Question

设x1、x2是一元二次方程2x²-5x-1=0的两根，求下列代数式的值

（1）（2）（3）（4）代表题目顺序

Answer 1

由韦达定理得
x1+x2=5/2
x1x2=-1/2
(1)
x1²x2+x2²x1
=x1x2(x1+x2)
=(-1/2)·(5/2)
=-5/4
(2)
x1²+x2²-3x1x2
=(x1+x2)²-5x1x2
=(5/2)²-5·(-1/2)
=25/4 +5/2
=35/4
(3)
x2/x1+x1/x2
=(x2²+x1²)/(x1x2)
=[(x1+x2)²-2x1x2]/(x1x2)
=[(5/2)²-2·(-1/2)]/(-1/2)
=-29/2
(4)
|x1-x2|
=√[(x1+x2)²-4x1x2]
=√[(5/2)²-4·(-1/2)]
=√(33/4)
=√33/2