f(x)=mx^2+3(m-4)x-9 若函数f(x)有两个零点x1,x2 求d=|x1-x2|最小值

△ = [3(m - 4)]² - 4m×(-9) = 9(m² - 8m + 16) + 36m = 9(m² - 4m + 16) = 9(m - 2)² + 108 > 0所以f（x）有两个零点设两个零点分别是 x1 和 x2 ,则x1 + x2 = 3(4 - m)/mx1 * x2 = -9/m (x1 - x2)²= (x1 + x2)² - 4x1*x2= 9(4 - m)²/m² + 36/m= 9(16/m² - 4/m + 1)= 9(16/m² - 4/m + 1/4) + 27/4= 9(4/m - 1/2)² + 27/4当 4/m = 1/2 ,即 m = 8 时 ,最小值是 27/4两个零点的距离的最小值= |x1 - x2|= √(x1 - x2)²= √(27/4)= 3√3/2