二元一次方程x^2+a^2x+b=0的解为x1,x2,y^2+5ay+7=0的解为y1,y2,已知x1-y1=x2-y2

x1+x2=-a²
y1+y2=-5a
相减,有x1-y1+x2-y2=5a-a²
∵x1-y1=x2-y2=2
∴上式化为2(x1-y1)=4=5a-a²
a=1或4
当a=1时,y²+5y+7=0无实数解
∴a=4,y²+20y+7=0
y1+y2=-20,y1y2=7
∵x1-y1=x2-y2,∴x1-x2=y1-y2=√[(y1+y2)²-4y1y2]=√372
x²+16x+b=0,x1+x2=-16,x1x2=b
4x1x2=4b=(x1+x2)²-(x1-x2)²=256-372=-116
∴b=-29