Question

已知x1、x2是方程x^2+6x+3=0的两实数根，试求下列代数式的值（1）x1^2+x2^2（2）x2/x1+x1/x2

(3)(x1+1)(x2+1)

Answer 1

解
x1.x2是方程的解
由韦达定理得：
x1+x2=-6，x1x2=3
∴x1²+x2²
=(x1+x2)²-2x1x2
=(-6)²-2×3
=36-6
=30

x2/x1+x1/x2
=(x2²+x1²)/(x1x2)
=(30)/(3)
=10

(x1+1)(x2+1)
=x1x2+(x1+x2)+1
=3-6+1
=-2