解有关一元二次方程的问题

x1+x2=ab,x1x2=1/2(a+b)x1+x2-x1x2=ab-1/2(a+b)=1/2(2ab-a-b)=1/2[a(b-1)+b(a-1)]>=0x1+x2>=x1x2不妨设x1>=x2>=12x1>=x1+x2>=x1x2,x2<=2若x2=2,x1+2>=2x1,x1<=2,x1=x2=2ab=4,a+b=8,无正整数解若x2=11-ab+1/2(a+b)=02ab-a-b-2=0a=(b+2)/(2b-1)>=1b+2>=2b-1,b<=3b=1,a=3b=2,a=4/3b=3,a=1∴a=1,b=3或a=3,b=1方程是x^2-3x+2=0,两个正整数解是1和2 收起