Question

数学题?己知函数fx=ax2+bx+c,且f1=-a/2(1)求证函数fx有两个不同的零点(2)设x1,x2是函数f(x

Answer 1

f(1) = a + b + c = - a/2 === b = -3a/2 - c

1.b^2 - 4ac = (-3a/2 - c)^2 - 4ac = 9a^2/4 - ac + c^2 =(c-a/2)^2+2a^2 >0

方程有两不同根,也就是函数有两不同交点.

2.设 c/a = m

|x1 -x2| ^2 = (x1+x2)^2 - 4x1x2 = (-b/a)^2 - 4c/a = (3/2 +m)^2 - 4m = m^2 -m + 9/4

= (m -1/2)^2 + 2

所以|x1-x2|>= 根2

3.