已知方程2x²+3x-1=0的两根为x1、x2,求1/x1+1/x2
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/21 11:01:33
![已知方程2x²+3x-1=0的两根为x1、x2,求1/x1+1/x2](/uploads/image/z/2764997-53-7.jpg?t=%E5%B7%B2%E7%9F%A5%E6%96%B9%E7%A8%8B2x%26%23178%3B%2B3x-1%3D0%E7%9A%84%E4%B8%A4%E6%A0%B9%E4%B8%BAx1%E3%80%81x2%2C%E6%B1%821%2Fx1%2B1%2Fx2)
已知方程2x²+3x-1=0的两根为x1、x2,求1/x1+1/x2
已知方程2x²+3x-1=0的两根为x1、x2,求1/x1+1/x2
已知方程2x²+3x-1=0的两根为x1、x2,求1/x1+1/x2
∵方程2x²+3x-1=0的两根为x1、x2
∴由根与系数关系得
x1+x2=-3/2 ①
x1x2=-1/2 ②
∵1/x1+1/x2=(x1+x2)/ x1x2 ③
∴将①②代入③得
1/x1+1/x2=(x1+x2)/ x1x2 =(-3/2 )/(-1/2 )=3
=x1x2/x1+x2=-3/2÷(-1)=3/2
令y=1/x, 带入并整理:y²-3y-2=0
而1/x1+1/x2 = y1+ y2
由根与系数关系得:
y1+ y2 = - b/a = 3/1=3
故1/x1+1/x2 = 3
同意一楼的答案