若抛物线y=2x²+4x+1与x轴的两个交点坐标分别是(a,0)和(b,0)则a²+b²=
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/21 03:43:57
![若抛物线y=2x²+4x+1与x轴的两个交点坐标分别是(a,0)和(b,0)则a²+b²=](/uploads/image/z/2523865-49-5.jpg?t=%E8%8B%A5%E6%8A%9B%E7%89%A9%E7%BA%BFy%EF%BC%9D2x%26%23178%3B%2B4x%2B1%E4%B8%8Ex%E8%BD%B4%E7%9A%84%E4%B8%A4%E4%B8%AA%E4%BA%A4%E7%82%B9%E5%9D%90%E6%A0%87%E5%88%86%E5%88%AB%E6%98%AF%EF%BC%88a%2C0%EF%BC%89%E5%92%8C%EF%BC%88b%2C0%EF%BC%89%E5%88%99a%26%23178%3B%2Bb%26%23178%3B%3D)
若抛物线y=2x²+4x+1与x轴的两个交点坐标分别是(a,0)和(b,0)则a²+b²=
若抛物线y=2x²+4x+1与x轴的两个交点坐标分别是(a,0)和(b,0)则a²+b²=
若抛物线y=2x²+4x+1与x轴的两个交点坐标分别是(a,0)和(b,0)则a²+b²=
答:
抛物线y=2x²+4x+1与x轴交点坐标为(a,0)和(b,0)
根据韦达定理有:
a+b=-4/2=-2
ab=1/2
所以:
a²+b²
=(a+b)²-2ab
=(-2)²-2*(1/2)
=4-1
=3