若方程7x²-(k+13)x+k²-k-2=0有两个不相等的实数根x1,x2,且0<x1<1<x2<2,求k取值范围
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 04:28:56
![若方程7x²-(k+13)x+k²-k-2=0有两个不相等的实数根x1,x2,且0<x1<1<x2<2,求k取值范围](/uploads/image/z/2485039-31-9.jpg?t=%E8%8B%A5%E6%96%B9%E7%A8%8B7x%26%23178%3B%EF%BC%8D%EF%BC%88k%EF%BC%8B13%EF%BC%89x%EF%BC%8Bk%26%23178%3B%EF%BC%8Dk%EF%BC%8D2%EF%BC%9D0%E6%9C%89%E4%B8%A4%E4%B8%AA%E4%B8%8D%E7%9B%B8%E7%AD%89%E7%9A%84%E5%AE%9E%E6%95%B0%E6%A0%B9x1%2Cx2%2C%E4%B8%940%EF%BC%9Cx1%EF%BC%9C1%EF%BC%9Cx2%EF%BC%9C2%2C%E6%B1%82k%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
若方程7x²-(k+13)x+k²-k-2=0有两个不相等的实数根x1,x2,且0<x1<1<x2<2,求k取值范围
若方程7x²-(k+13)x+k²-k-2=0有两个不相等的实数根x1,x2,且0<x1<1<x2<2,求k取值范围
若方程7x²-(k+13)x+k²-k-2=0有两个不相等的实数根x1,x2,且0<x1<1<x2<2,求k取值范围
记f(x)=7x²-(k+13)x+k²-k-2
则开口向上,由根的位置,则有f(0)>0,f(1)0,即有:
k²-k-2>0,得(k-2)(k+1)>0,得k>2或k