Question

设函数f(x)=x2+ax+b,集合A={x|f(x)=x}={a},求实数a,b的值

Answer 1

解答如下：
从集合可以知道f(x)-x=0有两个相同实数根并且为a
也就是x2+（a-1）x+b=0的判别式=（a-1）^2-4b=0
并且求得实数根为1-a /2 =a
解答得到a=1/3,b=1/9

Answer 2

x=f(x)
x=x^2+ax+b
x^2+(a-1)x+b = 0

a is only solution
then
△ = 0
(a-1)^2 - 4b = 0 => b =(a-1)^2/4
also a is solution, then
a^2 +(a-1)a + b = 0
a^2+ (a-1)a +(a-1)^2/4 =0
8a^2-4a +(a-1)^2 = 0
9a^2 -6a + 1 = 0
(3a-1)^2 =0
a = 1/3
b = 1/9

a = 1/3 and b=1/9 #

Answer 3

答案是 a=0 b=0
或a=1/3,b=1/9