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

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 #