Use the matching method. ax^2+bx+c=a[x^2+(b/a)x]+c=a{x^2+(b/a)x+[b/(2a)]^2}-a[b/(2a)]^2+c=a[x+b/(2a)]^2-b^2/(4a)+4ac/(4a)=a[x+b/(2a)]^2-(b^2-4ac)/(4a)=a[x+b/(2a)] ^2-a[√(b^2-4ac)]^2/(2a)^2=a{[x+b/(2a)]^2-[√(b^2-4ac)/(2a)]^2}=a[x+b/(2a)-√(b^2-4ac)/(2a)][x+b/(2a)+√(b^2-4ac)/(2a)]=a{x-[-b+√(b^2-4ac)]/(2a)}{x-[-b-√(b^2-4ac)]/(2a)}=a(x-r 1)(x-r2)。 Let R 1 and R2 be the relation on a in discrete mathematics, and prove that 20s (r1ur2) = s (r1) us (R2) r (r1ur2) = r (