已知x1和x2是方程ax^2+bx+c=0(a≠0)的两根,求x1^2+x2^2,x1^3+x2^3,x1^4+x2^4,x1^5+x2^5得值

Question

已知x1和x2是方程ax^2+bx+c=0(a≠0)的两根,求x1^2+x2^2,x1^3+x2^3,x1^4+x2^4,x1^5+x2^5得值

Answer 1

x1+x2=-b/a=p
x1x2=c/a=q
x1^2+x2^2=(x1+x2)^2-2x1x2=p^2-2q
x1^3+x2^3=(x1+x2)[(x1+x2)^2-3x1x2]=p(p^2-3q)=p^3-3pq
x1^4+x2^4=(x1^2+x2^2)^2-2(x1x2)^2=(p^2-2q)^2-2q^2=p^4-4p^2q+2q^2
x1^5+x2^5=(x1^2+x2^2)(x1^3+x2^3)-(x1x2)^2(x1+x2)=(p^2-2q)p(p^2-3q)-pq^2