已知x1,x2是方程2x^2+3x-4=0的两个根
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发布时间:2024-10-24 02:25
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热心网友
时间:2024-11-02 21:09
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
热心网友
时间:2024-11-02 21:09
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
热心网友
时间:2024-11-02 21:10
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