Question

求：韦达定理的详细推论。这，是如何得到x1+x2=-b/a x1x2=c/a

Answer 1

ax^2+bx+c=0
a(x^2+bx/a)=-c
a(x^2+bx/a+b^2/4a^2)=b^2/4a-c
a(x+b/2a)^2=(b^2-4ac)/4a
(x+b/2a)^2=(b^2-4ac)/4a^2
x+b/2a=±√(b^2-4ac)/2a
x=[-b±√(b^2-4ac)]/2a
x1=[-b+√(b^2-4ac)]/2a，x2=[-b-√(b^2-4ac)]/2a

x1+x2=[-b+√(b^2-4ac)]/2a +[-b-√(b^2-4ac)]/2a
=-2b/2a=-b/a
x1*x2=[-b+√(b^2-4ac)]/2a *[-b-√(b^2-4ac)]/2a
= [b^2-(b^2-4ac)]/4a^2
=(b^2-b^2+4ac)/4a^2
=c/a

Answer 2

解；方程ax²bx+c=0解为x1=(-b+√(b²-4ac))/2a, x2==(-b-√(b²-4ac))/2a
所以x1+x2==-b/a.x1x2=c/a

Answer 3

求根公式