Question

一道一元二次方程

a是X ^2-X-1的根，则a^4-3a-2的值为——————

Answer 1

x²-x-1=0有一根为a
则a²-a-1=0，a²=a+1
a^4-3a-2
=（a²）²-3a-2
=（a+1）²-3a-2
=a²-a-1
=0

Answer 2

a是X ^2-X-1=0的根，则a^2-a-1=0
a^2=a+1
a^4-3a-2
=(a+1)^2-3a-2
=a^2+2a+1-3a-2
=a+1+2a+1-3a-2
=0

Answer 3

-1或1/2

Answer 4

a是X ^2-X-1的根，所以a^2-a-1=0
a^2=a+1
a^4-3a-2=(a^2)^2-3a-2（将上式带入，降次）
有（a+1）^2-3a-2=a^2-a-1=0

Answer 5

由已知得a²-a-1=0,
所以a²=a+1
带入2式得到(a+1)²-3a-2=a²-a-1=0

Answer 6

因为a是X ^2-X-1=0的根，所以有a ^2-a-1=0
即a ^2=a+1
则a^4-3a-2=（a+1）^2-3a-2=a ^2-a-1=0