解方程：x^2-(a+2)x+2a=0,x^2-(a^2+2)x+a^2+1,ax^2-(a+b)x+b=0.(a为常数,要求用十字相乘发解决即韦达定理）麻烦大家了,感激不尽.

Question

解方程：x^2-(a+2)x+2a=0 ,x^2-(a^2+2)x+a^2+1,ax^2-(a+b)x+b=0.(a为常数,要求用十字相乘发解决即韦达定理）麻烦大家了,感激不尽.

Answer 1

x^2-(a+2)x+2a=0
(x-a)(x-2)=0
x=a,x=2
x^2-(a^2+2)x+a^2+1=0
(x-1)[x-(a^2+1)]=0
x=1,x=a^2+1
ax^2-(a+b)x+b=0
若a=0,b=0,恒成立,x取任意实数
若a=0,b≠0,则-bx+b=0,x=1
若a≠0,则(ax-b)(x-1)=0,x=b/a,x=1