一元二次方程ax²+bx+c=0(a不等于0)的两根为x1和x2(1)|x1-x2|和（x1+x2）/2(2)x1³+x2³

Question

一元二次方程ax²+bx+c=0(a不等于0)的两根为x1和x2
(1)|x1-x2|和（x1+x2）/2
(2)x1³+x2³

Answer 1

根据韦达定理
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)：
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/a)×(b²-3ac)/a²
=(3abc-b³)/a³