设y=a(x-1)^2+16,与x轴的交点横坐标:x1,x2
由:ax^2-2ax+(a+16)=0得:x1+x2=2,x1x2=(a+16)/a
而:(x1-x2)的绝对值=8,即:(x1-x2)^2=64,化为:(x1+x2)^2-4x1x2=64
4-4(a+16)/a=64
a=-1
y=-(x-1)^2+16即为所求
y=10时,(x-1)^2=6
x=1±根号6
设y=a(x-1)^2+16,与x轴的交点横坐标:x1,x2
由:ax^2-2ax+(a+16)=0得:x1+x2=2,x1x2=(a+16)/a
而:(x1-x2)的绝对值=8,即:(x1-x2)^2=64,化为:(x1+x2)^2-4x1x2=64
4-4(a+16)/a=64
a=-1
y=-(x-1)^2+16即为所求
y=10时,(x-1)^2=6
x=1±根号6