已知x1,x2是方程x²+4[kx+(1-4k)]²=4的两根,求(x1-x2)²化简是多少

已知x1,x2是方程x²+4[kx+(1-4k)]²=4的两根,求(x1-x2)²化简是多少
已知x1,x2是方程x²+4[kx+(1-4k)]²=4的两根,求(x1-x2)²化简是多少

已知x1,x2是方程x²+4[kx+(1-4k)]²=4的两根,求(x1-x2)²化简是多少
x²+4[kx+(1-4k)]²=4
x²+4[k²x²+2(1-4k)kx+(1-4k)²]=4
(4k²+1)x²-8(4k-1)kx+4(1-4k)²-4=0
(4k²+1)x²-8(4k-1)kx+16k²-8k=0
x1+x2=8(4k-1)k/(4k²+1) x1*x2=(16k²-8k)/(4k²+1)
(x1-x2)²
=(x1+x2)²-4x1*x2
=[8(4k-1)k/(4k²+1) ]²-4(16k²-8k)/(4k²+1)
=64k²(4k-1)²/(4k²+1)²-(64k²-32k)/(4k²+1)
=[64k²(4k-1)²-(64k²-32k)(4k²+1)]/(4k²+1)²
=[64k²(16k²-8k+1)-(256k⁴-128k³+64k²-32k)]/(4k²+1)²
=(1024k⁴-512k³+64k²-256k⁴+128k³-64k²+32k)/(4k²+1)²
=(768k⁴-384k³+32k)/(4k²+1)²
应该已经最简便了,题目是否还有其他条件,可以求出k的值啊