Question

一元二次方程ax^2+bx+c=0(a不等于0)的两根为x1和x2，求：(1)|x1-x2|和（x1+x2）/2；（2）x1^3+x2^3。

Answer 1

解：根据韦型昌达定理
x1+x2=-b/a
x1x2=c/a

(1)：
(x1-x2)²=(x1+x2)²-4x1x2
=(-b/a)²-(4c/a)
=(b²/a²)-(4ac/a²)
=(b²-4ac)/a²
当b²-4ac≥0，a≠0时
|x1-x2|=√[(b²-4ac)/a²]
=√(b²-4ac)/|a|

(x1+x2)/2=(-b/a)/2
=-b/(2a)

(2)：悉租迹
x1²+x2²=(x1+x2)²-2x1x2
=(-b/a)²-(2c/a)
=(b²/a²)-(2ac/a²)
=(b²-2ac)/a²
x1³+x2³=(x1+x2)(x1²-x1x2+x2²)
=(x1+x2)[(x1²+x2²)-x1x2]
=(-b/a)×[(b²-2ac)/a²-(c/a)]
=(-b/a)×[(b²-2ac)/a²-(ac/a²)]
=(-b/睁并a)×(b²-3ac)/a²
=(3abc-b³)/a³

Answer 2

第一题用伟达定理，第二题先用立升烂方和，等于（x1＋x2）（x1∧2-x1x2＋模拦x2∧2）,然后用伟旦笑胡达定理！