*i*671^(1/2))^(1/3)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+2*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)*(708+4*i*671^(1/2))^(2/3)+160*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+8*2^(1/2)*(708+4*i*671^(1/2))^(1/3))/(708+4*i*671^(1/2))^(1/3)/((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2))^(1/2)
-1/4*2^(1/2)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+1/4*(-(-32*(708+4*i*671^(1/2))^(1/3)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+2*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)*(708+4*i*671^(1/2))^(2/3)+160*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)-8*2^(1/2)*(708+4*i*671^(1/2))^(1/3))/(708+4*i*671^(1/2))^(1/3)/((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2))^(1/2)
-1/4*2^(1/2)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)-1/4*(-(-32*(708+4*i*671^(1/2))^(1/3)*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)+2*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)*(708+4*i*671^(1/2))^(2/3)+160*((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2)-8*2^(1/2)*(708+4*i*671^(1/2))^(1/3))/(708+4*i*671^(1/2))^(1/3)/((8*(708+4*i*671^(1/2))^(1/3)+(708+4*i*671^(1/2))^(2/3)+80)/(708+4*i*671^(1/2))^(1/3))^(1/2))^(1/2)
>> vpa(ans,5)
ans =
1.7982-.61035*i
1.7982+.61035*i
-1.7982-.30740*i
-1.7982+.30740*i
>> roots([1 0 -6 1 12])%当然熟悉Matlab的人,可以使用这个命令,求得的结果是一样的
ans =
1.7982 + 0.6103i
1.7982 - 0.6103i
-1.7982 + 0.3073i
-1.7982 - 0.3073i
可以看出该方程没有实根 希望能帮上你