Question

已知x1、x2是方程x²+x—4=0,求代数式x1³—5x2²+10的值.

Answer 1

已知x1、x2是方程x²+x—4=0

则x1²+x1-4=0 x1²=4-x1

x2²+x2-4=0 x2²=4-x2

由韦达定理x1+x2=-1

x1*x2=-4

所以x1³—5x2²+10=x1*(4-x1)-5(4-x2)+10

=4x1-x1²-20+5x2+10

=4x1-(4-x1)+5x2-10

=5x1-4+5x2-10

=5(x1+x2)-14

=5*(-1)-14

=-19