Question

已知x²=1+4x,求代数式2x四次方+8x³-4x²-8x+1的值.！！！拜托了！！！！大师们！！！！！

Answer 1

解：已知x²=1+4x,
有 x²-4x-1=0
a=1,b=-4,c=-1
b²-4ac=（-4）²-4×1×(-1)=20
当b²-4ac>0时，方程式有两个不相等的实数根x1,x2=-b±√b²-4ac/2a
x1=（-(-4)+2√5）/2=2+√5, x2=（-(-4)-2√5）/2=2-√5
则2x^4+8x³-4x²-8x+1
x1代入上式有：2×（2+√5）^4+8×（2+√5）³-4×（2+√5）²-8×（2+√5）+1
=322+144√5+184+136√5-36-16√5-16-8√5+1
=455+256√5
x2代入上式有：2×（2-√5）^4+8×（2-√5）³-4×（2-√5）²-8×（2-√5）+1
=322-144√5+184-136√5-36+16√5-16+8√5+1
=455-256√5
拿你这点真不容易啊......

Answer 2

x²=1+4x, x²-1=4x, x=2±√5
2x^4+8x³-4x²-8x+1
=2x²(x²-2)+8x(x²-1)+1
=2(4x+1)(4x-1)+8x*4x+1
=2(16x²-1)+32x+1
=32x²-2+32x+1
=32(1+4x)+32x-1
=32+128x+32x-1
=160x+31
=160(2±√5)+31
=320±160√5+31
=351±160√5