Question

设方程x^2-x-1=0的两根是p,q,则pq^4-p^5q+5q=

Answer 1

p,q是方程的两根,所以pq=-1,p+q=1

且 p^2-p-1=0,即 p^2=p+1

p^4=p^2+2p+1=3p+2

q^2=q+1

q^3=q^2+q=2q+1

所以

pq^4-p^5q+5q

=-q^3+p^4+5q

=-(2q+1)+3p+2+5q

=3(p+q)+1

=4