Question

已知X1,X2是方程X²+3X-5=0的两根,则 (X1+1)(X2+2)=? X1²+X2²=? (X1-X2)²=?

Answer 1

解：
由韦达定理得
x1+x2=-3
x1x2=-5
(x1+1)(x2+1)
你好像抄错题了。
=x1x2+x1+x2+1
=-5-3+1
=-7
x1²+x2²=(x1+x2)²-2x1x2
=(-3)²-2×(-5)
=9+10
=19
(x1-x2)²=(x1+x2)²-4x1x2
=(-3)²-4×(-5)
=9+20
=29

Answer 2

X1,X2是方程X畅攻扳纪殖慌帮苇爆俩178;+3X-5=0的两根
∴x1+x2=-3,
x1·x2=-5
∴
(X1+1)(X2+2)
=x1·x2+x1+x2+2
=-5-3+2
=-6
x1²+x2²
=(x1²+2x1·x2+x2²)-2x1·x2
=(x1+x2)²-2×(-5)
=9+10
=19
(x1-x2)²
=(x1+x2)²-4x1·x2
=9-4×(-5)
=29

Answer 3

两种方法：

1.认认真真算出x1,x2

2.因为1/x1
+1/x2=（x1+x2）/(x1*x2),运用方程两个根的和
积来算
x1+x2=
-b/a
x1*x2=c/a

思路就是这样