Question

已知x1和x2是方程x^2+6x+7的两根,求下列各式的值.1）（x1-x2)^2 2)x1/x2+x2/x1

Answer 1

由韦达定理

x1+x2=-6,x1*x2=7

1）（x1-x2)^2

=x1^2-2x1x2+x2^2

=x1^2+2x1x2+x2^2-4x1x2

=(x1+x2)^2-4x1x2

=(-6)^2-4*7

=8

2)x1/x2+x2/x1

=(x1^2+x2^2)/x1x2

=(x1^2+2x1x2+x2^2-2x1x2)/x1x2

=[(x1+x2)^2-2x1x2]/x1x2

=[(-6)^2-2*7]/7

=22/7