Question

已知x²+4x-1=0,求x⁴+8x³-4x²-8x+1的值

Answer 1

x⁴+8x³-4x²-8x+1＝x⁴+4x³+4x³-x²-3x²-8x+1＝x²（x²+4x-1）+4x³-3x²-8x+1＝4x³-3x²-8x+1
＝4x³-3x²+19x²-19x²-4x-4x+1＝4x（x²+4x-1）-19x²-4x+1=-19x²-4x+1=-19(x²+4x-1)+72x-18
=72x-18=72(x+2)-162
因为x²+4x-1=0，所以（x+2）²=2，所以x+2=正负根号2，所以原式＝-162加减根号2

Answer 2

已知x²+4x-1=0
所以x²=1-4x
所以x⁴+8x³-4x²-8x+1
=(1-4x)(1-4x)+8x(1-4x)-4x²-8x+1
=-20x²-8x+2
=-20(1-4x)-8x+2
=72x-18

因为²+4x-1=0
所以x=[-4±√(16+4)]/2=-2±√5
所以原式=72x-18=-162±72√5