Question

已知a,b是方程x平方-x-1=0的2个实数根,不解方程,试求a四次方+3b的值

神速来人！

Answer 1

a,b是一元二次方程x^2-x-1=0
即a^2-a-1=0
a^2-a=1
a^2=a+1
由韦达定理得
a+b=1

a^4+3b
=(a^2)^2+3b
=(a+1)^2+3b
=a^2+2a+1+3b
=a^2-a+3a+1+3b
=a^2-a+3a+3b+1
=a^2-a+3(a+b)+1
=1+3*1+1
=5

Answer 2

a,b是方程x平方-x-1=0的2个实数根
a²-a-1=0 即a²=a+1
a+b=1,ab=-1
a^4+3b
=﹙a²﹚²+3b
=(a+1)²+3b
=a²+2a+1+3b
=a+1+2a+1+3b
=3(a+b)+2
=3×1+2
=5