求一道数学题，要求用二次方程函数

Question

如题所述

Answer 1

设这两个数分别是 x、y，且 x-y = 8，
由于 xy = 1/4*[(x+y)^2 - (x-y)^2] = 1/4*(x+y)^2 - 16，
要使 xy 最小，就要使 (x+y)^2 取最小值，
由于 (x+y)^2 ≥ 0，因此 (x+y)^2 最小值为 0，此时 x+y = 0，
由 x-y = 8，x+y = 0 相加得 2x = 8，x = 4，因此 y = -4，
所以，这两个数是 4 与 -4 。

Answer 2

Suppose the two numbers we want to find are x and (x-8), then we need to minimize
x(x-8)
Obviously,
x(x-8) = x²-8x = (x²-8x+16)-16 = (x-4)²-16 ≥ -16
So the minimum value of x(x-8) is -16 with x=4，
hence the two numbers are x=4 and x-8=-4