Question

已知x1,x2是方程2x^2+3x-4=0的两个根

Answer 1

2x^2+3x-4=0

a=2,b=3,c=-4

x1+x2=-b/2=-3/2

x1*x2=c/a=-4/2=-2

1/x1+1/x2=(x1+x2)/(x1x2)=3/4

x1^2+x2^2=(x1+x2)^2-2x1x2=9/4+4=25/4

(x1+1)(x2+1)=x1*x2+(x1+x2)+1=-2+(-3/2)+1=-5/2

|x1-x2|=|√(b^2-4ac)/a|=√41/2