Question

设x1x2是方程x方-x-4=0的两根,求x1三次方+5x2方=10

Answer 1

因为x1,x2是x^2-x-4=0的根,

所以x1^2-x1-4=0,x2^2-x2-4=0

x1^2=x1+4,

x1^3=x1^2+4x,

x1^3+5x2^2+10

=(x1^2+4x1)+5x2^2+10

=x1^2+5x2^2+4x1+10

=x1^2+x^2+4x2^2+4x1+10

=(x1+x2)^2-2x1x2+(4x2+16)+4x1+10

=1+8+4(x1+x2)+26

=9+4+26

=39