已知方程x^2+px+q=0的两根是a,b.求证：一元二次方程qx^2+p(1+q)x+(1+q)^2=0的根为a+1/b和b+1/a

方程x^2+px+q=0的两根是a,b
所以a+b=-p,ab=q
则a+1/b+b+1/a
=a+b+(a+b)/ab
=-p-p/q
=-p(1+1/q)
=-p(q+1)/q
(a+1/b)(b+1/a)
=ab+1+1+1/ab
=q+1/q+2
=(q^2+2q+1)/q
=(q+1)^2/q
所以a+1/b和b+1/a是方程x^2+p(q+1)x/q+(q+1)^2/q的跟
即qx^2+p(1+q)x+(1+q)^2=0