已知a.b为常数,且a不等于零,f(x)=ax2+bx,且f(2)=0,方程f(x)=x有等根.求f(x)的解析式

f(2)=0
==>a*2^2+b*2=0
==>2a+b=0
==>a=-b/2
f(x)=x有等根
==>ax^2+bx=x有等根
==>ax^2+(b-1)x=0有等根
==>判别式(b-1)^2-4*a*0=0
==>(b-1)^2=0
==>b=1
==>a=-b/2=-1/2
==>f(x)=ax^2+bx=-x^2/2+x