题目

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

提问时间：2020-10-15

答案

x=f(x)x=x^2+ax+bx^2+(a-1)x+b = 0a is only solutionthen △ = 0(a-1)^2 - 4b = 0 => b =(a-1)^2/4also a is solution,thena^2 +(a-1)a + b = 0a^2+ (a-1)a +(a-1)^2/4 =08a^2-4a +(a-1)^2 = 09a^2 -6a + 1 = 0(3a-...