Question

设x1 x2是一元二次方程ax²+bx+c=0(a≠0)的两根 求x1+x2= - b/a x1*x2=c/a

设x1 x2是一元二次方程ax²+bx+c=0(a≠0)的两根

求x1+x2= - b/a x1*x2=c/a

Answer 1

由求根公式
x1=[-b-√(b²-4ac)]/2a
x2=[-b+√(b²-4ac)]/2a

所以x1+x2=[-b-√(b²-4ac)-b+√(b²-4ac)]/2a
=2b/2a
=-b/a

x1x2=[-b-√(b²-4ac)]/2a*[-b-√(b²-4ac)]/2a
分子是平方差
=[(-b)²-√(b²-4ac)²]/(2a)²
=[b²-(b²-4ac)]/4a²
=4ac/4a²
=c/a