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

解：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

Answer 2

1、
x1+x2=-b/a，x1x2=c/a
|x1-x2|
=√(x1-x2)²
=√(x1+x2)²-4x1x2
=√(b²/a²)-4c/a
=[√（b²-4ac)]/a

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

2、

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

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