Question

已知x1，x2是方程2x^2+3x-4=0的两个根

试求：x1+x2，x1.x2，1/x1+1/x2,x1^2+x2^2,(x1+1)(x2+1),x1-x2绝对值，的值。

希望步骤详细·

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

Answer 2

x1+x2=-3/2，
x1.x2=-4/2=-2，
1/x1+1/x2=(x1+x2)/x1*x2=(-3/2)/(-2)=3/4,
x1^2+x2^2=(x1+x2)^2-2x1*x2=9/4+4=25/4,
(x1+1)(x2+1)=x1*x2+x1+x2+1=-2-3/2+1=-5/2,
(x1-x2)^2=x1^2-2x1*x2+x2^2=(x1+x2)^2-4x1*x2=9/4+8=41/4
｜x1-x2｜=√41/2

Answer 3

x1+x2＝ －b/a=-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=25/4
(x1+1)(x2+1)＝x1x2+x1+x2+1=-5/2
x1-x2绝对值＝根号下41/2