实系数一元二次方程 x^2-(2a+1)x+a+2=0有虚根 且这个根的立方是实数 求a的值

求详解过程
2024年12月02日 13:40
有4个网友回答
网友(1):

有虚根,首先判别式=4a^2-7<0

解得 -根号7/2
所以所得的虚根为x1=(2a+1)/2+根号(4a^2-7)/2
x2=(2a+1)/2-根号(4a^2-7)/2

其中含有虚数的部分都是 根号(4a^2-7)/2

为了简便,设实部=p,含有虚数的部分=q

则根的平方=(p+q)^3=p^3+3p^2q+3pq^2+q^3

或=(p-q)^3=p^3-3p^2q+3pq^2-q^3

你应该学过,虚数的偶次幂是实数,奇次幂仍未虚数,要想根为实数,必须把展开式中的虚数部分(奇次幂部分)消去才可以

根据上述两个式子,不难看出,需要3p^2q+q^3=0

化简3p^2=-q^2

带入

3(2a+1)^2/4=(7-4a^2)/4

解得a=-1或1/4

千万别忘了检验判别式

最后a=-1或1/4

反正就是这个思路,如果我算错了你再算一遍

网友(2):

先证明一个一般规律
设z=a+bi
则z³=a³+3a²bi+3a(bi)²+(bi)³
=a³-3ab²+(3a²b-b³)i
如果z³是实数,那么3a²b-b³=0,b=0或者3a²=b²
x^2-(2a+1)x+a+2=0
△=(2a+1)²-4(a+2)=4a²-7<0
x=(2a+1)/2±[√(7-4a²)/2]i
所以3[(2a+1)/2]²=7-4a²/4
4a²+3a-1=0
a=1/4或者a=1

网友(3):

设该虚根为:r(cosθ+isinθ)(r>0)
代入x²-(2a+1)x+a+2=0有:
r²(cos2θ+isin2θ)-(2a+1)(cosθ+isinθ)+a+2=0
[r²cos2θ-(2a+1)cosθ+a+2] + i[r²sin2θ-(2a+1)sinθ]=0
∴r²cos2θ-(2a+1)cosθ+a+2=0
∴r²(2cos²θ-1)-(2a+1)cosθ+a+2=0。。。。。1
r²sin2θ-(2a+1)sinθ=0
2r²sinθcosθ-(2a+1)sinθ=0
又根r(cosθ+isinθ)是虚根,sinθ≠0
∴2r²cosθ-(2a+1)=0。。。。。。。。。。。。2
而[r(cosθ+isinθ)]³=r³cos3θ +r³isin3θ是个实数,
∴r³sin3θ=0即:sin3θ=0
而sin3θ=sin(θ+2θ)=sinθcos2θ+cosθsin2θ=sinθ(2cos²θ-1)+2sinθcos²θ
=sinθ(4cos²θ-1)
∴sinθ(4cos²θ-1)=0
又sinθ≠0
∴4cos²θ-1=0
cosθ=±1/2
cosθ=1/2时,代入1式,2式子有:
-r²+3=0
r²-(2a+1)=0
解:a=1
cosθ=-1/2时,代入1式子,2式子有:
-r²+4a+5=0
-r²-(2a+1)=0
解:a=-1
∴a的值为:a=1或a=-1

网友(4):

由题意可知该方程的两个根分别是x1=(2*a+1)/2+(1-a^2)^(1/2)*i,x2=(2*a+1)/2-(1-a^2)^(1/2)*i,
假设这两个根有字母表示即:x=k×i+m
则x^3=(-k^3+3*k*m^2)*i+(m^3-3*m*k^2)有题意知X的三次方是实数,故虚部的系数为零从而得出
-k^3+3*K*m^2=0即k^2=3*m^2带入即1-a^2=3*(a+1/2)^2从而解得a=(-3+(13)^1/2)/8或a=(-3+(13)^1/2)/8完毕,希望对你有所帮助

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