Question

已知X1，X2是方程4X平方-4mx+m+2=0的两个实根，当实数m为何值时，X1的平方+X2的平方取最小值

Answer 1

解：由已知可得： x1+x2 = m x1*x2 = (m+2)/4 所以，f(m) = x1�0�5 + x2�0�5 = （x1+x2)�0�5 - 2x1*x2 = m�0�5 - 2（m+2)/4 = m�0�5 - 1/2 m -1 = (m - 1/4)�0�5 - 17/16 根据判别式△ ≥0，即：( -4m)�0�5 - 4*4(m+2) = 16m�0�5 - 16m - 32 ≥0 解，得：m≤ -1 或 m≥2 ①当m≤ -1时，f(m)为减函数，所以，f(m)min = f(-1) = 1/2 ②当m≥ 2 时，f(m)为增函数，所以，f(m)max = f(2) = 2 综上所述，x1�0�5+x2�0�5的最小值为 1/2