Question

已知x1，x2是方程2x²-3x-1=0的两根，求下列个式的值

（1）x1²+x2²；（2）1/x1+1/x2；（3）x1四次方+x1²x2²+x2四次方；（4）（x1+1/x2）（x2+1/x1）

Answer 1

解
x1，x2是方程的根
∴x1+x2=3/2
x1x2=-1/2
∴x1²+x2²
=(x1+x2)²-2x1x2
=(3/2)²+2×(1/2)
=9/4+1
=13/4
(2)
1/x1+1/x2
=(x1+x2)/(x1x2)
=(3/2)×(-2)
=-3
(3)
x1^4+x1²x2²+x2^4
=(x1²+x2²)²-2x1²x2²+x1²x2²
=(x1²+x2²)²-(x1x2)²
=(13/4)²-(-1/2)²
=169/16-1/4
=165/4
(4)
(x1+1/x2)(x2+1/x1)
=x1x2+1+1+1/x1x2)
=(-1/2+2-2)
=-1/2